latexindent (13nov13)

git-svn-id: svn://tug.org/texlive/trunk@32150 c570f23f-e606-0410-a88d-b1316a301751

36 files changed, 16166 insertions, 0 deletions

diff --git a/Master/texmf-dist/doc/support/latexindent/README b/Master/texmf-dist/doc/support/latexindent/README
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@@ -0,0 +1,47 @@
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+    PERL script to indent code within environments, and align delimited 
+    environments in .tex files.
+
+    This program is free software: you can redistribute it and/or modify
+    it under the terms of the GNU General Public License as published by
+    the Free Software Foundation, either version 3 of the License, or
+    (at your option) any later version.
+    
+    This program is distributed in the hope that it will be useful,
+    but WITHOUT ANY WARRANTY; without even the implied warranty of
+    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+    GNU General Public License for more details.
+    
+    See http://www.gnu.org/licenses/
+
+    Dr. C. M. Hughes
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+
+
+FOR COMPLETE DETAILS, PLEASE SEE documentation/manual.pdf
+
+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+*+
+
+USAGE
+    You'll need
+
+        latexindent.pl
+        defaultSettings.yaml
+
+    in the same directory. Windows users might prefer to grab latexindent.exe
+
+* IMPORTANT *
+
+This script may not work for your style of formatting; I highly 
+recommend comparing the outputfile.tex to make sure that 
+nothing has been changed (or removed) in a way that will damage
+your file.
+
+I recommend both using the following:
+    - a visual check, at the very least, make sure that 
+      each file has the same number of lines
+    - a check using latexdiff inputfile.tex outputfile.tex
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
diff --git a/Master/texmf-dist/doc/support/latexindent/documentation/manual.pdf b/Master/texmf-dist/doc/support/latexindent/documentation/manual.pdf
new file mode 100644
index 00000000000..16a709c50e9
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/documentation/manual.pdf
diff --git a/Master/texmf-dist/doc/support/latexindent/documentation/manual.tex b/Master/texmf-dist/doc/support/latexindent/documentation/manual.tex
new file mode 100644
index 00000000000..6f6c76f25e8
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/documentation/manual.tex
@@ -0,0 +1,1042 @@
+% arara: pdflatex
+% !arara: bibtex
+% !arara: pdflatex
+% !arara: pdflatex
+% !arara: indent: {overwrite: yes, trace: yes, localSettings: no, silent: yes}
+\begin{filecontents}{mybib.bib}
+	@online{cmh:videodemo,
+		title="Video demonstration of latexindet.pl on youtube",
+		url="http://www.youtube.com/watch?v=s_AMmNVg5WM"}
+	@online{cpan,
+		title="CPAN: Comprehensive Perl Archive Network",
+		url="http://www.cpan.org/"}
+	@online{strawberryperl,
+		title="Strawberry Perl",
+		url="http://strawberryperl.com/"}
+	@online{cmhblog,
+		title="A Perl script for indenting tex files",
+		url="http://tex.blogoverflow.com/2012/08/a-perl-script-for-indenting-tex-files/"}
+\end{filecontents}
+\documentclass[11pt]{article}
+%   This program is free software: you can redistribute it and/or modify
+%   it under the terms of the GNU General Public License as published by
+%   the Free Software Foundation, either version 3 of the License, or
+%   (at your option) any later version.
+%   
+%   This program is distributed in the hope that it will be useful,
+%   but WITHOUT ANY WARRANTY; without even the implied warranty of
+%   MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+%   GNU General Public License for more details.
+%   
+%   See <http://www.gnu.org/licenses/>.
+\usepackage[left=4.5cm,right=2.5cm,showframe=false,
+top=2cm,bottom=1.5cm]{geometry}                      % page setup
+\usepackage{parskip}                                 % paragraph skips
+\usepackage{booktabs}                                % beautiful tables
+\usepackage{listings}                                % nice verbatim environments
+\usepackage{titlesec}                                % customize headings
+\usepackage{changepage}                              % adjust width of page
+\usepackage{fancyhdr}                                % headers & footers
+\usepackage[sc,format=hang,font=small]{caption}      % captions
+\usepackage[backend=bibtex]{biblatex}                % bibliography
+\usepackage{mdframed}                                % framed environments
+\usepackage[charter]{mathdesign}                     % changes font
+\usepackage[expansion=false,kerning=true]{microtype} % better kerning
+\usepackage{enumitem}                                % custom lists
+\usepackage{tikz}                                    % so so much
+\usetikzlibrary{positioning}
+\usepackage{varioref}                                % clever referencing
+\usepackage[colorlinks=true,linkcolor=blue,citecolor=black]{hyperref}
+\usepackage{cleveref}
+
+\addbibresource{mybib}
+
+\newmdenv[linecolor=red,innertopmargin=.5cm,linewidth=3pt,
+	splittopskip=\topskip,skipbelow=0pt,%
+]{warning}
+
+\lstset{%
+	basicstyle=\small\ttfamily,language={[LaTeX]TeX},   
+	numbers=left,
+	numberstyle=\ttfamily\small,
+	breaklines=true,frame=single,framexleftmargin=8mm, xleftmargin=8mm,
+	prebreak = \raisebox{0ex}[0ex][0ex]{\ensuremath{\hookrightarrow}},
+	backgroundcolor=\color{green!5},frameround=fttt,
+	rulecolor=\color{blue!70!black},
+	keywordstyle=\color{blue},                    % keywords
+	commentstyle=\color{purple},    % comments
+	tabsize=3,
+	%columns=fullflexible   
+}%
+\lstdefinestyle{demo}{numbers=none,xleftmargin=0mm,framexleftmargin=0mm,linewidth=1.25\textwidth}
+\newcommand{\verbitem}[1]{\small\ttfamily{#1}}
+% stolen from arara.sty http://mirrors.med.harvard.edu/ctan/support/arara/doc/arara.sty
+\lstnewenvironment{yaml}[1][]{\lstset{%
+	basicstyle=\ttfamily,
+	numbers=left,
+	xleftmargin=1.5em,
+	breaklines=true,
+	numberstyle=\ttfamily\small,
+	columns=flexible,
+	mathescape=false,
+	#1,
+}}
+{}
+
+\newcommand{\fixthis}[1]
+{%
+	\marginpar{\huge \color{red} \framebox{FIX}}%
+	\typeout{FIXTHIS: p\thepage : #1^^J}%
+}
+% custom section
+\titleformat{\section}
+{\normalfont\Large\bfseries}
+{\llap{\thesection\hskip.5cm}}
+{0pt}
+{}
+% custom subsection
+\titleformat{\subsection}
+{\normalfont\bfseries}
+{\llap{\thesubsection\hskip.5cm}}
+{0pt}
+{}
+% custom subsubsection
+\titleformat{\subsubsection}
+{\normalfont\bfseries}
+{\llap{\thesubsubsection\hskip.5cm}}
+{0pt}
+{}
+
+
+\titlespacing\section{0pt}{12pt plus 4pt minus 2pt}{-5pt plus 2pt minus 2pt}
+\titlespacing\subsection{0pt}{12pt plus 4pt minus 2pt}{-6pt plus 2pt minus 2pt}
+\titlespacing\subsubsection{0pt}{12pt plus 4pt minus 2pt}{-6pt plus 2pt minus 2pt}
+
+
+% cleveref settings
+\crefname{table}{Table}{Tables}
+\Crefname{table}{Table}{Tables}
+\crefname{figure}{Figure}{Figures}
+\Crefname{figure}{Figure}{Figures}
+\crefname{section}{Section}{Sections}
+\Crefname{section}{Section}{Sections}
+\crefname{lstlisting}{Listing}{Listings}
+\Crefname{lstlisting}{Listing}{Listings}
+
+\begin{document}
+
+% \begin{noindent}
+	\title{\lstinline[basicstyle=\huge\ttfamily]!latexindent.pl!\\[1cm]
+		Version 1.1R}
+% \end{noindent}
+\author{Chris Hughes \footnote{smr01cmh AT users.sourceforge.net}}
+\maketitle
+\begin{abstract}
+	\lstinline!latexindent.pl! is a \lstinline!Perl! script that indents \lstinline!.tex!
+	files according to an indentation scheme that the user can modify to suit their 
+	taste. Environments, including those with alignment delimiters (such as \lstinline!tabular!), 
+	and commands, including those that can split braces and brackets across lines, 
+	are \emph{usually} handled correctly by the script. Options for \lstinline!verbatim!-like
+	environments and indentation after headings (such as \lstinline!\chapter!, \lstinline!\section!, etc) 
+	are also available.
+\end{abstract}
+			
+\tableofcontents
+\lstlistoflistings
+			
+\section{Before we begin}
+\subsection{Thanks}
+I first created \lstinline!latexindent.pl! to help me format chapter files 
+in a big project. After I blogged about it on the 
+\TeX{} stack exchange \cite{cmhblog} I received some positive feedback and 
+follow-up feature requests. A big thank you to Harish Kumar who has really 
+helped to drive the script forward and has put it through a number of challenging
+tests-- I look forward to more challenges in the future Harish!
+			 	 	 	 	 	
+The \lstinline!yaml!-based interface of \lstinline!latexindent.pl! was inspired 
+by the wonderful \lstinline!arara! tool; any similarities are deliberate, and 
+I hope that it is perceived as the compliment that it is. Thank you to Paulo Cereda and the 
+team for releasing this awesome tool; I initially worried that I was going to 
+have to make a GUI for \lstinline!latexindent.pl!, but the release of \lstinline!arara! 
+has meant there is no need. Thank you to Paulo for all of your advice and 
+encouragement.
+			 	 	 	 	 	
+\subsection{License}
+\lstinline!latexindent.pl! is free and open source, and it always will be.
+Before you start using it on any important files, bear in mind that \lstinline!latexindent.pl! has the option to overwrite your \lstinline!.tex! files.
+It will always make at least one backup (you can choose how many it makes, see \cpageref{page:onlyonebackup})
+but you should still be careful when using it. The script has been tested on many
+files, but there are some known limitations (see \cref{sec:knownlimitations}). 
+You, the user, are responsible for ensuring that you maintain backups of your files
+before running \lstinline!latexindent.pl! on them. I think it is important at this
+stage to restate an important part of the license here:
+\begin{quote}\itshape
+	This program is distributed in the hope that it will be useful,
+	but WITHOUT ANY WARRANTY; without even the implied warranty of
+	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+	GNU General Public License for more details.
+\end{quote}
+There is certainly no malicious intent in releasing this script, and I do hope
+that it works as you expect it to-- if it does not, please first of all 
+make sure that you have the correct settings, and then feel free to let me know with a 
+complete minimum working example as I would like to improve the code as much as possible. 
+			 	 	 	 	 	
+\begin{warning}
+	Before you try the script on anything important (like your thesis), test it 
+	out on the sample files that come with it in the \lstinline!success! directory. 
+\end{warning}
+			 	 	 	 	 
+			 	 	 	 	 	
+\section{Demonstration: before and after}
+Let's give a demonstration of some before and after code-- after all, you probably
+won't want to try the script if you don't much like the results. You might also 
+like to watch the video demonstration I made on youtube \cite{cmh:videodemo}
+			     
+As you look at \crefrange{lst:filecontentsbefore}{lst:pstricksafter}, remember
+that \lstinline!latexindent.pl! is just following its rules-- there is nothing 
+particular about these code snippets. All of the rules can be modified 
+so that each user can personalize their indentation scheme. 
+			
+In each of the samples given in \crefrange{lst:filecontentsbefore}{lst:pstricksafter}
+the `before' case is a `worst case scenario' with no effort to make indentation. The `after'
+result would be the same, regardless of the leading white space at the beginning of 
+each line which is stripped by \lstinline!latexindent.pl! (unless a \lstinline!verbatim!-like
+environment or \lstinline!noIndentBlock! is specified-- more on this in \cref{sec:defuseloc}).
+			
+\begin{adjustwidth}{-2cm}{2cm}
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[style=demo,caption={\lstinline!filecontents! before},label={lst:filecontentsbefore}]
+\begin{filecontents}{mybib.bib}
+@online{strawberryperl,
+title="Strawberry Perl",
+url="http://strawberryperl.com/"}
+@online{cmhblog,
+title="A Perl script ...
+url="...
+\end{filecontents}
+		\end{lstlisting}
+	\end{minipage}%
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[style=demo,caption={\lstinline!filecontents! after}]
+\begin{filecontents}{mybib.bib}
+	@online{strawberryperl,
+		title="Strawberry Perl",
+		url="http://strawberryperl.com/"}
+	@online{cmhblog,
+		title="A Perl script for ...
+		url="...
+\end{filecontents}
+		\end{lstlisting}
+	\end{minipage}
+																					 	 	 	 	 				
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[style=demo,caption={\lstinline!tikzset! before}]
+\tikzset{
+shrink inner sep/.code={
+\pgfkeysgetvalue...
+\pgfkeysgetvalue...
+}
+}
+		\end{lstlisting}
+	\end{minipage}%
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[style=demo,caption={\lstinline!tikzset! after}]
+\tikzset{
+	shrink inner sep/.code={
+		\pgfkeysgetvalue...
+		\pgfkeysgetvalue...
+	}
+}
+		\end{lstlisting}
+	\end{minipage}
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[style=demo,caption={\lstinline!pstricks! before}]
+\def\Picture#1{%
+\def\stripH{#1}%
+\begin{pspicture}[showgrid...
+\psforeach{\row}{%
+{{3,2.8,2.7,3,3.1}},% <=== Only this 
+{2.8,1,1.2,2,3},%
+...
+}{%
+\expandafter...
+}
+\end{pspicture}}
+		\end{lstlisting}
+	\end{minipage}%
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[style=demo,caption={\lstinline!pstricks! after},label={lst:pstricksafter}]
+\def\Picture#1{%
+	\def\stripH{#1}%
+	\begin{pspicture}[showgrid...
+		\psforeach{\row}{%
+			{{3,2.8,2.7,3,3.1}},% <=== 
+			{2.8,1,1.2,2,3},%
+            ...
+			}{%
+			\expandafter...
+		}
+	\end{pspicture}}
+		\end{lstlisting}
+	\end{minipage}
+\end{adjustwidth}
+			     
+\section{How to use the script}
+There are two ways to use \lstinline!latexindent.pl!: from the command line, 
+and using \lstinline!arara!.  We will discuss how to change the settings and behaviour 
+of the script in \cref{sec:defuseloc}.
+			
+\lstinline!latexindent.pl! ships with \lstinline!latexindent.exe! for Windows 
+users, so that you can use the script with or without a Perl distribution. 
+If you plan to use \lstinline!latexindent.pl! (i.e, the original Perl script) then you will 
+need a few standard Perl modules-- see \vref{sec:requiredmodules} for details.
+			
+In what follows, we will always refer to \lstinline!latexindent.pl!, but depending on 
+your operating system and preference, you might substitute \lstinline!latexindent.exe! or
+simply \lstinline!latexindent!.
+			     
+\subsection{From the command line}\label{sec:commandline}
+\lstinline!latexindent.pl! has a number of different switches/flags/options, which 
+can be combined in any way that you like. \lstinline!latexindent.pl! 
+produces a \lstinline!.log! file, \lstinline!indent.log! every time it
+is run. There is a base of information that is written to \lstinline!indent.log!,
+but other additional information will be written depending 
+on which of the following options are used.
+			 	 	 	 	 	
+\begin{itemize}[labelsep=.5cm]
+	\item[] \lstinline!latexindent.pl!
+																					
+	This will output a welcome message to the terminal, including the version number
+	and available options.
+	\item[\verbitem{-h}] \lstinline!latexindent.pl -h!
+																					
+	As above this will output a welcome message to the terminal, including the version number
+	and available options.
+	\item[] \lstinline!latexindent.pl myfile.tex!
+																						 		 		 		 		 					
+	This will operate on \lstinline!myfile.tex!, but will simply output to your terminal; \lstinline!myfile.tex! will	not be changed in any way using this command. 
+	\item[\verbitem{-w}] \lstinline!latexindent.pl -w myfile.tex!
+																						 		 		 		 		 					
+	This \emph{will} overwrite \lstinline!myfile.tex!, but it will
+	make a copy of \lstinline!myfile.tex! first. You can control the name of 
+	the extension (default is \lstinline!.bak!), and how many different backups are made-- 
+	more on this in \cref{sec:defuseloc}; see \lstinline!backupExtension! and \lstinline!onlyOneBackUp!.
+																						 		 		 		 		 					
+	Note that if \lstinline!latexindent.pl! can not create the backup, then it 
+	will exit without touching your original file; an error message will be given
+	asking you to check the permissions of the backup file.
+	\item[\verbitem{-o}] \lstinline!latexindent.pl -o myfile.tex outputfile.tex!
+																						 		 		 		 		 					
+	This will indent \lstinline!myfile.tex! and output it to \lstinline!outputfile.tex!, 
+	overwriting it (\lstinline!outputfile.tex!) if it already exists. Note that if \lstinline!latexindent.pl! is called with both
+	the \lstinline!-w! and \lstinline!-o! switches, then \lstinline!-w! will
+	be ignored and \lstinline!-o! will take priority (this seems safer than the 
+	other way round).
+																						 		 		 		 		 					
+	Note that using \lstinline!-o! is equivalent to using \lstinline!latexindent.pl myfile.tex > outputfile.tex!
+	\item[\verbitem{-s}] \lstinline!latexindent.pl -s myfile.tex!
+																						 		 		 		 		 					
+	Silent mode: no output will be given to the terminal.
+	\item[\verbitem{-t}] \lstinline!latexindent.pl -t myfile.tex!
+																						 		 		 		 		 					
+	Tracing mode: verbose output will be given to \lstinline!indent.log!. This 
+	is useful if \lstinline!latexindent.pl! has made a mistake and you're
+	trying to find out where and why. You might also be interested in learning
+	about \lstinline!latexindent.pl!'s thought process-- if so, this 
+	switch is for you.
+	\item[\verbitem{-l}] \lstinline!latexindent.pl -l myfile.tex!
+																						 		 		 		 		 					
+	\label{page:localswitch}
+	Local settings: you might like to read \cref{sec:defuseloc} before 
+	using this switch. \lstinline!latexindent.pl! will always load \lstinline!defaultSettings.yaml!
+	and if it is called with the \lstinline!-l! switch and it finds \lstinline!localSettings.yaml! 
+	in the same directory as \lstinline!myfile.tex! then these settings will be 
+	added to the indentation scheme. Information will be given in \lstinline!indent.log! on 
+	the success or failure of loading \lstinline!localSettings.yaml!.
+	\item[\verbitem{-d}] \lstinline!latexindent.pl -d myfile.tex!
+									
+	Only \lstinline!defaultSettings.yaml!: you might like to read \cref{sec:defuseloc} before 
+	using this switch. By default, \lstinline!latexindent.pl! will always search for 
+	\lstinline!indentconfig.yaml! in your home directory. If you would prefer it not to do so
+	then (instead of deleting or renaming \lstinline!indentconfig.yaml!) you can simply 
+	call the script with the \lstinline!-d! switch; note that this will also tell 
+	the script to ignore \lstinline!localSettings.yaml! even if it has been called with the 
+	\lstinline!-l! switch.
+								
+	\item[\verbitem{-c}]\lstinline!latexindent.pl -c=/path/to/directory/ myfile.tex!
+								
+	If you wish to have backup files and \lstinline!indent.log! written to a directory
+	other than the current working directory, then you can send these `cruft' files
+	to another directory. 
+    % this switch was made as a result of http://tex.stackexchange.com/questions/142652/output-latexindent-auxiliary-files-to-a-different-directory
+\end{itemize}
+			
+\subsection{From \lstinline!arara!}
+Using \lstinline!latexindent.pl! from the command line is fine for some folks, but
+others may find it easier to use from \lstinline!arara!. \lstinline!latexindent.pl!
+ships with an \lstinline!arara! rule, \lstinline!indent.yaml!, which can be copied 
+to the directory of
+your other \lstinline!arara! rules; otherwise  you can add the directory in which \lstinline!latexindent.pl!
+resides to your \lstinline!araraconfig.yaml! file.
+			 	 	 	 	 	
+Once you have told \lstinline!arara! where to find your \lstinline!indent! rule, 
+you can use it any of the ways described in \cref{lst:arara} (or combinations thereof). 
+In fact, \lstinline!arara! allows yet greater flexibility-- you can use \lstinline!yes/no!, \lstinline!true/false!, or \lstinline!on/off! to toggle the various options.
+			 	 	 	 	 	
+\begin{lstlisting}[caption={\lstinline!arara! sample usage},label={lst:arara},escapeinside={(*@}{@*)}]
+%(*@@*) arara: indent
+%(*@@*) arara: indent: {overwrite: yes}
+%(*@@*) arara: indent: {output: myfile.tex}
+%(*@@*) arara: indent: {silent: yes}
+%(*@@*) arara: indent: {trace: yes}
+%(*@@*) arara: indent: {localSettings: yes}
+%(*@@*) arara: indent: {onlyDefault: on}
+%(*@@*) arara: indent: { cruft: /home/cmhughes/Desktop }
+\documentclass{article}
+...
+\end{lstlisting}
+			 	 	 	 	 	
+Hopefully the use of these rules is fairly self-explanatory, but for completeness
+\cref{tab:orbsandswitches} shows the relationship between \lstinline!arara! directive arguments and the 
+switches given in \cref{sec:commandline}.
+			 	 	 	 	 	
+\begin{table}[!ht]
+	\centering
+	\caption{\lstinline!arara! directive arguments and corresponding switches}
+	\label{tab:orbsandswitches}
+	\begin{tabular}{lc}
+		\toprule
+		\lstinline!arara! directive argument & switch         \\
+		\midrule
+		\lstinline!overwrite!                & \lstinline!-w! \\
+		\lstinline!output!                   & \lstinline!-o! \\
+		\lstinline!silent!                   & \lstinline!-s! \\
+		\lstinline!trace!                    & \lstinline!-t! \\
+		\lstinline!localSettings!            & \lstinline!-l! \\
+		\lstinline!onlyDefault!              & \lstinline!-d! \\
+		\lstinline!cruft!                    & \lstinline!-c! \\
+		\bottomrule
+	\end{tabular}
+\end{table}
+		
+The \lstinline!cruft! directive does not work well when used with 
+directories that contain spaces.
+		
+\section{default, user, and local settings}\label{sec:defuseloc}
+\lstinline!latexindent.pl! loads its settings from \lstinline!defaultSettings.yaml! 
+(rhymes with camel). The idea is to separate the behaviour of the script 
+from the internal working-- this is very similar to the way that we separate content
+from form when writing our documents in \LaTeX.
+			     
+\subsection{\lstinline!defaultSettings.yaml!}
+If you look in \lstinline!defaultSettings.yaml! you'll find the switches 
+that govern the behaviour of \lstinline!latexindent.pl!. If you're not sure where
+\lstinline!defaultSettings.yaml! resides on your computer, don't worry as \lstinline!indent.log!
+will tell you where to find it.
+\lstinline!defaultSettings.yaml! is commented, 
+but here is a description of what each switch is designed to do. The default 
+value is given in each case.
+			 	 	 	 	 	
+You can certainly feel free to edit \lstinline!defaultSettings.yaml!, but 
+this is not ideal as it may be overwritten when you update your distribution--
+all of your hard work tweaking the script would be undone! Don't worry, 
+there's a solution-- feel free to peek ahead to \cref{sec:indentconfig} if you like.
+\begin{itemize}
+	\item[\verbitem{defaultIndent}] \lstinline!"\t"!
+																						 		 		 		 		 					
+	This is the default indentation (\lstinline!\t! means a tab) used in the absence of other details 
+	for the command or environment we are working with-- see \lstinline!indentRules!
+	for more details (\cpageref{page:indentRules}).
+																						 		 		 		 		 					
+	If you're interested in experimenting with \lstinline!latexindent.pl! then you 
+	can \emph{remove} all indentation by setting \lstinline!defaultIndent: ""!
+	\item[\verbitem{backupExtension}] \lstinline!.bak!
+																						 		 		 		 		 					
+	If you call \lstinline!latexindent.pl! with the \lstinline!-w! switch (to overwrite
+	\lstinline!myfile.tex!) then it will create a backup file before doing 
+	any indentation: \lstinline!myfile.bak0! 
+																						 		 		 		 		 					
+	By default, every time you call \lstinline!latexindent.pl! after this with 
+	the \lstinline!-w! switch it will create \lstinline!myfile.bak1!, \lstinline!myfile.bak2!, 
+	etc.
+	\item[\verbitem{onlyOneBackUp}] \lstinline!0!
+																						 		 		 		 		 					
+	\label{page:onlyonebackup}
+	If you don't want a backup for every time that you call \lstinline!latexindent.pl! (so 
+	you don't want \lstinline!myfile.bak1!, \lstinline!myfile.bak2!, etc) and you simply
+	want \lstinline!myfile.bak! (or whatever you chose \lstinline!backupExtension! to be)
+	then change \lstinline!onlyOneBackUp! to \lstinline!1!.
+								
+	\item[\verbitem{maxNumberOfBackUps}]\lstinline!0!
+								
+	Some users may only want a finite number of backup files, 
+	say at most $3$, in which case, they can change this switch.
+	The smallest value of \lstinline!maxNumberOfBackUps! is $0$ which will \emph{not} 
+	prevent back up files being made-- in this case, the behaviour will be dictated 
+    entirely by \lstinline!onlyOneBackUp!. 
+    %\footnote{This was a feature request made on \href{https://github.com/cmhughes/latexindent.plx}{github}}
+								
+	\item[\verbitem{indentPreamble}] \lstinline!0!
+																						 		 		 		 		 					
+	The preamble of a document can sometimes contain some trickier code 
+	for \lstinline!latexindent.pl! to work with. By default, \lstinline!latexindent.pl!
+	won't try to operate on the preamble, but if you'd like it to try then 
+	change \lstinline!indentPreamble! to \lstinline!1!.
+	\item[\verbitem{alwaysLookforSplitBraces}] \lstinline!1!
+																						 		 		 		 		 					
+	This switch tells \lstinline!latexindent.pl! to look for commands that 
+	can split \emph{braces} across lines, such as \lstinline!parbox!, \lstinline!tikzset!, etc. In older
+	versions of \lstinline!latexindent.pl! you had to specify each one in \lstinline!checkunmatched!-- this 
+	clearly became tedious, hence the introduction of \lstinline!alwaysLookforSplitBraces!. 
+																						 		 		 		 		 					
+	\emph{As long as you leave this switch on (set to 1) you don't need to specify which 
+		commands can split braces across lines-- you can ignore the 
+		fields \lstinline!checkunmatched! and \lstinline!checkunmatchedELSE! described later}.
+	\item[\verbitem{alwaysLookforSplitBrackets}] \lstinline!1!
+																						 		 		 		 		 					
+	This switch tells \lstinline!latexindent.pl! to look for commands that 
+	can split \emph{brackets} across lines, such as \lstinline!psSolid!, \lstinline!pgfplotstabletypeset!, 
+	etc. In older versions of \lstinline!latexindent.pl! you had to specify each one in \lstinline!checkunmatchedbracket!-- 
+	this clearly became tedious, hence the introduction of \lstinline!alwaysLookforSplitBraces!. 
+																						 		 		 		 		 					
+	\emph{As long as you leave this switch on (set to 1) you don't need to specify which 
+		commands can split brackets across lines-- you can ignore \lstinline!checkunmatchedbracket! described later}.
+							
+	\item[\verbitem{removeTrailingWhitespace}] \lstinline!0!
+	
+	By default \lstinline!latexindent.pl! indents every line (including empty lines)
+	which creates `trailing whitespace' feared by most version control systems. If
+	this option is set to \lstinline!1!, trailing whitespace is removed from all
+	lines, also non-empty ones. In general this should not create any problems, but
+    by precaution this option is turned off by default. \footnote{Thanks to \href{https://github.com/vosskuhle}{vosskuhle} for 
+	providing this feature.}
+																 		 		 		 		 					
+	\item[\verbitem{lookForAlignDelims}] This is the first example of a field
+	in \lstinline!defaultSettings.yaml! that has more than one line; \cref{lst:aligndelims}
+	shows more details.
+																						 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!lookForAlignDelims!},label={lst:aligndelims}]
+lookForAlignDelims:
+   tabular: 1
+   array: 1
+   matrix: 1
+   bmatrix: 1
+   pmatrix: 1
+   align: 1
+   align*: 1
+   alignat: 1
+   alignat*: 1
+   aligned: 1
+   cases: 1
+   dcases: 1
+   pmatrix: 1
+   listabla: 1
+	\end{yaml}
+																						 		 		 		 		 					
+	The environments specified in this field will be operated on in a special way  by \lstinline!latexindent.pl!. In particular, it will try and align each column by its alignment
+	tabs. It does have some limitations (discussed further in \cref{sec:knownlimitations}), 
+	but in many cases it will produce results such as those in \cref{lst:tabularbefore,lst:tabularafter}. 
+																						 		 		 		 		 					
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[caption={\lstinline!tabular! before},label={lst:tabularbefore}]
+\begin{tabular}{cccc}
+1&	2 &3       &4\\
+5& &6       &\\
+\end{tabular}
+		\end{lstlisting}
+	\end{minipage}
+	\begin{minipage}{.5\textwidth}
+		\begin{lstlisting}[caption={\lstinline!tabular! after},label={lst:tabularafter}]
+\begin{tabular}{cccc}
+ 1 & 2 & 3 & 4 \\
+ 5 &   & 6 &   \\
+\end{tabular}
+		\end{lstlisting}
+	\end{minipage}
+																					 		 		 		 		 					
+	If you find that \lstinline!latexindent.pl! does not perform satisfactorily on such 
+	environments then you can either remove them from \lstinline!lookForAlignDelims! altogether, or set the relevant key to \lstinline!0!, for example \lstinline!tabular: 0!, or if you just want to ignore \emph{specific} 
+	instances of the environment, you could wrap them in something from \lstinline!noIndentBlock! (see \cref{lst:noIndentBlock}).
+																					 		 		 		 		 					
+	\item[\verbitem{verbatimEnvironments}] A field that contains a list of environments
+	that you would like left completely alone-- no indentation will be done
+	to environments that you have specified in this field-- see \cref{lst:verbatimEnvironments}.
+																					 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!verbatimEnvironments!},label={lst:verbatimEnvironments}]
+verbatimEnvironments:
+    verbatim: 1
+    lstlisting: 1
+	\end{yaml}
+	Note that if  you put an environment in \lstinline!verbatimEnvironments! 
+	and in other fields such as \lstinline!lookForAlignDelims! or \lstinline!noAdditionalIndent! 
+	then \lstinline!latexindent.pl! will \emph{always} prioritize \lstinline!verbatimEnvironments!.
+																					 		 		 		 		 					
+	\item[\verbitem{noIndentBlock}] If you have a block of code that you don't 
+	want \lstinline!latexindent.pl! to touch (even if it is \emph{not} a verbatim-like
+	environment) then you can wrap it in an environment from \lstinline!noIndentBlock!;
+	you can use any name you like for this, provided you populate it as demonstrate in 
+	\cref{lst:noIndentBlock}.
+																					 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!noIndentBlock!},label={lst:noIndentBlock}]
+noIndentBlock:
+    noindent: 1
+    cmhtest: 1
+	\end{yaml}
+																					 		 		 		 		 					
+	Of course, you don't want to have to specify these as null environments
+	in your code, so you use them with a comment symbol, \lstinline!%!, followed 
+	by as many spaces (possibly none) as you like; see \cref{lst:noIndentBlockdemo} for 
+	example.
+	\begin{lstlisting}[caption={\lstinline!noIndentBlock! demonstration},label={lst:noIndentBlockdemo},escapeinside={(*@}{@*)}]
+%(*@@*) \begin{noindent} 
+        this code
+                won't 
+     be touched
+                    by 
+             latexindent.pl!
+%(*@@*)\end{noindent} 
+	\end{lstlisting}
+																					 		 		 		 		 					
+	\item[\verbitem{noAdditionalIndent}] If you would prefer some of your
+	environments or commands not to receive any additional indent, then 
+	populate \lstinline!noAdditionalIndent!; see \cref{lst:noAdditionalIndent}. 
+	Note that these environments will still receive the \emph{current} level
+	of indentation unless they belong to \lstinline!verbatimEnvironments!, or \lstinline!noIndentBlock!.
+																					 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!noAdditionalIndent!},label={lst:noAdditionalIndent}]
+noAdditionalIndent:
+    document: 1
+    myexample: 1
+    mydefinition: 1
+    problem: 1
+    exercises: 1
+    mysolution: 1
+    foreach: 0
+    widepage: 1
+    comment: 1
+    \[: 1
+    \]: 1
+    frame: 0
+	\end{yaml}
+	Note in particular from \cref{lst:noAdditionalIndent} that if you wish content within 
+	\lstinline!\[!  and \lstinline!\]! to receive no additional content then 
+	you have to specify \emph{both} as \lstinline!1! (the default is \lstinline!0!). 
+	If you do not specify both as the same value you may get some interesting results!
+	\item[\verbitem{indentRules}] If\label{page:indentRules} you would prefer to specify 
+	individual rules for certain environments or commands, just
+	populate \lstinline!indentRules!; see \cref{lst:indentRules}
+																					 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!indentRules!},label={lst:indentRules}]
+indentRules:
+   myenvironment: "\t\t"
+   anotherenvironment: "\t\t\t\t"
+   \[: "\t"
+	\end{yaml}      %%%%%\] just here to stop vim from colouring the rest of my code
+	Note that in contrast to \lstinline!noAdditionalIndent! you do \emph{not} 
+	need to specify both \lstinline!\[! and \lstinline!\]! in this field. 
+																					 		 		 		 		 					
+	If you put an environment in both \lstinline!noAdditionalIndent! and in 
+	\lstinline!indentRules! then \lstinline!latexindent.pl! will resolve the conflict 
+	by ignoring \lstinline!indentRules! and prioritizing \lstinline!noAdditionalIndent!.
+	You will get a warning message in \lstinline!indent.log!; note that you will only 
+	get one warning message per command or environment. Further discussion 
+	is given in \cref{sec:fieldhierachy}.
+																					 		 		 		 		 			
+	\item[\verbitem{indentAfterHeadings}] This field enables the user to specify 
+	indentation rules that take effect after heading commands such as \lstinline!\part!, \lstinline!\chapter!, 
+	\lstinline!\section!, \lstinline!\subsection*! etc. This field is slightly different from all 
+	of the fields that we have considered previously, because each element is 
+	itself a field which has two elements: \lstinline!indent! and \lstinline!level!.
+	\begin{yaml}[caption={\lstinline!indentAfterHeadings!},label={lst:indentAfterHeadings}]
+indentAfterHeadings:
+    part:
+       indent: 0
+       level: 1
+    chapter: 
+       indent: 0
+       level: 2
+    section:
+       indent: 0
+       level: 3
+    ...
+	\end{yaml}
+	The default settings do \emph{not} place indentation after a heading-- you 
+	can easily switch them on by changing \lstinline!indent: 0! to \lstinline!indent: 1!.
+	The \lstinline!level! field tells \lstinline!latexindent.pl! the hierarchy of the heading 
+	structure in your document. You might, for example, like to have both \lstinline!section!
+	and \lstinline!subsection! set with \lstinline!level: 3! because you do not want the indentation to go too deep. 
+																					 		 		 		
+	You can add any of your own custom heading commands to this field, specifying the \lstinline!level!
+	as appropriate.  You can also specify your own indentation in \lstinline!indentRules!-- 
+	you will find the default \lstinline!indentRules! contains \lstinline!chapter: " "! which 
+	tells \lstinline!latexindent.pl! simply to use a space character after \lstinline!\chapter! headings 
+	(once \lstinline!indent! is set to \lstinline!1! for \lstinline!chapter!).
+	\begin{warning}
+		\emph{The following fields are marked in red, as they are not necessary
+			unless you wish to micro-manage your indentation scheme.
+			Note that in each case, you should \emph{not} use the backslash.}
+	\end{warning}
+																					 		 		 		 		 					
+	% to anyone reading the source code- I know the next line isn't the
+	% correct way to do it :)
+	\item[\color{red}\verbitem{checkunmatched}] Assuming you keep \lstinline!alwaysLookforSplitBraces! set to \lstinline!1! (which
+	is the default) then you don't need to worry about \lstinline!checkunmatched!. 
+																					 		 		 		 		 					
+	Should you wish to deactivate \lstinline!alwaysLookforSplitBraces! by setting it to \lstinline!0!, then 
+	you can populate \lstinline!checkunmatched! with commands that can split braces across 
+	lines-- see \cref{lst:checkunmatched}.
+																					 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!checkunmatched!},label={lst:checkunmatched}]
+checkunmatched:
+    parbox: 1
+    vbox: 1
+	\end{yaml}
+	\item[\color{red}\verbitem{checkunmatchedELSE}] Similarly, assuming you keep \lstinline!alwaysLookforSplitBraces! set to \lstinline!1! (which
+	is the default) then you don't need to worry about \lstinline!checkunmatchedELSE!. 
+																					 		 		 		 		 					
+	As in \lstinline!checkunmatched!, should you wish to deactivate \lstinline!alwaysLookforSplitBraces! by setting it to \lstinline!0!, then 
+	you can populate \lstinline!checkunmatchedELSE! with commands that can split braces across 
+	lines \emph{and} have an `else' statement-- see \cref{lst:checkunmatchedELSE}.
+																					 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!checkunmatchedELSE!},label={lst:checkunmatchedELSE}]
+checkunmatchedELSE:
+    pgfkeysifdefined: 1
+    DTLforeach: 1
+    ifthenelse: 1
+	\end{yaml}
+	\item[\color{red}\verbitem{checkunmatchedbracket}] Assuming you keep \lstinline!alwaysLookforSplitBrackets! 
+	set to \lstinline!1! (which is the default) then you don't need to worry about \lstinline!checkunmatchedbracket!. 
+																					 		 		 		 		 					
+	Should you wish to deactivate \lstinline!alwaysLookforSplitBrackets! by setting it 
+	to \lstinline!0!, then you can populate \lstinline!checkunmatchedbracket! with commands that can 
+	split \emph{brackets} across lines-- see \cref{lst:checkunmatchedbracket}.
+																					 		 		 		 		 					
+	\begin{yaml}[caption={\lstinline!checkunmatchedbracket!},label={lst:checkunmatchedbracket}]
+checkunmatchedbracket:
+    psSolid: 1
+    pgfplotstablecreatecol: 1
+    pgfplotstablesave: 1
+    pgfplotstabletypeset: 1
+    mycommand: 1
+	\end{yaml}
+\end{itemize}
+			 	 	 	 	 	
+\subsubsection{Hierarchy of fields}\label{sec:fieldhierachy}
+After reading the previous section, it should sound reasonable that 
+\lstinline!noAdditionalIndent!, \lstinline!indentRules!, and 
+\lstinline!verbatim! all serve mutually exclusive tasks. Naturally, you may 
+well wonder what happens if you choose to ask \lstinline!latexindent.pl! to 
+prioritize one above the other.
+			 	 	 	 	 	
+For example, let's say that you put the fields in \cref{lst:conflict} into 
+one of your settings files.  
+\begin{yaml}[caption={Conflicting ideas},label={lst:conflict}]
+indentRules:
+   myenvironment: "\t\t"
+noAdditionalIndent:
+   myenvironment: 1
+\end{yaml}
+			 	 	 	 	 	
+Clearly these fields conflict: first of all 
+you are telling \lstinline!latexindent.pl! that \lstinline!myenvironment! should 
+receive two tabs of indentation, and then you are telling it 
+not to put any indentation in the environment. \lstinline!latexindent.pl!
+will always make the decision to prioritize \lstinline!noAdditionalIndent! above
+\lstinline!indentRules! regardless of the order that you load them in 
+your settings file. The first 
+time it encounters \lstinline!myenvironment! it will put a warning in \lstinline!indent.log!
+and delete the offending key from \lstinline!indentRules! so that any future 
+conflicts won't have to be addressed.
+			 	 	 	 	 	
+Let's consider another conflicting example in \cref{lst:bigconflict}
+\begin{yaml}[caption={More conflicting ideas},label={lst:bigconflict}]
+lookForAlignDelims:
+   myenvironment: 1
+verbatimEnvironments:
+   myenvironment: 1
+\end{yaml}
+This is quite a significant conflict-- we are first of all telling \lstinline!latexindent.pl!
+to look for alignment delimiters in \lstinline!myenvironment! and then 
+telling it that actually we would like \lstinline!myenvironment! to be considered 
+as a \lstinline!verbatim!-like environment. Regardless of the order that we 
+state \cref{lst:bigconflict} the \lstinline!verbatim! instruction will always win.
+As in \cref{lst:conflict} you will only receive a warning in \lstinline!indent.log! the 
+first time \lstinline!latexindent.pl! encounters \lstinline!myenvironment! as the 
+offending key is deleted from \lstinline!lookForAlignDelims!.
+			 	 	 	 	 	
+To summarize, \lstinline!latexindent.pl! will prioritize the various fields in the 
+following order:
+\begin{enumerate}
+	\item \lstinline!verbatimEnvironments!
+	\item \lstinline!noAdditionalIndent!
+	\item \lstinline!indentRules!
+\end{enumerate}
+\subsection{\lstinline!indentconfig.yaml! (for user settings)}\label{sec:indentconfig}
+Editing \lstinline!defaultSettings.yaml! is not ideal as it may be overwritten when 
+updating your distribution-- a better way to customize the settings to your liking 
+is to set up your own settings file, 
+\lstinline!mysettings.yaml! (or any name you like, provided it ends with \lstinline!.yaml!). 
+The only thing you have to do is tell \lstinline!latexindent.pl! where to find it. 
+			 	 	 	 	 	
+\lstinline!latexindent.pl! will always check your home directory for \lstinline!indentconfig.yaml! (unless 
+it is called with the \lstinline!-d! switch), 
+which is a plain text file you can create that contains the \emph{absolute}
+paths for any settings files that you wish \lstinline!latexindent.pl! to load.
+Note that Mac and Linux users home directory is \lstinline!~/username! while
+Windows (Vista onwards) is \lstinline!C:\Users\username! \footnote{If you're not sure 
+	where to put \lstinline!indentconfig.yaml!, don't 
+	worry \lstinline!latexindent.pl! will tell you in the log file exactly where to 
+put it assuming it doesn't exist already.}
+\Cref{lst:indentconfig} shows a sample \lstinline!indentconfig.yaml! file.
+			 	 	 	 	 	
+\begin{yaml}[caption={\lstinline!indentconfig.yaml! (sample)},label={lst:indentconfig}]
+# Paths to user settings for latexindent.pl
+#
+# Note that the settings will be read in the order you 
+# specify here- each successive settings file will overwrite
+# the variables that you specify
+
+paths:
+- /home/cmhughes/Documents/yamlfiles/mysettings.yaml
+- /home/cmhughes/folder/othersettings.yaml
+- /some/other/folder/anynameyouwant.yaml
+- C:\Users\chughes\Documents\mysettings.yaml
+- C:\Users\chughes\Desktop\test spaces\more spaces.yaml
+\end{yaml}
+			 	 	 	 	 	
+Note that the \lstinline!.yaml! files you specify in \lstinline!indentconfig.yaml!
+will be loaded in the order that you write them in. Each file doesn't have 
+to have every switch from \lstinline!defaultSettings.yaml!; in fact, I recommend 
+that you only keep the switches that you want to \emph{change} in these 
+settings files.
+			 	 	 	 	 	
+To get started with your own settings file, you might like to save a copy of 
+\lstinline!defaultSettings.yaml! in another directory and call it, for 
+example, \lstinline!mysettings.yaml!. Once you have added the path to \lstinline!indentconfig.yaml!
+feel free to start changing the switches and adding more environments to it 
+as you see fit-- have a look at \cref{lst:mysettings} for an example 
+that uses four tabs for the default indent, and adds the \lstinline!tabbing!
+environment to the list of environments that contains alignment delimiters.
+			 	 	 	 	 	
+\begin{yaml}[caption={\lstinline!mysettings.yaml! (example)},label={lst:mysettings}]
+# Default value of indentation
+defaultIndent: "\t\t\t\t"
+
+# environments that have tab delimiters, add more 
+# as needed
+lookForAlignDelims:
+   tabbing: 1
+\end{yaml}
+			 	 	 	 	 	
+You can make sure that your settings are loaded by checking \lstinline!indent.log!
+for details-- if you have specified a path that \lstinline!latexindent.pl! doesn't 
+recognize then you'll get a warning, otherwise you'll get confirmation that 
+\lstinline!latexindent.pl! has read your settings file \footnote{Windows users 
+	may find that they have to end \lstinline!.yaml! files with a blank line}.
+			 	 	 	 	 	
+\begin{warning}
+	When editing \lstinline!.yaml! files it is \emph{extremely} important 
+	to remember how sensitive they are to spaces. I highly recommend copying 
+	and pasting from \lstinline!defaultSettings.yaml! when you create your
+	first \lstinline!whatevernameyoulike.yaml! file.
+																					
+	If \lstinline!latexindent.pl! can not read your \lstinline!.yaml! file it 
+	will tell you so in \lstinline!indent.log!.
+\end{warning}
+			 	 	 	 	 	
+\subsection{\lstinline!localSettings.yaml!}
+You may remember on \cpageref{page:localswitch} we discussed the \lstinline!-l! switch
+that tells \lstinline!latexindent.pl! to look for \lstinline!localSettings.yaml! in the 
+\emph{same directory} as \lstinline!myfile.tex!. This settings file will 
+be read \emph{after} \lstinline!defaultSettings.yaml! and, assuming they exist, 
+user settings. 
+			 	 	 	 	 	
+In contrast to the \emph{user} settings which can be named anything you like (provided that
+they are detailed in \lstinline!indentconfig.yaml!), the \emph{local} settings file
+must be called \lstinline!localSettings.yaml!. It can contain any switches that you'd
+like to change-- a sample is shown in \cref{lst:localSettings}.
+			 	 	 	 	 	
+\begin{yaml}[caption={\lstinline!localSettings.yaml! (example)},label={lst:localSettings}]
+# Default value of indentation
+defaultIndent: " "
+
+# environments that have tab delimiters, add more 
+# as needed
+lookForAlignDelims:
+   tabbing: 0
+
+#  verbatim environments- environments specified 
+#  in this hash table will not be changed at all!
+verbatimEnvironments:
+    cmhenvironment: 0
+\end{yaml}
+			 	 	 	 	 	
+You can make sure that your local settings are loaded by checking \lstinline!indent.log!
+for details-- if \lstinline!localSettings.yaml! can not be read then you will
+get a warning, otherwise you'll get confirmation that 
+\lstinline!latexindent.pl! has read \lstinline!localSettings.yaml!.
+			 	 	 	 	 	
+\subsection{Settings load order}\label{sec:loadorder}
+\lstinline!latexindent.pl! loads the settings files in the following order:
+\begin{enumerate}
+	\item \lstinline!defaultSettings.yaml! (always loaded, can not be renamed)
+	\item \lstinline!anyUserSettings.yaml! (and any other arbitrarily-named files specified in \lstinline!indentconfig.yaml!)
+	\item \lstinline!localSettings.yaml! (if found in same directory as \lstinline!myfile.tex! and called
+	with \lstinline!-l! switch; can not be renamed)
+\end{enumerate}
+A visual representation of this is given in \cref{fig:loadorder}.
+			
+\begin{figure}
+	\centering
+	\begin{tikzpicture}[
+			needed/.style={very thick, draw=blue,fill=blue!20,
+				text centered, minimum height=2.5em,rounded corners=1ex},
+			optional/.style={draw=black, very thick,scale=0.8,
+				text centered, minimum height=2.5em,rounded corners=1ex},
+			optionalfill/.style={fill=black!10},
+			connections/.style={draw=black!30,dotted,line width=3pt,text=red},
+		]
+		% Draw diagram elements
+		\node (latexindent) [needed,circle]  {\lstinline!latexindent.pl!};
+		\node (default) [needed,above right=.5cm of latexindent]  {\lstinline!defaultSettings.yaml!};
+		\node (indentconfig) [optional,right=of latexindent]  {\lstinline!indentconfig.yaml!};
+		\node (any) [optional,optionalfill,above right=of indentconfig]  {\lstinline!any.yaml!};
+		\node (name) [optional,optionalfill,right=of indentconfig]  {\lstinline!name.yaml!};
+		\node (you) [optional,optionalfill,below right=of indentconfig]  {\lstinline!you.yaml!};
+		\node (want) [optional,optionalfill,below=of indentconfig]  {\lstinline!want.yaml!};
+		\node (local) [optional,below=of latexindent]  {\lstinline!localSettings.yaml!};
+		% Draw arrows between elements
+		\draw[connections,solid] (latexindent) to[in=-90]node[pos=0.5,anchor=north]{1} (default.south) ;
+		\draw[connections,optional] (latexindent) -- node[pos=0.5,anchor=north]{2} (indentconfig) ;
+		\draw[connections,optional] (indentconfig) to[in=-90] (any.south) ;
+		\draw[connections,optional] (indentconfig) -- (name) ;
+		\draw[connections,optional] (indentconfig) to[out=-45,in=90] (you) ;
+		\draw[connections,optional] (indentconfig) -- (want) ;
+		\draw[connections,optional] (latexindent) -- node[pos=0.5,anchor=west]{3} (local) ;
+	\end{tikzpicture}
+	\caption{Schematic of the load order described in \cref{sec:loadorder}; solid lines represent 
+		mandatory files, dotted lines represent optional files. \lstinline!indentconfig.yaml! can 
+		contain as many files as you like-- the files will be loaded in order; if you specify 
+	settings for the same field in more than one file, the most recent takes priority. }
+	\label{fig:loadorder}
+\end{figure}
+			
+\subsection{An important example}
+I was working on a document that had the text shown in \cref{lst:casestudy}. 
+\begin{lstlisting}[caption={When to set \lstinline!alwaysLookforSplitBrackets=0!},label={lst:casestudy},escapeinside={(*@}{@*)}]
+Hence determine how many zeros the function $h(x)=f(x)-g(x)$
+has on the interval $[0,9)$.(*@\label{line:interval1}@*)
+\begin{shortsolution}
+	The function $h$ has $10$ zeros on the interval $[0,9)$.(*@\label{line:interval2}@*)
+\end{shortsolution}
+\end{lstlisting}
+I had allowed \lstinline!alwaysLookforSplitBrackets=1!, which is the default setting. 
+Unfortunately, this caused undesired results, as \lstinline!latexindent.pl! thought that the opening 
+\lstinline![!  in the interval notation (\cref{line:interval1,line:interval2}) 
+was an opening brace that needed to be closed (with a corresponding \lstinline!]!). Clearly 
+this was inappropriate, but also expected since \lstinline!latexindent.pl! was simply
+following its matching rules.
+			
+In this particular instance, I set up \lstinline!localSettings.yaml! 
+to contain \lstinline!alwaysLookforSplitBrackets: 0! and then specified the commands 
+that could split brackets across lines (such as \lstinline!begin{axis}!) individually
+in \lstinline!checkunmatchedbracket!.  Another option would have been to wrap the 
+the line in an environment from \lstinline!noIndentBlock! which treats its contents
+as a verbatim environment.
+			
+			 	 	 	 	 	
+\section{Known limitations}\label{sec:knownlimitations}
+There are a number of known limitations of the script, and almost certainly quite a
+few that are \emph{unknown}!
+			      
+The main limitation is to do with the alignment routine of environments that contain 
+delimiters-- in other words, environments that are entered in \lstinline!lookForAlignDelims!.
+Indeed, this is the only part of the script that can \emph{potentially} remove 
+lines from \lstinline!myfile.tex!. Note that \lstinline!indent.log! will always
+finish with a comparison of line counts before and after. 
+			      
+The routine works well for `standard' blocks of code that have the same number of \lstinline!&!
+per line, but it will not do anything for blocks that do not-- such examples 
+include \lstinline!tabular! environments that use \lstinline!\multicolumn! or 
+perhaps spread cell contents across multiple lines.  For each alignment block (\lstinline!tabular!, 
+\lstinline!align!, etc) \lstinline!latexindent.pl! first of all makes a record 
+of the maximum number of \lstinline!&!; if each row does not have that 
+number of \lstinline!&! then it will not try to format that row. Details 
+will be given in \lstinline!indent.log! assuming that \lstinline!trace! mode
+is active.
+			
+If you have a \lstinline!verbatim!-like environment inside a \lstinline!tabular!-like
+environment, the \lstinline!verbatim! environment \emph{will} be formatted, which 
+is probably not what you want. I hope to address this in future versions, but for the 
+moment wrap it in a \lstinline!noIndentBlock! (see \cpageref{lst:noIndentBlockdemo}).
+			      
+I hope that this script is useful to some-- if you find an example where the 
+script does not behave as you think it should, feel free to e-mail me or else
+come and find me on the \url{http://tex.stackexchange.com} site; I'm often around 
+and in the chat room.
+			      
+\printbibliography[heading=bibintoc]
+			
+\appendix
+\section{Required \lstinline!Perl! modules}\label{sec:requiredmodules}
+If you intend to use \lstinline!latexindent.pl! and \emph{not} one of the supplied standalone executable files, then you will need a few standard Perl modules-- if you can run the 
+minimum code in \cref{lst:helloworld} (\lstinline!perl helloworld.pl!) then you will be able to run \lstinline!latexindent.pl!, otherwise you may 
+need to install the missing modules.
+			 	 	 	 	 	
+\begin{lstlisting}[language=Perl,caption={\lstinline!helloworld.pl!},label={lst:helloworld}]
+#!/usr/bin/perl
+
+use strict;
+use warnings;           
+use FindBin;            
+use YAML::Tiny;         
+use File::Copy;         
+use File::Basename;     
+use Getopt::Std;        
+use File::HomeDir;      
+
+print "hello world";
+exit;
+\end{lstlisting}
+My default installation on Ubuntu 12.04 did \emph{not} come
+with all of these modules as standard, but Strawberry Perl for Windows \cite{strawberryperl}
+did.
+				
+Installing the modules given in \cref{lst:helloworld} will vary depending on your 
+operating system and \lstinline!Perl! distribution. For example, Ubuntu users 
+might visit the software center, and Strawberry Perl users on Windows might use 
+\lstinline!CPAN client!. All of the modules are readily available on CPAN \cite{cpan}.
+			
+\section{The \lstinline!arara! rule}
+The \lstinline!arara! rule (\lstinline!indent.yaml!) contains lines such as those
+given in \cref{lst:arararule}. With this setup, the user \emph{always} has 
+to specify whether or not they want (in this example) to use the \lstinline!trace!
+identifier.
+\begin{yaml}[caption={The \lstinline!arara! rule},label={lst:arararule},numbers=none]
+...
+arguments:
+- identifier: trace
+  flag: <arara> @{ isTrue( parameters.trace, "-t" ) }
+...
+\end{yaml}
+			
+If you would like to have the \lstinline!trace! option on by default every time you 
+call \lstinline!latexindent.pl! from \lstinline!arara! (without having to write \lstinline!% arara: indent: {trace: yes}!), then simply
+amend \cref{lst:arararule} so that it looks like \cref{lst:arararulemod}.
+\begin{yaml}[caption={The \lstinline!arara! rule (modified)},label={lst:arararulemod},numbers=none]
+...
+arguments:
+- identifier: trace
+  flag: <arara> @{ isTrue( parameters.trace, "-t" ) }
+  default: "-t"
+...
+\end{yaml}
+			
+With this modification in place, you now simply to write \lstinline!% arara: indent! and 
+\lstinline!trace! mode will be activated by default. If you wish to turn off \lstinline!trace!
+mode then you can write \lstinline!% arara: indent: {trace: off}!.
+			
+Of course, you can apply these types of modifications to \emph{any} of the identifiers, 
+but proceed with caution if you intend to do this for \lstinline!overwrite!.
+			
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/indent.yaml b/Master/texmf-dist/doc/support/latexindent/indent.yaml
new file mode 100644
index 00000000000..0db81a066f2
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/indent.yaml
@@ -0,0 +1,38 @@
+!config
+# indent rule for arara
+# author: Paulo Cereda, Chris Hughes
+# last updated: 11/9/2013
+# requires arara 3.0+
+#
+# Sample usage:
+#
+# % arara: indent
+# % arara: indent: {overwrite: yes}
+# % arara: indent: {output: myfile.tex, silent: no}
+# % arara: indent: {output: myfile.tex, silent: yes, overwrite: yes}
+# % arara: indent: {trace: true}
+# % arara: indent: {localSettings: true}
+# % arara: indent: {onlyDefault: on}
+# % arara: indent: { cruft: /home/cmhughes/Desktop }
+#
+# Directories with spaces will cause issues in the cruft call.
+# 
+# Note: output will take priority above overwrite
+identifier: indent
+name: Indent
+command: <arara> @{ isWindows( "cmd /c latexindent.exe", "latexindent.pl" ) } @{silent} @{trace} @{localSettings} @{cruft}@{ isNotEmpty( cruft, '="'.concat(parameters.cruft).concat('"') ) } @{overwrite}  @{onlyDefault} @{output} "@{file}" @{ isNotEmpty( output, '"'.concat(parameters.output).concat('"') ) }
+arguments:
+- identifier: overwrite
+  flag: <arara> @{ isTrue( parameters.overwrite, "-w" ) }
+- identifier: silent
+  flag: <arara> @{ isTrue( parameters.silent, "-s" ) }
+- identifier: trace
+  flag: <arara> @{ isTrue( parameters.trace, "-t" ) }
+- identifier: localSettings
+  flag: <arara> @{ isTrue( parameters.localSettings, "-l" ) }
+- identifier: output
+  flag: <arara> @{ isNotEmpty( parameters.output, "-o" ) }
+- identifier: onlyDefault
+  flag: <arara> @{ isTrue( parameters.onlyDefault, "-d" ) }
+- identifier: cruft
+  flag: <arara> @{ isNotEmpty( parameters.cruft, "-c" ) }
diff --git a/Master/texmf-dist/doc/support/latexindent/success/bigTest.tex b/Master/texmf-dist/doc/support/latexindent/success/bigTest.tex
new file mode 100644
index 00000000000..0e47c4a2cf2
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/bigTest.tex
@@ -0,0 +1,293 @@
+% arara: indent: {trace: true}
+\documentclass[10pt,twoside]{report}
+\begin{document}
+
+% needed for the mini-tableofcontents
+\dominitoc
+\faketableofcontents
+
+\fancyhf{} % delete current header and footer
+\fancyhead[LE,RO]{\bfseries\thepage}
+\fancyhead[LO,RE]{\tiny\rightmark}
+\fancyheadoffset[LE,LO]{4cm}
+
+\pagestyle{fancy}
+%\include{coverpage}
+\include{functions}
+%\include{exponentialfunctions}
+%\include{functions2}
+%\include{logarithms}
+%\include{polyrat}
+%\include{ideas}
+
+%=======================
+%   BEGIN SOLUTIONS
+%=======================
+
+% change the page geometry using \newgeometry
+%\cleardoublepage
+\clearpage
+%\setbool{@twoside}{false} 
+\fancyheadoffset[RE,RO]{2cm}
+\fancyheadoffset[LE,LO]{1cm}
+\renewcommand{\rightmark}{Solutions to Section \thesection}
+\fancyhead[LO,RE]{\rightmark}
+\newgeometry{left=4cm,right=4cm,showframe=true,
+	marginratio=1:1,
+	top=1.5cm,bottom=1.5cm,bindingoffset=0cm}
+
+% finish the php file
+\Writetofile{crossrefsWEB}{?>}
+
+% close the solutions files
+\Closesolutionfile{shortsolutions}
+\Closesolutionfile{longsolutions}
+%\Closesolutionfile{hints}
+\Closesolutionfile{crossrefsWEB}
+
+% when itemized lists are used in the solutions, they
+% are actually at 2nd depth because the solution environment 
+% uses an \itemize environment to get the indendation correct
+\setlist[itemize,2]{label=\textbullet}
+
+% SHORT solution to problem (show only odd, even, all)
+% Note: this renewenvironment needs to go here
+%       so that the answers package can still
+%       display correctly to the page if needed
+\newbool{oddproblemnumber}
+\renewenvironment{shortSoln}[1]{%
+	\exploregroups % needed to ignore {}
+	% before the environment starts - this is a stretchable space
+	\vskip 0.1cm plus 2cm minus 0.1cm%
+	\fullexpandarg % need this line so that '.' are counted
+	%
+	% either problems, or subproblems, e.g: 3.1 or 3.1.4 respectively
+	% determine which one by counting the '.'
+	\StrCount{#1}{.}[\numberofdecimals]
+	%
+	% find the problem number by splitting the string
+	\ifnumequal{\numberofdecimals}{0}%
+	{%
+		% problems, such as 4, 5, 6, ...
+		\def\problemnumber{#1}%
+	}%
+	{}%
+	\ifnumequal{\numberofdecimals}{1}%
+	{%
+		% subproblems, such as 4.3, 1.2, 10.5
+		\StrBefore{#1}{.}[\problemnumber]%
+	}%
+	{}%
+	\ifnumequal{\numberofdecimals}{2}%
+	{%
+		% subproblems such as 1.3.1, 1.2.4, 7.5.6
+		% note that these aren't currently used, but maybe someday
+		\StrBehind{#1}{.}[\newbit]%
+		\StrBefore{\newbit}{.}[\problemnumber]%
+	}%
+	{}%
+	%
+	% determine if the problem number is odd or even
+	% and depending on our choices above, display or not
+	\ifnumodd{\problemnumber}%
+	{%
+		% set a boolean that says the problem number is odd (used later)
+		\setbool{oddproblemnumber}{true}%
+		% display or not 
+		\ifbool{showoddsolns}%
+		{%
+			% if we want to show the odd problems
+			\ifbool{coreproblemYesNo}%
+			{%  Core problem
+				\expandafter\itemize[label=\llap{$\bigstar$ }\bfseries \hyperlink{prob:#1:\thechapter:\thesection}{#1}.]\item%
+			}%
+			{%  NOT Core problem
+				\expandafter\itemize[label=\bfseries \hyperlink{prob:#1:\thechapter:\thesection}{#1}.]\item%
+			}%
+		}%
+		{%
+			% otherwise don't show them!
+			\expandafter\comment%          
+		}%
+	}%
+	{%
+		% even numbered problem, set the boolean (used later)
+		\setbool{oddproblemnumber}{false}%
+		\ifbool{showevensolns}%
+		{%
+			% if we want to show the even problems
+			\ifbool{coreproblemYesNo}%
+			{%  Core problem
+				\expandafter\itemize[label=\llap{$\bigstar$ }\itshape \hyperlink{prob:#1:\thechapter:\thesection}{#1}.]\item%
+			}%
+			{%  NOT Core problem
+				\expandafter\itemize[label=\itshape \hyperlink{prob:#1:\thechapter:\thesection}{#1}.]\item%
+			}%
+		}%
+		{%
+			% otherwise don't show them!
+			\expandafter\comment%          
+		}%
+	}%
+}%
+{%
+	% after the environment finishes
+	\ifbool{oddproblemnumber}%
+	{%
+		% odd numbered problems 
+		\ifbool{showoddsolns}%
+		{%
+			% if we want to show the odd problems
+			% then the environment is finished
+			\enditemize%
+		}%
+		{%
+			% otherwise we need to finish the comment
+			\expandafter\endcomment%
+		}%
+	}%
+	{%
+		% even numbered problems
+		\ifbool{showevensolns}%
+		{%
+			% if we want to show the even problems
+			% then the environment is finished
+			\enditemize%
+		}%
+		{%
+			% otherwise we need to finish the comment
+			\expandafter\endcomment%
+		}%
+	}%
+}
+
+% LONG solution to problem (show only odd, even, all)
+% Note: this renewenvironment needs to go here
+%       so that the answers package can still
+%       display correctly to the page if needed
+\renewenvironment{longSoln}[1]{%
+	\exploregroups % needed to ignore {}
+	% before the environment starts - this is a stretchable space
+	\vskip 0.1cm plus 2cm minus 0.1cm%
+	\fullexpandarg % need this line so that '.' are counted
+	%
+	% either problems, or subproblems, e.g: 3.1 or 3.1.4 respectively
+	% determine which one by counting the '.'
+	\StrCount{#1}{.}[\numberofdecimals]
+	%
+	% find the problem number by splitting the string
+	\ifnumequal{\numberofdecimals}{0}%
+	{%
+		% problems, such as 4, 5, 6, ...
+		\def\problemnumber{#1}%
+	}%
+	{}%
+	\ifnumequal{\numberofdecimals}{1}%
+	{%
+		% problems, such as 4.3, 1.2, 10.5
+		\StrBefore{#1}{.}[\problemnumber]%
+	}%
+	{}%
+	\ifnumequal{\numberofdecimals}{2}%
+	{%
+		% subproblems such as 1.3.1, 1.2.4, 7.5.6
+		\StrBehind{#1}{.}[\newbit]%
+		\StrBefore{\newbit}{.}[\problemnumber]%
+	}%
+	{}%
+	%
+	% determine if the problem number is odd or even
+	% and depending on our choices above, display or not
+	\ifnumodd{\problemnumber}%
+	{%
+		% set a boolean that says the problem number is odd (used later)
+		\setbool{oddproblemnumber}{true}%
+		% display or not 
+		\ifbool{showoddsolns}%
+		{%
+			% if we want to show the odd problems
+			{\bfseries \hyperlink{prob:#1:\thechapter:\thesection}{#1}.}%
+		}%
+		{%
+			% otherwise don't show them!
+			\expandafter\comment%          
+		}%
+	}%
+	{%
+		% even numbered problem, set the boolean (used later)
+		\setbool{oddproblemnumber}{false}%
+		\ifbool{showevensolns}%
+		{%
+			% if we want to show the even problems
+			{\itshape \hyperlink{prob:#1:\thechapter:\thesection}{#1}.}%
+		}%
+		{%
+			% otherwise don't show them!
+			\expandafter\comment%          
+		}%
+	}%
+}%
+{%
+	% after the environment finishes
+	\ifbool{oddproblemnumber}%
+	{%
+		% odd numbered problems 
+		\ifbool{showoddsolns}%
+		{%
+			% if we want to show the odd problems
+			% then the environment is finished
+		}%
+		{%
+			% otherwise we need to finish the comment
+			\expandafter\endcomment%
+		}%
+	}%
+	{%
+		% even numbered problems
+		\ifbool{showevensolns}%
+		{%
+			% if we want to show the even problems
+			% then the environment is finished
+		}%
+		{%
+			% otherwise we need to finish the comment
+			\expandafter\endcomment%
+		}%
+	}%
+}
+
+% renew tikzpicture environment to make it use valign=t
+% on every one, which fixes vertical alignment of tikzpicture
+% with the solution label: http://tex.stackexchange.com/questions/30367/aligning-enumerate-labels-to-top-of-image
+\BeforeBeginEnvironment{tikzpicture}{\begin{adjustbox}{valign=t}}
+\AfterEndEnvironment{tikzpicture}{\end{adjustbox}}
+
+% do the same for the tabular environment
+\BeforeBeginEnvironment{tabular}{\begin{adjustbox}{valign=t}}
+\AfterEndEnvironment{tabular}{\end{adjustbox}}
+
+% set every picture in the solutions to have \solutionfigurewidth
+\pgfplotsset{
+	every axis/.append style={%
+		width=\solutionfigurewidth}}
+
+% input the SHORT solutions file
+\IfFileExists{shortsolutions.tex}{\input{shortsolutions.tex}}{}
+
+\clearpage
+% input the LONG solutions file
+%\IfFileExists{longsolutions.tex}{\input{longsolutions.tex}}{}
+
+\clearpage
+% input the HINTS file
+%\IfFileExists{hints.tex}{\input{hints.tex}}{}
+%=======================
+%   END SOLUTIONS
+%=======================
+
+%=======================
+%   INDEX
+%=======================
+\printindex
+
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/braceTest.tex b/Master/texmf-dist/doc/support/latexindent/success/braceTest.tex
new file mode 100644
index 00000000000..2fd9e34a10f
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/braceTest.tex
@@ -0,0 +1,57 @@
+% arara: indent: { overwrite: false, output: outputfile.tex }
+
+\hypersetup{%
+pdfstartview={%
+FitH \hypercalcbp{\paperheight-\topmargin-0in
+-\headheight-\headsep
+}
+}%
+%---------------------------------------------------------------------------
+}
+
+some other text
+some other text
+some other text
+
+\hypersetup{%
+pdfstartview={%
+FitH \hypercalcbp{\paperheight-\topmargin-0in
+-\headheight-\headsep
+}
+}}
+
+some other text
+some other text
+some other text
+
+\hypersetup{%
+pdfstartview={%
+FitH \hypercalcbp{\paperheight-\topmargin-0in
+-\headheight-\headsep
+}}}
+
+some other text
+some other text
+some other text
+
+
+\hypersetup{%
+pdfstartview={%
+FitH \hypercalcbp{\paperheight-\topmargin-0in
+-\headheight-\headsep
+}}
+}
+
+some other text
+some other text
+some other text
+
+\parbox{
+\begin{something}
+
+\end{something}
+}
+
+some other text
+some other text
+some other text
diff --git a/Master/texmf-dist/doc/support/latexindent/success/braceTestsmall.tex b/Master/texmf-dist/doc/support/latexindent/success/braceTestsmall.tex
new file mode 100644
index 00000000000..48ddbf68ae6
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/braceTestsmall.tex
@@ -0,0 +1,31 @@
+% arara: indent: {trace: on}
+\hypersetup{%
+pdfstartview={%
+FitH \hypercalcbp{\paperheight-\topmargin-0in
+-\headheight-\headsep }%
+}%
+%---------------------------------------------------------------------------
+}
+
+some other text
+some other text
+some other text
+
+\hypersetup{%
+pdfstartview={%
+FitH \hypercalcbp{\paperheight-\topmargin-0in
+-\headheight-\headsep }}%
+}
+
+some other text
+some other text
+some other text
+
+\hypersetup{%
+pdfstartview={\someothercommand{here}%
+FitH \hypercalcbp{\paperheight-\topmargin-0in
+-\headheight-\headsep }}}
+
+some other text
+some other text
+some other text
diff --git a/Master/texmf-dist/doc/support/latexindent/success/environments.tex b/Master/texmf-dist/doc/support/latexindent/success/environments.tex
new file mode 100644
index 00000000000..b33d9982896
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/environments.tex
@@ -0,0 +1,38 @@
+% arara: indent: { cruft: /home/cmhughes/Desktop/tmp/, overwrite: on, trace: yes, silent: yes, onlyDefault: no}
+some 
+other 
+text
+\begin{enumerate}
+	\item $3x^3y(-4x^3+y) = -12x^6y+3x^3y^2$
+	\item $3x^3+y(-4x^3+y) = 3x^3-4x^3y+y^2$
+	\item \begin{align*}
+	{\color{red}(3x^3+y)}(-4x^3+y) & =	{\color{red}(3x^3+y)}(-4x^3)+{\color{red}(3x^3+y)}(y) \\
+	& =	-12x^6-4x^3y+3x^3y+y^2                                \\
+	& =	-12x^6-x^3y+y^2                                       
+	\end{align*}
+\end{enumerate}
+\begin{enumerate}
+	\item $3x^3y(-4x^3+y) = -12x^6y+3x^3y^2$
+	\item $3x^3+y(-4x^3+y) = 3x^3-4x^3y+y^2$
+	\item 
+	\begin{align*}
+		{\color{red}(3x^3+y)}(-4x^3+y) & =	{\color{red}(3x^3+y)}(-4x^3)+{\color{red}(3x^3+y)}(y) \\
+		                               & =	-12x^6-4x^3y+3x^3y+y^2                                \\
+		                               & =	-12x^6-x^3y+y^2                                       
+	\end{align*}
+\end{enumerate}
+	
+\begin{something}
+	\begin{else}
+		again
+	\end{else}
+\end{something}
+no
+environments
+here
+\[ x^2+ 3x\]
+other text
+\[ 
+	x^2+ 3x
+\]
+other text
diff --git a/Master/texmf-dist/doc/support/latexindent/success/filecontents.tex b/Master/texmf-dist/doc/support/latexindent/success/filecontents.tex
new file mode 100644
index 00000000000..d6132684232
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/filecontents.tex
@@ -0,0 +1,73 @@
+% arara: indent: {overwrite: true, trace: false, localSettings: yes}
+
+% used with localSettings.yaml as follows
+%indentPreamble: 1
+%indentRules:
+%   @online: "\t\t\t\t"
+%   #myenvironment: "\t\t"
+%   myotherenvironment: "\t\t"
+%   \[: "\t\t"
+%   tabular: "\t\t\t"
+%noAdditionalIndent:
+%   @online: 0
+%   myotherenvironment: 1
+%   \[: 0
+%   \]: 0
+%   tabular: 0
+%   something: 0
+%   parbox: 1
+%verbatimEnvironments:
+%   myotherenvironment: 1
+%   tabular: 0
+%   someothername: 0
+
+
+%        \begin{noindent}
+here we are in a block
+% \end{noindent}
+some more
+\begin{tabular}{cccc}
+			1 & 2 & 3 & 4 \\
+			5 &   & 6 &   \\
+\end{tabular}
+
+another test
+\begin{tabular}{cccc}
+			1 & 2 & 3 & 4 \\
+			5 &   & 6 &   \\
+\end{tabular}
+
+\begin{something}
+	\parbox{something
+	else
+	goes 
+	here
+	}
+	some text some text
+	some text some text
+	some text some text
+	\[
+			x^2+2x
+	\]
+	some text some text
+	some text some text
+	some text some text
+	some text some text
+	some text some text
+\end{something}
+\begin{filecontents}{mybib.bib}
+	@online{strawberryperl,
+					title="Strawberry Perl",
+					url="http://strawberryperl.com/"}
+	@online{cmhblog,
+					title="A Perl script for indenting tex files",
+					url="http://tex.blogoverflow.com/2012/08/a-perl-script-for-indenting-tex-files/"}
+\end{filecontents}
+
+\begin{myotherenvironment}
+	some text goes here
+	some text goes here
+	some text goes here
+	some text goes here
+\end{myotherenvironment}
+
diff --git a/Master/texmf-dist/doc/support/latexindent/success/matrix.tex b/Master/texmf-dist/doc/support/latexindent/success/matrix.tex
new file mode 100644
index 00000000000..3d60ebc3905
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/matrix.tex
@@ -0,0 +1,14 @@
+% arara: indent: {overwrite: true, trace: true}
+\[ 
+	\begin{matrix}[cc|cccccc|c]  
+		                &   &                       &                       &           &                       & {\color{blue}\downarrow} & {\color{blue}\downarrow} & S                       \\\hline
+		                & 6 & {\color{red}\newmoon} &                       &           &                       &                          &                          & {\color{red}\leftarrow} \\
+		{\color{blue}*} & 5 &                       &                       &           &                       &                          & {\color{blue}\newmoon}   &                         \\
+		                & 4 &                       &                       &           & {\color{red}\newmoon} &                          &                          & {\color{red}\leftarrow} \\
+		                & 3 &                       & {\color{red}\newmoon} &           &                       &                          &                          & {\color{red}\leftarrow} \\
+		                & 2 &                       &                       & \fullmoon &                       &                          &                          &                         \\
+		{\color{blue}*} & 1 &                       &                       &           &                       & {\color{blue}\newmoon}   &                          &                         \\\hline
+		                &   & 1                     & 2                     & 3         & 4                     & 5                        & 6                        &                         \\
+		\%              &   & {\color{red}*}        & {\color{red}*}        &           & {\color{red}*}        &                          &                          &                         
+	\end{matrix}
+\]
diff --git a/Master/texmf-dist/doc/support/latexindent/success/nestedalignment.tex b/Master/texmf-dist/doc/support/latexindent/success/nestedalignment.tex
new file mode 100644
index 00000000000..2f6afd65169
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/nestedalignment.tex
@@ -0,0 +1,33 @@
+% arara: indent: { overwrite: on, trace: yes}
+\begin{tabular}{p{3cm}|c|p{8cm}}
+	Example & Choice                         & Why                                                                      \\ \hline \hline
+	\(
+	\left\{
+	\begin{array}{rl}
+	x+y & =6	\\
+	2x+y & =8 
+	\end{array}
+	\right.
+	\)
+	        & Substitution {\em or }Addition & Because it is easy to solve for $x$ in the 1st equation                  
+	{\em or}
+	Because it is easy to multiply the first equation by -1 \\ \hline
+	\(
+	\left\{
+	\begin{array}{rl}
+	3x-7y & =13 \\
+	6x+5y & =7  
+	\end{array}
+	\right.
+	\)
+	        & Addition                       & Because there is no obvious way to use substitution                      \\ \hline
+	\(
+	\left\{
+	\begin{array}{rl}
+	x-7y  & =13 \\
+	6x+5y & =7  
+	\end{array}
+	\right.
+	\)
+	        & Substitution                   & Because the first equation can easily be solved for one of the variables 
+\end{tabular}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/nestedalignment1.tex b/Master/texmf-dist/doc/support/latexindent/success/nestedalignment1.tex
new file mode 100644
index 00000000000..54a012f8e2b
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/nestedalignment1.tex
@@ -0,0 +1,25 @@
+% arara: indent: { overwrite: on, silent: no, trace: yes}
+\begin{tabular}{p{3cm}|c|p{8cm}}
+	x+y       & = & 6     \\
+	2x+y & \&\%=8 % \% &  
+	\%\&\%\%  &   &       \\ % &   & 2x+y & =8   
+	x+y       & = & 6     \\
+	2x+y & =8       
+	2x+y \&\& & = & 8 % trailine comment            
+\end{tabular}
+
+here's another line $\{ x^2 + 5x \}$
+\begin{minipage}{\textwidth}
+	content
+	content
+	content
+	content
+	content
+\end{minipage}\\[3cm]
+\begin{minipage}{\textwidth}
+	content
+	content
+	content
+	content
+	content
+\end{minipage}\\[3cm]
diff --git a/Master/texmf-dist/doc/support/latexindent/success/outputfile.tex b/Master/texmf-dist/doc/support/latexindent/success/outputfile.tex
new file mode 100644
index 00000000000..71278475a00
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/outputfile.tex
@@ -0,0 +1,57 @@
+% arara: indent: { overwrite: false, output: outputfile.tex }
+
+\hypersetup{%
+	pdfstartview={%
+		FitH \hypercalcbp{\paperheight-\topmargin-0in
+			-\headheight-\headsep
+		}
+	}%
+	%---------------------------------------------------------------------------
+}
+
+some other text
+some other text
+some other text
+
+\hypersetup{%
+	pdfstartview={%
+		FitH \hypercalcbp{\paperheight-\topmargin-0in
+			-\headheight-\headsep
+		}
+}}
+
+some other text
+some other text
+some other text
+
+\hypersetup{%
+	pdfstartview={%
+		FitH \hypercalcbp{\paperheight-\topmargin-0in
+			-\headheight-\headsep
+}}}
+
+some other text
+some other text
+some other text
+
+
+\hypersetup{%
+	pdfstartview={%
+		FitH \hypercalcbp{\paperheight-\topmargin-0in
+			-\headheight-\headsep
+	}}
+}
+
+some other text
+some other text
+some other text
+
+\parbox{
+	\begin{something}
+				
+	\end{something}
+}
+
+some other text
+some other text
+some other text
diff --git a/Master/texmf-dist/doc/support/latexindent/success/preamble.tex b/Master/texmf-dist/doc/support/latexindent/success/preamble.tex
new file mode 100644
index 00000000000..666277aa88f
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/preamble.tex
@@ -0,0 +1,29 @@
+% arara: indent: {trace: on}
+\documentclass[10pt,twoside]{report}
+\newenvironment{widepage}{\begin{adjustwidth}{-\offsetpage}{}%
+	\addtolength{\textwidth}{\offsetpage}}%
+{\end{adjustwidth}}
+
+% Define fix command
+% 	- it puts a comment in the margin
+% 	- it writes to a file with a list of things that need fixing
+\newcommand{\fixthis}[1]
+{%
+	\marginpar{\huge \color{red} \framebox{FIX}}%
+	\typeout{FIXTHIS: p\thepage : #1^^J}%
+}
+
+% Define pccname command
+% 	- it writes to the log file with a detail of the name-
+%     this is useful for tracking names for diversity purposes
+\newcommand{\pccname}[1]
+{%
+	#1%
+	\typeout{PCCNAME: p\thepage : #1}%
+}
+
+\begin{document}
+
+some text
+
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/pstricks1.tex b/Master/texmf-dist/doc/support/latexindent/success/pstricks1.tex
new file mode 100644
index 00000000000..2dfed9849f1
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/pstricks1.tex
@@ -0,0 +1,25 @@
+% arara: indent: {overwrite: true, silent: on}
+\documentclass[pstricks]{standalone}
+\usepackage{pstricks,multido}
+
+\def\Bottle#1{{\pscustom[linewidth=2pt]{%
+	\rotate{#1} 
+	\psline(-1,3.5)(-1,4)(1,4)(1,3.5)
+	\pscurve(3,2)(1,0)\psline(-1,0)
+	\pscurve(-3,2)(-1,3.5)}}}
+
+\def\BottleWithWater(#1)#2{%
+	\rput[c]{#2}(#1){%
+		\rput{*0}(0,0){%
+			\psclip{\Bottle{#2}}
+			\psframe*[linecolor=gray](-6,-2)(6,2)
+			\endpsclip}\rput{*0}(0,0){\Bottle{#2}}}}
+
+\begin{document}
+
+\multido{\iA=-45+5}{19}{%
+	\begin{pspicture}(-2.5,-0.5)(6,5.5)
+		\BottleWithWater(1.5,1){\iA}
+	\end{pspicture}}
+
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/pstricks2.tex b/Master/texmf-dist/doc/support/latexindent/success/pstricks2.tex
new file mode 100644
index 00000000000..7690da820e7
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/pstricks2.tex
@@ -0,0 +1,68 @@
+% arara: indent: {overwrite: true, trace: true, localSettings: off}
+\documentclass[pstricks,border=12pt]{standalone}
+\usepackage{pst-eucl}
+\usepackage[nomessages]{fp}
+
+\definecolor{myblue}{RGB}{37,111,197}
+\definecolor{mybrown}{RGB}{211,200,134}
+
+\def\rOne{0.75}
+\def\rTwo{0.75}
+\def\tOne{20}
+\def\tTwo{-45}
+\FPset\RowMaxIndex{4}% because zero based index
+\FPset\ColMaxIndex{4}% because zero based index
+
+
+\psset
+{
+	PointName=none,
+	PointSymbol=none,
+	fillstyle=solid,
+	linejoin=1,
+}
+
+\def\Bar(#1,#2)#3{%
+	\rput(!#2 \rTwo\space mul \tTwo\space PtoC){%
+		\rput(!#1 \rOne\space mul \tOne\space 180 add PtoC){%
+			\pstGeonode{O}(\rOne;\tOne){One}(\rTwo;\tTwo){Two}
+			\pstTranslation{O}{Two}{One}[Three]
+			\pnode(0,\stripH){O'}
+			\pstTranslation{O}{O'}{One,Two,Three}
+			\pnode(0,#3){O''}
+			\pstTranslation{O}{O''}{One,Two,Three}[One'',Two'',Three'']
+			\psset{fillcolor=mybrown}
+			\pspolygon(O'')(O)(Two)(Two'')
+			\pspolygon(Two'')(Two)(Three)(Three'')
+			\pspolygon(One'')(O'')(Two'')(Three'')
+			\psset{fillcolor=myblue,opacity=0.75,linestyle=none,linewidth=0}
+			\FPifeq{#1}{\RowMaxIndex}\pspolygon(O')(O)(Two)(Two')\fi
+			\FPifeq{#2}{\ColMaxIndex}\pspolygon(Two')(Two)(Three)(Three')\fi
+			\FPiflt{#3}{\stripH}\pspolygon(One')(O')(Two')(Three')\fi
+		}%
+	}%
+}
+
+\newcount\OuterIndex
+\def\SaveListContents#1\relax{\def\Contents{#1}}
+
+\def\Picture#1{%
+	\def\stripH{#1}%
+	\begin{pspicture}[showgrid=false](-2.5,-3.35)(3.05,3.05)
+		\psforeach{\row}{%
+			{{3,2.8,2.7,3,3.1}},% <=== Only this row must use double curly braces. It is a feature!
+			{2.8,1,1.2,2,3},%
+			{2.8,1,1.2,2,2.8},%
+			{2.6,1.6,1.8,1.9,1.8},%
+			{2.4,1.5,1.7,1.9,1.5}%
+			}{%
+			\expandafter\SaveListContents\row\relax
+			\OuterIndex=\psLoopIndex\relax
+			\psforeach{\col}{\Contents}{\Bar(\the\OuterIndex,\the\psLoopIndex){\col}}%
+			\psLoopIndex=\OuterIndex\relax
+		}
+	\end{pspicture}}
+
+\begin{document}
+\multido{\n=0.0+0.2}{17}{\Picture{\n}}
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/pstricks3.tex b/Master/texmf-dist/doc/support/latexindent/success/pstricks3.tex
new file mode 100644
index 00000000000..09c132f5c73
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/pstricks3.tex
@@ -0,0 +1,49 @@
+% arara: indent: {overwrite: yes}
+\documentclass[pstricks,border=12pt]{standalone}
+\usepackage{pst-node}
+\addtopsstyle{gridstyle}{gridlabels=0pt,strokeopacity=.25}
+
+\begin{document}
+\begin{pspicture}[showgrid=top](8,8)
+	\multips(0,.5)(0,1){8}{%
+		\multips(.5,0)(1,0){8}{%
+			\psline[linecolor=red](6pt;-135)(6pt;45)
+			\psline[linecolor=red](6pt;135)(6pt;-45)}}
+	\pscustom
+	[
+		dimen=middle,
+		fillstyle=eovlines*,
+		fillcolor=white,
+		hatchcolor=blue,
+		linecolor=blue,
+	]
+	{
+		\psframe(8,8)
+		\pspolygon
+		(3,1)
+		(3,4)
+		(1,4)
+		(1,6)
+		(2,6)
+		(2,7)
+		(7,7)
+		(7,4)
+		(6,4)
+		(6,6)
+		(5,6)
+		(5,4)
+		(6,4)
+		(6,3)
+		(7,3)
+		(7,1)
+		(6,1)
+		(6,2)
+		(5,2)
+		(5,1)
+	}
+	\psset{linecolor=blue,nodesep=7pt}
+	\pscircle*(4,2){2pt}
+	\pscircle*(4,3){2pt}
+	\pcline(4,2)(4,3)
+\end{pspicture}
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/sampleAFTER.tex b/Master/texmf-dist/doc/support/latexindent/success/sampleAFTER.tex
new file mode 100644
index 00000000000..cc7a2b6c4d7
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/sampleAFTER.tex
@@ -0,0 +1,5818 @@
+% A sample chapter file- it contains a lot of 
+% environments, including tabulars, align, etc
+%   
+% Don't try and compile this file using pdflatex etc, just 
+% compare the *format* of it to the format of the
+% sampleAFTER.tex
+%
+% In particular, compare the tabular and align-type
+% environments before and after running the script
+
+\section{Polynomial functions}
+\reformatstepslist{P} % the steps list should be P1, P2, \ldots
+In your previous mathematics classes you have studied \emph{linear} and 
+\emph{quadratic} functions. The most general forms of these types of 
+functions can be represented (respectively) by  the functions $f$ 
+and $g$ that have formulas
+\begin{equation}\label{poly:eq:linquad}
+	f(x)=mx+b, \qquad g(x)=ax^2+bx+c
+\end{equation}
+We know that $m$ is the slope of $f$, and that $a$ is the \emph{leading coefficient} 
+of $g$. We also know that the \emph{signs} of $m$ and $a$ completely 
+determine the behavior of the functions $f$ and $g$. For example, if $m>0$
+then $f$ is an \emph{increasing} function, and if $m<0$ then $f$ is 
+a \emph{decreasing} function.  Similarly, if $a>0$ then $g$ is 
+\emph{concave up} and if $a<0$ then $g$ is \emph{concave down}. Graphical 
+representations of these statements are given in \cref{poly:fig:linquad}.
+
+\begin{figure}[!htb]
+	\setlength{\figurewidth}{.2\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-10:8]{(x+2)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$m>0$}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-10:8]{-(x+2)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$m<0$}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-4:4]{(x^2-6)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$a>0$}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-4:4]{-(x^2-6)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$a<0$}
+	\end{subfigure}
+	\caption{Typical graphs of linear and quadratic functions.}
+	\label{poly:fig:linquad}
+\end{figure}
+
+Let's look a little more closely at the formulas for $f$ and $g$ in 
+\cref{poly:eq:linquad}. Note that the \emph{degree} 
+of $f$ is $1$ since the highest power of $x$ that is present in the 
+formula for $f(x)$ is $1$. Similarly, the degree of $g$ is $2$ since
+the highest power of $x$ that is present in the formula for $g(x)$ 
+is $2$.
+
+In this section we will build upon our knowledge of these elementary
+functions. In particular, we will generalize the functions $f$ and $g$ to a function $p$ that has 
+any degree that we wish.
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{essentialskills}
+	%===================================
+	%   Author: Hughes
+	%   Date:   March 2012
+	%===================================
+	\begin{problem}[Quadratic functions]
+	Every quadratic function has the form $y=ax^2+bx+c$; state the value 
+	of $a$ for each of the following functions, and hence decide if the 
+	parabola that represents the function opens upward or downward.
+	\begin{multicols}{2}
+		\begin{subproblem}
+			$F(x)=x^2+3$ 
+			\begin{shortsolution}
+				$a=1$; the parabola opens upward. 
+			\end{shortsolution}
+		\end{subproblem}
+		\begin{subproblem}
+			$G(t)=4-5t^2$ 
+			\begin{shortsolution}
+				$a=-5$; the parabola opens downward. 
+			\end{shortsolution}
+		\end{subproblem}
+		\begin{subproblem}
+			$H(y)=4y^2-96y+8$ 
+			\begin{shortsolution}
+				$a=4$; the parabola opens upward. 
+			\end{shortsolution}
+		\end{subproblem}
+		\begin{subproblem}
+			$K(z)=-19z^2$ 
+			\begin{shortsolution}
+				$m=-19$; the parabola opens downward. 
+			\end{shortsolution}
+		\end{subproblem}
+	\end{multicols}
+	Now let's generalize our findings for the most general quadratic function $g$
+	that has formula $g(x)=a_2x^2+a_1x+a_0$. Complete the following sentences.
+	\begin{subproblem}
+		When $a_2>0$, the parabola that represents $y=g(x)$ opens $\ldots$ 
+		\begin{shortsolution}
+			When $a_2>0$, the parabola that represents the function opens upward.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		When $a_2<0$, the parabola that represents $y=g(x)$ opens $\ldots$ 
+		\begin{shortsolution}
+			When $a_2<0$, the parabola that represents the function opens downward.
+		\end{shortsolution}
+	\end{subproblem}
+	\end{problem}
+\end{essentialskills}
+
+\subsection*{Power functions with positive exponents}
+The study of polynomials will rely upon a good knowledge 
+of power functions| you may reasonably ask, what is a power function?
+\begin{pccdefinition}[Power functions]
+Power functions have the form
+\[
+	f(x) = a_n x^n
+\]
+where $n$ can be any real number.
+
+Note that for this section we will only be concerned with the 
+case when $n$ is a positive integer.
+\end{pccdefinition}
+
+You may find assurance in the fact that you are already very comfortable 
+with power functions that have $n=1$ (linear) and $n=2$ (quadratic). Let's 
+explore some power functions that you might not be so familiar with.
+As you read \cref{poly:ex:oddpow,poly:ex:evenpow}, try and spot 
+as many patterns and similarities as you can.
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{pccexample}[Power functions with odd positive exponents]
+\label{poly:ex:oddpow}
+Graph each of the following functions, state their domain, and their 
+long-run behavior as $x\rightarrow\pm\infty$
+\[
+	f(x)=x^3,   \qquad  g(x)=x^5, \qquad h(x)=x^7
+\]
+\begin{pccsolution}
+The functions $f$, $g$, and $h$ are plotted in \cref{poly:fig:oddpow}.
+The domain of each of the functions $f$, $g$, and $h$ is $(-\infty,\infty)$. Note that 
+the long-run behavior of each of the functions is the same, and in particular
+\begin{align*}
+	f(x)\rightarrow\infty                           & \text{ as } x\rightarrow\infty  \\
+	\mathllap{\text{and }}   f(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty 
+\end{align*}
+The same results hold for $g$ and $h$.
+\end{pccsolution}
+\end{pccexample}
+
+\begin{figure}[!htb]
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-1.5,xmax=1.5,
+			   ymin=-5,ymax=5,
+			   xtick={-1.0,-0.5,...,1.0},
+			   minor ytick={-3,-1,...,3},
+			   grid=both,
+			   width=\textwidth,
+			   legend pos=north west,
+			   ]
+			   \addplot expression[domain=-1.5:1.5]{x^3};
+			   \addplot expression[domain=-1.379:1.379]{x^5};
+			   \addplot expression[domain=-1.258:1.258]{x^7};
+			   \addplot[soldot]coordinates{(-1,-1)} node[axisnode,anchor=north west]{$(-1,-1)$};
+			   \addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=south east]{$(1,1)$};
+			   \legend{$f$,$g$,$h$}
+			\end{axis}
+		\end{tikzpicture}
+		\caption{Odd power functions}
+		\label{poly:fig:oddpow}
+	\end{minipage}%
+	\hfill
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-2.5,xmax=2.5,
+			   ymin=-5,ymax=5,
+			   xtick={-2.0,-1.5,...,2.0},
+			   minor ytick={-3,-1,...,3},
+			   grid=both,
+			   width=\textwidth,
+			   legend pos=south east,
+			   ]
+			   \addplot expression[domain=-2.236:2.236]{x^2};
+			   \addplot expression[domain=-1.495:1.495]{x^4};
+			   \addplot expression[domain=-1.307:1.307]{x^6};
+			   \addplot[soldot]coordinates{(-1,1)} node[axisnode,anchor=east]{$(-1,1)$};
+			   \addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=west]{$(1,1)$};
+			   \legend{$F$,$G$,$H$}
+			\end{axis}
+		\end{tikzpicture}
+		\caption{Even power functions}
+		\label{poly:fig:evenpow}
+	\end{minipage}%
+\end{figure}
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{pccexample}[Power functions with even positive exponents]\label{poly:ex:evenpow}%
+Graph each of the following functions, state their domain, and their 
+long-run behavior as $x\rightarrow\pm\infty$
+\[
+	F(x)=x^2, \qquad G(x)=x^4, \qquad H(x)=x^6
+\]
+\begin{pccsolution}
+The functions $F$, $G$, and $H$ are plotted in \cref{poly:fig:evenpow}. The domain
+of each of the functions is $(-\infty,\infty)$. Note that the long-run behavior 
+of each of the functions is the same, and in particular
+\begin{align*}
+	F(x)\rightarrow\infty                          & \text{ as } x\rightarrow\infty  \\
+	\mathllap{\text{and }}   F(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty 
+\end{align*}
+The same result holds for $G$ and $H$.
+\end{pccsolution}
+\end{pccexample}
+
+\begin{doyouunderstand}
+	\begin{problem}
+	Repeat \cref{poly:ex:oddpow,poly:ex:evenpow} using (respectively)
+	\begin{subproblem}
+		$f(x)=-x^3,   \qquad  g(x)=-x^5, \qquad h(x)=-x^7$
+		\begin{shortsolution}
+			The functions $f$, $g$, and $h$ have domain $(-\infty,\infty)$ and 
+			are graphed below.
+			
+			\begin{tikzpicture}
+				\begin{axis}[
+				   framed,
+				   xmin=-1.5,xmax=1.5,
+				   ymin=-5,ymax=5,
+				   xtick={-1.0,-0.5,...,0.5},
+				   minor ytick={-3,-1,...,3},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   legend pos=north east,
+				   ]
+				   \addplot expression[domain=-1.5:1.5]{-x^3};
+				   \addplot expression[domain=-1.379:1.379]{-x^5};
+				   \addplot expression[domain=-1.258:1.258]{-x^7};
+				   \legend{$f$,$g$,$h$}
+				\end{axis}
+			\end{tikzpicture}
+			
+			Note that
+			\begin{align*}
+				f(x)\rightarrow-\infty                         & \text{ as } x\rightarrow\infty  \\
+				\mathllap{\text{and }}   f(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty 
+			\end{align*}
+			The same is true for $g$ and $h$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$F(x)=-x^2,   \qquad  G(x)=-x^4, \qquad H(x)=-x^6$
+		\begin{shortsolution}
+			The functions $F$, $G$, and $H$ have domain $(-\infty,\infty)$ and 
+			are graphed below.
+			
+			\begin{tikzpicture}
+				\begin{axis}[
+				   framed,
+				   xmin=-2.5,xmax=2.5,
+				   ymin=-5,ymax=5,
+				   xtick={-1.0,-0.5,...,0.5},
+				   minor ytick={-3,-1,...,3},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   legend pos=north east,
+				   ]
+				   \addplot expression[domain=-2.236:2.236]{-x^2};
+				   \addplot expression[domain=-1.495:1.495]{-x^4};
+				   \addplot expression[domain=-1.307:1.307]{-x^6};
+				   \legend{$F$,$G$,$H$}
+				\end{axis}
+			\end{tikzpicture}
+			
+			Note that
+			\begin{align*}
+				F(x)\rightarrow-\infty                          & \text{ as } x\rightarrow\infty  \\
+				\mathllap{\text{and }}   F(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty 
+			\end{align*}
+			The same is true for $G$ and $H$.
+		\end{shortsolution}
+	\end{subproblem}
+	\end{problem}
+\end{doyouunderstand}
+
+\subsection*{Polynomial functions}
+Now that we have a little more familiarity with power functions, 
+we can define polynomial functions. Provided that you were comfortable
+with our opening discussion about linear and quadratic functions (see 
+$f$ and $g$ in \cref{poly:eq:linquad}) then there is every chance 
+that you'll be able to master polynomial functions as well; just remember
+that polynomial functions are a natural generalization of linear
+and quadratic functions. Once you've studied the examples and problems
+in this section, you'll hopefully agree that polynomial functions
+are remarkably predictable.
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccdefinition}[Polynomial functions]
+Polynomial functions have the form
+\[
+	p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0
+\]
+where $a_n$, $a_{n-1}$, $a_{n-2}$, \ldots, $a_0$ are real numbers.
+\begin{itemize}
+	\item We call $n$ the degree of the polynomial, and require that $n$
+	is a non-negative integer;
+	\item $a_n$, $a_{n-1}$, $a_{n-2}$, \ldots, $a_0$ are called the coefficients;
+	\item We typically write polynomial functions in descending powers of $x$.
+\end{itemize}
+In particular, we call $a_n$ the \emph{leading} coefficient, and $a_nx^n$ the 
+\emph{leading term}.
+
+Note that if a polynomial is given in factored form, then the degree can be found 
+by counting the number of linear factors.
+\end{pccdefinition}
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{pccexample}[Polynomial or not]
+Identify the following functions as polynomial or not; if the function 
+is a polynomial, state its degree.
+\begin{multicols}{3}
+	\begin{enumerate}
+		\item $p(x)=x^2-3$     
+		\item $q(x)=-4x^{\nicefrac{1}{2}}+10$     
+		\item $r(x)=10x^5$
+		\item $s(x)=x^{-2}+x^{23}$
+		\item $f(x)=-8$
+		\item $g(x)=3^x$
+		\item $h(x)=\sqrt[3]{x^7}-x^2+x$
+		\item $k(x)=4x(x+2)(x-3)$
+		\item $j(x)=x^2(x-4)(5-x)$
+	\end{enumerate}
+\end{multicols}
+\begin{pccsolution}
+\begin{enumerate}
+	\item $p$ is a polynomial, and its degree is $2$.
+	\item $q$ is \emph{not} a polynomial, because $\frac{1}{2}$ is not an integer.
+	\item $r$ is a polynomial, and its degree is $5$.
+	\item $s$ is \emph{not} a polynomial, because $-2$ is not a positive integer.
+	\item $f$ is a polynomial, and its degree is $0$.
+	\item $g$ is \emph{not} a polynomial, because the independent 
+	variable, $x$, is in the exponent.
+	\item $h$ is \emph{not} a polynomial, because $\frac{7}{3}$ is not an integer.
+	\item $k$ is a polynomial, and its degree is $3$.
+	\item $j$ is a polynomial, and its degree is $4$.
+\end{enumerate}
+\end{pccsolution}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{pccexample}[Typical graphs]\label{poly:ex:typical}
+\Cref{poly:fig:typical} shows graphs of some polynomial functions;
+the ticks have deliberately been left off the axis to allow us to concentrate
+on the features of each graph. Note in particular that:
+\begin{itemize}
+	\item \cref{poly:fig:typical1} shows a degree-$1$ polynomial (you might also 
+	classify the function as linear) whose leading coefficient, $a_1$, is positive.
+	\item \cref{poly:fig:typical2} shows a degree-$2$ polynomial (you might also
+	classify the function as quadratic) whose leading coefficient, $a_2$, is positive.
+	\item \cref{poly:fig:typical3} shows a degree-$3$ polynomial whose leading coefficient, $a_3$,
+	is positive| compare its overall
+	shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}.
+	\item \cref{poly:fig:typical4} shows a degree-$4$ polynomial whose leading coefficient, $a_4$,
+	is positive|compare its overall shape and long-run behavior to the functions described in \cref{poly:ex:evenpow}.
+	\item \cref{poly:fig:typical5} shows a degree-$5$ polynomial whose leading coefficient, $a_5$,
+	is positive| compare its overall
+	shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}.
+\end{itemize}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{figure}[!htb]
+	\begin{widepage}
+	\setlength{\figurewidth}{\textwidth/6}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-10:8]{(x+2)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$a_1>0$}
+		\label{poly:fig:typical1}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-4:4]{(x^2-6)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$a_2>0$}
+		\label{poly:fig:typical2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-7.5:7.5]{0.05*(x+6)*x*(x-6)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$a_3>0$}
+		\label{poly:fig:typical3}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-2.35:5.35,samples=100]{0.2*(x-5)*x*(x-3)*(x+2)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$a_4>0$}
+		\label{poly:fig:typical4}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\textwidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-5.5:6.3,samples=100]{0.01*(x+2)*x*(x-3)*(x+5)*(x-6)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$a_5>0$}
+		\label{poly:fig:typical5}
+	\end{subfigure}
+	\end{widepage}
+	\caption{Graphs to illustrate typical curves of polynomial functions.}
+	\label{poly:fig:typical}
+\end{figure}
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{doyouunderstand}
+	\begin{problem}
+	Use \cref{poly:ex:typical} and \cref{poly:fig:typical} to help you sketch 
+	the graphs of polynomial functions that have negative leading coefficients| note
+	that there are many ways to do this! The intention with this problem
+	is to use your knowledge of transformations- in particular, \emph{reflections}- 
+	to guide you.
+	\begin{shortsolution}
+		$a_1<0$:
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\solutionfigurewidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-10:8]{-(x+2)};
+			\end{axis}
+		\end{tikzpicture}
+		
+		$a_2<0$
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\solutionfigurewidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-4:4]{-(x^2-6)};
+			\end{axis}
+		\end{tikzpicture}
+		
+		$a_3<0$
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\solutionfigurewidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-7.5:7.5]{-0.05*(x+6)*x*(x-6)};
+			\end{axis}
+		\end{tikzpicture}
+		
+		$a_4<0$
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\solutionfigurewidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-2.35:5.35,samples=100]{-0.2*(x-5)*x*(x-3)*(x+2)};
+			\end{axis}
+		\end{tikzpicture}
+		
+		$a_5<0$
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   width=\solutionfigurewidth,
+			   xtick={-11},
+			   ytick={-11},
+			   ]
+			   \addplot expression[domain=-5.5:6.3,samples=100]{-0.01*(x+2)*x*(x-3)*(x+5)*(x-6)};
+			\end{axis}
+		\end{tikzpicture}
+	\end{shortsolution}
+	\end{problem}
+\end{doyouunderstand}
+
+\fixthis{poly: Need a more basic example here- it can have a similar
+format to the multiple zeros example, but just keep it simple; it should
+be halfway between the 2 examples surrounding it}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Multiple zeros]
+Consider the polynomial functions $p$, $q$, and $r$ which are 
+graphed in \cref{poly:fig:moremultiple}. 
+The formulas for $p$, $q$, and $r$ are as follows
+\begin{align*}
+	p(x) & =(x-3)^2(x+4)^2       \\
+	q(x) & =x(x+2)^2(x-1)^2(x-3) \\
+	r(x) & =x(x-3)^3(x+1)^2      
+\end{align*}
+Find the degree of $p$, $q$, and $r$, and decide if the functions bounce off or cut 
+through the horizontal axis at each of their zeros.
+\begin{pccsolution}
+The degree of $p$ is 4. Referring to \cref{poly:fig:bouncep}, 
+the curve bounces off the horizontal axis at both zeros, $3$ and $4$.
+
+The degree of $q$ is 6. Referring to \cref{poly:fig:bounceq},
+the curve bounces off the horizontal axis at $-2$ and $1$, and cuts 
+through the horizontal axis at $0$ and $3$.
+
+The degree of $r$ is 6. Referring to \cref{poly:fig:bouncer},
+the curve bounces off the horizontal axis at $-1$, and cuts through 
+the horizontal axis at $0$ and at $3$, although is flattened immediately to the left and right of $3$.
+\end{pccsolution}
+\end{pccexample}
+
+\setlength{\figurewidth}{0.25\textwidth}
+\begin{figure}[!htb]
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-6,xmax=5,
+			   ymin=-30,ymax=200,
+			   xtick={-4,-2,...,4},
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-5.63733:4.63733,samples=50]{(x-3)^2*(x+4)^2};
+			   \addplot[soldot]coordinates{(3,0)(-4,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=p(x)$}
+		\label{poly:fig:bouncep}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-3,xmax=4,
+			   xtick={-2,...,3},
+			   ymin=-60,ymax=40,
+			   width=\textwidth,
+			   ]
+			   \addplot+[samples=50] expression[domain=-2.49011:3.11054]{x*(x+2)^2*(x-1)^2*(x-3)};
+			   \addplot[soldot]coordinates{(-2,0)(0,0)(1,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=q(x)$}
+		\label{poly:fig:bounceq}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-2,xmax=4,
+			   xtick={-1,...,3},
+			   ymin=-40,ymax=40,
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-1.53024:3.77464,samples=50]{x*(x-3)^3*(x+1)^2};
+			   \addplot[soldot]coordinates{(-1,0)(0,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=r(x)$}
+		\label{poly:fig:bouncer}
+	\end{subfigure}
+	\caption{}
+	\label{poly:fig:moremultiple}
+\end{figure}
+
+\begin{pccdefinition}[Multiple zeros]\label{poly:def:multzero}
+Let $p$ be a polynomial that has a repeated linear factor $(x-a)^n$. Then we say
+that $p$ has a multiple zero at $a$ of multiplicity $n$ and 
+\begin{itemize}
+	\item if the factor $(x-a)$ is repeated an even number of times, the graph of $y=p(x)$ does not
+	cross the $x$ axis at $a$, but `bounces' off the horizontal axis at $a$.
+	\item if the factor $(x-a)$ is repeated an odd number of times, the graph of $y=p(x)$ crosses the
+	horizontal axis at $a$, but it looks `flattened' there
+\end{itemize}
+If $n=1$, then we say that $p$ has a \emph{simple} zero at $a$.
+\end{pccdefinition}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Find a formula]
+Find formulas for the polynomial functions, $p$ and $q$, graphed in \cref{poly:fig:findformulademoboth}.
+\begin{figure}[!htb]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[framed,
+			   xmin=-5,xmax=5,
+			   ymin=-10,ymax=10,
+			   xtick={-4,-2,...,4},
+			   minor xtick={-3,-1,...,3},
+			   ytick={-8,-6,...,8},
+			   width=\textwidth,
+			   grid=both]
+			   \addplot expression[domain=-3.25842:2.25842,samples=50]{-x*(x-2)*(x+3)*(x+1)};
+			   \addplot[soldot]coordinates{(1,8)}node[axisnode,inner sep=.35cm,anchor=west]{$(1,8)$};
+			   \addplot[soldot]coordinates{(-3,0)(-1,0)(0,0)(2,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$p$}
+		\label{poly:fig:findformulademo}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			            \begin{axis}[framed,
+			               	xmin=-5,xmax=5,
+			               	ymin=-10,ymax=10,
+			               	xtick={-4,-2,...,4},
+			   minor xtick={-3,-1,...,3},
+			               	ytick={-8,-6,...,8},
+			   width=\textwidth,
+			   grid=both]
+			   \addplot expression[domain=-4.33:4.08152]{-.25*(x+2)^2*(x-3)};
+			   \addplot[soldot]coordinates{(2,4)}node[axisnode,anchor=south west]{$(2,4)$};
+			   \addplot[soldot]coordinates{(-2,0)(3,0)};
+			            \end{axis}
+		\end{tikzpicture}
+		\caption{$q$}
+		\label{poly:fig:findformulademo1}
+	\end{subfigure}
+	\caption{}
+	\label{poly:fig:findformulademoboth}
+\end{figure}
+\begin{pccsolution}
+\begin{enumerate}
+	\item We begin by noting that the horizontal intercepts of $p$ are $(-3,0)$, $(-1,0)$, $(0,0)$ and $(2,0)$. 
+	We also note that each zero is simple (multiplicity $1$).
+	If we assume that $p$ has no other zeros, then we can start by writing
+	\begin{align*}
+		p(x) & =(x+3)(x+1)(x-0)(x-2) \\
+		     & =x(x+3)(x+1)(x-2)     \\
+	\end{align*}
+	According to \cref{poly:fig:findformulademo}, the point $(1,8)$ lies 
+	on the curve $y=p(x)$.
+	Let's check if the formula we have written satisfies this requirement
+	\begin{align*}
+		p(1) & = (1)(4)(2)(-1) \\
+		     & = -8            
+	\end{align*}
+	which is clearly not correct| it is close though. We can correct this by 
+	multiplying $p$ by a constant $k$; so let's assume that
+	\[
+		p(x)=kx(x+3)(x+1)(x-2)
+	\]
+	Then $p(1)=-8k$, and if this is to equal $8$, then $k=-1$. Therefore
+	the formula for $p(x)$ is
+	\[
+		p(x)=-x(x+3)(x+1)(x-2)
+	\]
+	\item The function $q$ has a zero at $-2$ of multiplicity $2$, and zero of 
+	multiplicity $1$ at $3$ (so $3$ is a simple zero of $q$); we can therefore assume that $q$ has the form
+	\[
+		q(x)=k(x+2)^2(x-3)
+	\]
+	where $k$ is some real number. In order to find $k$, we use the given ordered pair, $(2,4)$, and
+	evaluate $p(2)$
+	\begin{align*}
+		p(2) & =k(4)^2(-1) \\
+		     & =-16k       
+	\end{align*}
+	We solve the equation $4=-8k$ and obtain $k=-\frac{1}{4}$ and conclude that the 
+	formula for $q(x)$ is
+	\[
+		q(x)=-\frac{1}{4}(x+2)^2(x-3)
+	\]
+\end{enumerate}
+\end{pccsolution}
+\end{pccexample}
+
+
+\fixthis{Chris: need sketching polynomial problems}
+\begin{pccspecialcomment}[Steps to follow when sketching polynomial functions]
+	\begin{steps}
+		\item \label{poly:step:first} Determine the degree of the polynomial, 
+		its leading term and leading coefficient, and hence determine
+		the long-run behavior of the polynomial| does it behave like $\pm x^2$ or $\pm x^3$ 
+		as $x\rightarrow\pm\infty$?
+		\item Determine the zeros and their multiplicity. Mark all zeros 
+		and the vertical intercept on the graph using solid circles $\bullet$.
+		\item \label{poly:step:last}  Deduce the overall shape of the curve, and sketch it. If there isn't
+		enough information from the previous steps, then construct a table of values.
+	\end{steps}
+	Remember that until we have the tools of calculus, we won't be able to 
+	find the exact coordinates of local minimums, local maximums, and points
+	of inflection.
+\end{pccspecialcomment}
+Before we demonstrate some examples, it is important to remember the following:
+\begin{itemize}
+	\item our sketches will give a good representation of the overall 
+	shape of the graph, but until we have the tools of calculus (from MTH 251)
+	we can not find local minimums, local maximums, and inflection points algebraically. This
+	means that we will make our best guess as to where these points are.
+	\item we will not concern ourselves too much with the vertical scale (because of 
+	our previous point)| we will, however, mark the vertical intercept (assuming there is one), 
+	and any horizontal asymptotes.
+\end{itemize}
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{poly:ex:simplecubic}
+Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $p$ 
+that has formula
+\[
+	p(x)=\frac{1}{2}(x-4)(x-1)(x+3)
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $p$ has degree $3$. The leading term of $p$ is $\frac{1}{2}x^3$, so the leading coefficient of $p$ 
+	is $\frac{1}{2}$. The long-run behavior of $p$ is therefore similar to that of $x^3$.
+	\item The zeros of $p$ are $-3$, $1$, and $4$; each zero is simple (i.e, it has multiplicity $1$).
+	This means that the curve of $p$ cuts the horizontal axis at each zero. The vertical 
+	intercept of $p$ is $(0,6)$.
+	\item We draw the details we have obtained so far on \cref{poly:fig:simplecubicp1}. Given 
+	that the curve of $p$ looks like the curve of $x^3$ in the long-run, we are able to complete a sketch of the 
+	graph of $p$ in \cref{poly:fig:simplecubicp2}.
+	
+	Note that we can not find the coordinates of the local minimums, local maximums, and inflection
+	points| for the moment we make reasonable guesses as to where these points are (you'll find how
+	to do this in calculus).
+\end{steps}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=15,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:simplecubicp1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=15,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$};
+			   \addplot[pccplot] expression[domain=-3.57675:4.95392,samples=100]{.5*(x-4)*(x-1)*(x+3)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:simplecubicp2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{1}{2}(x-4)(x-1)(x+3)$}
+	\label{poly:fig:simplecubic}
+\end{figure}
+\end{pccsolution}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{poly:ex:degree5}
+Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $q$ 
+that has formula
+\[
+	q(x)=\frac{1}{200}(x+7)^2(2-x)(x-6)^2
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $q$ has degree $4$. The leading term of $q$ is 
+	\[
+		-\frac{1}{200}x^5
+	\]
+	so the leading coefficient of $q$ is $-\frac{1}{200}$. The long-run behavior of $q$
+	is therefore similar to that of $-x^5$.
+	\item The zeros of $q$ are $-7$ (multiplicity 2), $2$ (simple), and $6$ (multiplicity $2$).
+	The curve of $q$ bounces off the horizontal axis at the zeros with multiplicity $2$ and
+	cuts the horizontal axis at the simple zeros. The vertical intercept of $q$ is $\left( 0,\frac{441}{25} \right)$.
+	\item We mark the details we have found so far on \cref{poly:fig:degree5p1}. Given that 
+	the curve of $q$ looks like the curve of $-x^5$ in the long-run, we can complete \cref{poly:fig:degree5p2}.
+\end{steps}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=40,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,0,...,35},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\left( 0, \frac{441}{25} \right)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:degree5p1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=40,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,0,...,35},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\left( 0, \frac{441}{25} \right)$};
+			   \addplot[pccplot] expression[domain=-8.83223:7.34784,samples=50]{1/200*(x+7)^2*(2-x)*(x-6)^2};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:degree5p2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{1}{200}(x+7)^2(2-x)(x-6)^2$}
+	\label{poly:fig:degree5}
+\end{figure}
+\end{pccsolution}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}
+Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $r$ 
+that has formula
+\[
+	r(x)=\frac{1}{100}x^3(x+4)(x-4)(x-6)
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $r$ has degree $6$. The leading term of $r$ is 
+	\[
+		\frac{1}{100}x^6
+	\]
+	so the leading coefficient of $r$ is $\frac{1}{100}$. The long-run behavior of $r$
+	is therefore similar to that of $x^6$.
+	\item The zeros of $r$ are $-4$ (simple), $0$ (multiplicity $3$), $4$ (simple), 
+	and $6$ (simple). The vertical intercept of $r$ is $(0,0)$. The curve of $r$
+	cuts the horizontal axis at the simple zeros, and goes through the axis
+	at $(0,0)$, but does so in a flattened way.
+	\item We mark the zeros and vertical intercept on \cref{poly:fig:degree6p1}. Given that
+	the curve of $r$ looks like the curve of $x^6$ in the long-run, we complete the graph
+	of $r$ in \cref{poly:fig:degree6p2}.
+\end{steps}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=10,
+			   ymin=-20,ymax=10,
+			   xtick={-4,-2,...,8},
+			   ytick={-15,-10,...,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:degree6p1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=10,
+			   ymin=-20,ymax=10,
+			   xtick={-4,-2,...,8},
+			   ytick={-15,-10,...,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)};
+			   \addplot[pccplot] expression[domain=-4.16652:6.18911,samples=100]{1/100*(x+4)*x^3*(x-4)*(x-6)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:degree6p2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{1}{100}(x+4)x^3(x-4)(x-6)$}
+\end{figure}
+\end{pccsolution}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{pccexample}[An open-topped box] 
+A cardboard company makes open-topped boxes for their clients. The specifications
+dictate that the box must have a square base, and that it must be open-topped. 
+The company uses sheets of cardboard that are $\unit[1200]{cm^2}$. Assuming that 
+the base of each box has side $x$ (measured in cm), it can be shown that the volume of each box, $V(x)$,
+has formula
+\[
+	V(x)=\frac{x}{4}(1200-x^2)
+\]
+Find the dimensions of the box that maximize the volume.
+\begin{pccsolution}
+We graph $y=V(x)$ in \cref{poly:fig:opentoppedbox}. Note that because 
+$x$ represents the length of a side, and $V(x)$ represents the volume
+of the box, we necessarily require both values to be positive; we illustrate
+the part of the curve that applies to this problem using a solid line. 
+
+\begin{figure}[!htb]
+	\centering
+	\begin{tikzpicture}
+		\begin{axis}[framed,
+		   xmin=-50,xmax=50,
+		   ymin=-5000,ymax=5000,
+		   xtick={-40,-30,...,40},
+		   minor xtick={-45,-35,...,45},
+		   minor ytick={-3000,-1000,1000,3000},
+		   width=.75\textwidth,
+		   height=.5\textwidth,
+		   grid=both]
+		   \addplot[pccplot,dashed,<-] expression[domain=-40:0,samples=50]{x/4*(1200-x^2)};
+		   \addplot[pccplot,-] expression[domain=0:34.64,samples=50]{x/4*(1200-x^2)};
+		   \addplot[pccplot,dashed,->] expression[domain=34.64:40,samples=50]{x/4*(1200-x^2)};
+		   \addplot[soldot] coordinates{(20,4000)};
+		\end{axis}
+	\end{tikzpicture}
+	\caption{$y=V(x)$}
+	\label{poly:fig:opentoppedbox}
+\end{figure}
+
+According to \cref{poly:fig:opentoppedbox}, the maximum volume of such a box is 
+approximately $\unit[4000]{cm^2}$, and we achieve it using a base of length 
+approximately $\unit[20]{cm}$. Since the base is square and each sheet of cardboard
+is $\unit[1200]{cm^2}$, we conclude that the dimensions of each box are $\unit[20]{cm}\times\unit[20]{cm}\times\unit[30]{cm}$.
+\end{pccsolution}
+\end{pccexample}
+
+\subsection*{Complex zeros}
+There has been a pattern to all of the examples that we have seen so far|
+the degree of the polynomial has dictated the number of \emph{real} zeros that the 
+polynomial has. For example, the function $p$ in \cref{poly:ex:simplecubic}
+has degree $3$, and $p$ has $3$ real zeros; the function $q$ in \cref{poly:ex:degree5}
+has degree $5$ and $q$ has $5$ real zeros.
+
+You may wonder if this result can be generalized| does every polynomial that
+has degree $n$ have $n$ real zeros? Before we tackle the general result, 
+let's consider an example that may help motivate it.
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{pccexample}\label{poly:ex:complx}
+Consider the polynomial function $c$ that has formula
+\[
+	c(x)=x(x^2+1)
+\]
+It is clear that $c$ has degree $3$, and that $c$ has a (simple) zero at $0$. Does
+$c$ have any other zeros, i.e, can we find any values of $x$ that satisfy the equation
+\begin{equation}\label{poly:eq:complx}
+	x^2+1=0
+\end{equation}
+The solutions to \cref{poly:eq:complx} are $\pm i$.
+
+We conclude that $c$ has $3$ zeros: $0$ and $\pm i$; we note that \emph{not 
+all of them are real}.
+\end{pccexample}
+\Cref{poly:ex:complx} shows that not every degree-$3$ polynomial has $3$
+\emph{real} zeros; however, if we are prepared to venture into the complex numbers, 
+then we can state the following theorem.
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{pccspecialcomment}[The fundamental theorem of algebra]
+	Every polynomial function of degree $n$ has $n$ roots, some of which may 
+	be complex, and some may be repeated.
+\end{pccspecialcomment}
+\fixthis{Fundamental theorem of algebra: is this wording ok? do we want
+it as a theorem?}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{pccexample}
+Find all the zeros of the polynomial function $p$ that has formula
+\[
+	p(x)=x^4-2x^3+5x^2
+\]
+\begin{pccsolution}
+We begin by factoring $p$
+\begin{align*}
+	p(x) & =x^4-2x^3+5x^2 \\
+	     & =x^2(x^2-2x+5) 
+\end{align*}
+We note that $0$ is a zero of $p$ with multiplicity $2$. The other zeros of $p$ 
+can be found by solving the equation
+\[
+	x^2-2x+5=0
+\]
+This equation can not be factored, so we use the quadratic formula
+\begin{align*}
+	x & =\frac{2\pm\sqrt{(-2)^2}-20}{2(1)} \\
+	  & =\frac{2\pm\sqrt{-16}}{2}          \\
+	  & =1\pm 2i                           
+\end{align*}
+We conclude that $p$ has $4$ zeros: $0$ (multiplicity $2$), and $1\pm 2i$ (simple).
+\end{pccsolution}
+\end{pccexample}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{pccexample}
+Find a polynomial that has zeros at $2\pm i\sqrt{2}$. 
+\begin{pccsolution}
+We know that the zeros of a polynomial can be found by analyzing the linear
+factors. We are given the zeros, and have to work backwards to find the 
+linear factors. 
+
+We begin by assuming that $p$ has the form
+\begin{align*}
+	p(x) & =(x-(2-i\sqrt{2}))(x-(2+i\sqrt{2}))                           \\
+	     & =x^2-x(2+i\sqrt{2})-x(2-i\sqrt{2})+(2-i\sqrt{2})(2+i\sqrt{2}) \\
+	     & =x^2-4x+(4-2i^2)                                              \\
+	     & =x^2-4x+6                                                     
+\end{align*}
+We conclude that a possible formula for a polynomial function, $p$, 
+that has zeros at $2\pm i\sqrt{2}$ is
+\[
+	p(x)=x^2-4x+6
+\]
+Note that we could multiply $p$ by any real number and still ensure
+that $p$ has the same zeros.
+\end{pccsolution}
+\end{pccexample}
+\investigation*{}
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a graph]
+For each of the polynomials in \cref{poly:fig:findformula}
+\begin{enumerate}
+	\item count the number of times the curve turns round, and cuts/bounces off the $x$ axis;
+	\item approximate the degree of the polynomial;
+	\item use your information to find the linear factors of each polynomial, and therefore write a possible formula for each;
+	\item make sure your polynomial goes through the given ordered pair.
+\end{enumerate}
+\begin{shortsolution}
+	\Vref{poly:fig:findformdeg2}: 
+	\begin{enumerate}
+		\item the curve turns round once;
+		\item the degree could be 2;
+		\item based on the zeros, the linear factors are $(x+5)$ and $(x-3)$; since the 
+		graph opens downwards, we will assume the leading coefficient is negative: $p(x)=-k(x+5)(x-3)$;
+		\item $p$ goes through $(2,2)$, so we need to solve $2=-k(7)(-1)$ and therefore $k=\nicefrac{2}{7}$, so
+		\[
+			p(x)=-\frac{2}{7}(x+5)(x-3)
+		\]
+	\end{enumerate}
+	\Vref{poly:fig:findformdeg3}:
+	\begin{enumerate}
+		\item the curve turns around twice;
+		\item the degree could be 3;
+		\item based on the zeros, the linear factors are $(x+2)^2$, and $(x-1)$;
+		based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+2)^2(x-1)$;
+		\item $p$ goes through $(0,-2)$, so we need to solve $-2=k(4)(-1)$ and therefore $k=\nicefrac{1}{2}$, so
+		\[
+			p(x)=\frac{1}{2}(x+2)^2(x-1)
+		\]
+	\end{enumerate}
+	\Vref{poly:fig:findformdeg5}:
+	\begin{enumerate}
+		\item the curve turns around 4 times;
+		\item the degree could be 5;
+		\item based on the zeros, the linear factors are $(x+5)^2$, $(x+1)$, $(x-2)$, $(x-3)$;
+		based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+5)^2(x+1)(x-2)(x-3)$;
+		\item $p$ goes through $(-3,-50)$, so we need to solve $-50=k(64)(-2)(-5)(-6)$ and therefore $k=\nicefrac{5}{384}$, so
+		\[
+			p(x)=\frac{5}{384}(x+5)^2(x+1)(x-2)(x-3)
+		\]
+	\end{enumerate}
+\end{shortsolution}
+\end{problem}
+
+
+\begin{figure}[!htb]
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-2,ymax=5,
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-4.5:3.75]{-1/3*(x+4)*(x-3)};
+			   \addplot[soldot] coordinates{(-4,0)(3,0)(2,2)} node[axisnode,above right]{$(2,2)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:findformdeg2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-3,xmax=2,
+			   ymin=-2,ymax=4,
+			   xtick={-2,...,1},
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-2.95:1.75]{1/3*(x+2)^2*(x-1)};
+			   \addplot[soldot]coordinates{(-2,0)(1,0)(0,-1.33)}node[axisnode,anchor=north west]{$(0,-2)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:findformdeg3}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-100,ymax=150,
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-4.5:3.4,samples=50]{(x+4)^2*(x+1)*(x-2)*(x-3)};
+			   \addplot[soldot]coordinates{(-4,0)(-1,0)(2,0)(3,0)(-3,-60)}node[axisnode,anchor=north]{$(-3,-50)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:findformdeg5}
+	\end{subfigure}
+	\caption{}
+	\label{poly:fig:findformula}
+\end{figure}
+
+
+
+
+\begin{exercises}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Prerequisite classifacation skills]
+Decide if each of the following functions are linear or quadratic.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$f(x)=2x+3$ 
+		\begin{shortsolution}
+			$f$ is linear. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$g(x)=10-7x$ 
+		\begin{shortsolution}
+			$g$ is linear 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$h(x)=-x^2+3x-9$ 
+		\begin{shortsolution}
+			$h$ is quadratic. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$k(x)=-17$ 
+		\begin{shortsolution}
+			$k$ is linear.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$l(x)=-82x^2-4$
+		\begin{shortsolution}
+			$l$ is quadratic 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$m(x)=6^2x-8$ 
+		\begin{shortsolution}
+			$m$ is linear. 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Prerequisite slope identification]
+State the slope of each of the following linear functions, and 
+hence decide if each function is increasing or decreasing.
+\begin{multicols}{4}
+	\begin{subproblem}
+		$\alpha(x)=4x+1$ 
+		\begin{shortsolution}
+			$m=4$; $\alpha$ is increasing. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$\beta(x)=-9x$ 
+		\begin{shortsolution}
+			$m=-9$; $\beta$ is decreasing. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$\gamma(t)=18t+100$ 
+		\begin{shortsolution}
+			$m=18$; $\gamma$ is increasing.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$\delta(y)=23-y$ 
+		\begin{shortsolution}
+			$m=-1$; $\delta$ is decreasing. 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+Now let's generalize our findings for the most general linear function $f$
+that has formula $f(x)=mx+b$. Complete the following sentences.
+\begin{subproblem}
+	When $m>0$, the function $f$ is $\ldots$  
+	\begin{shortsolution}
+		When $m>0$, the function $f$ is $\ldots$  \emph{increasing}.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	When $m<0$, the function $f$ is $\ldots$  
+	\begin{shortsolution}
+		When $m<0$, the function $f$ is $\ldots$  \emph{decreasing}.
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Polynomial or not?]
+Identify whether each of the following functions is a polynomial or not. 
+If the function is a polynomial, state its degree.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$p(x)=2x+1$
+		\begin{shortsolution}
+			$p$ is a polynomial (you might also describe $p$ as linear). The degree of $p$ is 1.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=7x^2+4x$
+		\begin{shortsolution}
+			$p$ is a polynomial (you might also describe $p$ as quadratic). The degree of $p$ is 2.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=\sqrt{x}+2x+1$
+		\begin{shortsolution}
+			$p$ is not a polynomial; we require the powers of $x$ to be integer values.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=2^x-45$
+		\begin{shortsolution}
+			$p$ is not a polynomial; the $2^x$ term is exponential.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=6x^4-5x^3+9$
+		\begin{shortsolution}
+			$p$ is a polynomial, and the degree of $p$ is $6$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=-5x^{17}+9x+2$
+		\begin{shortsolution}
+			$p$ is a polynomial, and the degree of $p$ is 17.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=4x(x+7)^2(x-3)^3$ 
+		\begin{shortsolution}
+			$p$ is a polynomial, and the degree of $p$ is $6$.     
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=4x^{-5}-x^2+x$ 
+		\begin{shortsolution}
+			$p$ is not a polynomial because $-5$ is not a positive integer. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=-x^6(x^2+1)(x^3-2)$ 
+		\begin{shortsolution}
+			$p$ is a polynomial, and the degree of $p$ is $11$.
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Polynomial graphs]
+Three polynomial functions $p$, $m$, and $n$ are shown in \crefrange{poly:fig:functionp}{poly:fig:functionn}.
+The functions have the following formulas
+\begin{align*}
+	p(x) & = (x-1)(x+2)(x-3)           \\ 
+	m(x) & = -(x-1)(x+2)(x-3)          \\
+	n(x) & = (x-1)(x+2)(x-3)(x+1)(x+4) 
+\end{align*}
+Note that for our present purposes we are not concerned with the vertical scale of the graphs.
+\begin{subproblem}
+	Identify both on the graph {\em and} algebraically, the zeros of each polynomial.
+	\begin{shortsolution}
+		$y=p(x)$ is shown below.
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-10,ymax=10,
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)};
+			   \addplot[soldot] coordinates{(-2,0)(1,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		
+		$y=m(x)$ is shown below.
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-10,ymax=10,
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)};
+			   \addplot[soldot] coordinates{(-2,0)(1,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		
+		$y=n(x)$ is shown below.
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-90,ymax=70,
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot expression[domain=-4.15:3.15,samples=50]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)};
+			   \addplot[soldot] coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		
+		The zeros of $p$ are $-2$, $1$, and $3$; the zeros of $m$ are $-2$, $1$, and $3$; the zeros of $n$ are 
+		$-4$, $-2$, $-1$, and $3$.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Write down the degree, how many times the curve of each function `turns around', 
+	and how many zeros it has
+	\begin{shortsolution}
+		\begin{itemize}
+			\item The degree of $p$ is 3, and the curve $y=p(x)$ turns around twice. 
+			\item The degree of $q$ is also 3, and the curve $y=q(x)$ turns around twice. 
+			\item The degree of $n$ is $5$, and the curve $y=n(x)$ turns around 4 times.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+\begin{figure}[!htb]
+	\begin{widepage}
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-10,ymax=10,
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)};
+			   \addplot[soldot]coordinates{(-2,0)(1,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		    \caption{$y=p(x)$}
+		\label{poly:fig:functionp}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-10,ymax=10,
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)};
+			   \addplot[soldot]coordinates{(-2,0)(1,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=m(x)$}
+		\label{poly:fig:functionm}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-90,ymax=70,
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-4.15:3.15,samples=100]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)};
+			   \addplot[soldot]coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		        \caption{$y=n(x)$}
+		\label{poly:fig:functionn}
+	\end{subfigure}
+	\caption{}
+	\end{widepage}
+\end{figure}
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Horizontal intercepts]\label{poly:prob:matchpolys}%
+State the horizontal intercepts (as ordered pairs) of the following polynomials.
+\begin{multicols}{2}
+	\begin{subproblem}\label{poly:prob:degree5}
+		$p(x)=(x-1)(x+2)(x-3)(x+1)(x+4)$
+		\begin{shortsolution}
+			$(-4,0)$, $(-2,0)$, $(-1,0)$, $(1,0)$, $(3,0)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$q(x)=-(x-1)(x+2)(x-3)$
+		\begin{shortsolution}
+			$(-2,0)$, $(1,0)$, $(3,0)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$r(x)=(x-1)(x+2)(x-3)$
+		\begin{shortsolution}
+			$(-2,0)$, $(1,0)$, $(3,0)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}\label{poly:prob:degree2}
+		$s(x)=(x-2)(x+2)$
+		\begin{shortsolution}
+			$(-2,0)$, $(2,0)$
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Minimums, maximums, and concavity]\label{poly:prob:incdec}
+Four polynomial functions are graphed in \cref{poly:fig:incdec}. The formulas
+for these functions are (not respectively)
+\begin{gather*}
+	p(x)=\frac{x^3}{6}-\frac{x^2}{4}-3x, \qquad q(x)=\frac{x^4}{20}+\frac{x^3}{15}-\frac{6}{5}x^2+1\\
+	r(x)=-\frac{x^5}{50}-\frac{x^4}{40}+\frac{2x^3}{5}+6, \qquad s(x)=-\frac{x^6}{6000}-\frac{x^5}{2500}+\frac{67x^4}{4000}+\frac{17x^3}{750}-\frac{42x^2}{125}
+\end{gather*}
+\begin{figure}[!htb]
+	\begin{widepage}
+	\setlength{\figurewidth}{.23\textwidth}
+	\centering
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   width=\textwidth,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-8,-6,...,8},
+			   grid=major,
+			   ]
+			   \addplot expression[domain=-5.28:4.68,samples=50]{-x^5/50-x^4/40+2*x^3/5+6};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:incdec3}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   width=\textwidth,
+			   xmin=-10,xmax=10,ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-8,-6,...,8},
+			   grid=major,
+			   ]
+			   \addplot expression[domain=-6.08:4.967,samples=50]{x^4/20+x^3/15-6/5*x^2+1};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:incdec2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   width=\textwidth,
+			   xmin=-6,xmax=8,ymin=-10,ymax=10,
+			   xtick={-4,-2,...,6},
+			   ytick={-8,-4,4,8},
+			   minor ytick={-6,-2,...,6},
+			   grid=both,
+			   ]
+			   \addplot expression[domain=-4.818:6.081,samples=50]{x^3/6-x^2/4-3*x};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:incdec1}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   width=\textwidth,
+			   xmin=-10,xmax=10,ymin=-10,ymax=10,
+			   xtick={-8,-4,4,8},
+			   ytick={-8,-4,4,8},
+			   minor xtick={-6,-2,...,6},
+			   minor ytick={-6,-2,...,6},
+			   grid=both,
+			   ]
+			   \addplot expression[domain=-9.77:8.866,samples=50]{-x^6/6000-x^5/2500+67*x^4/4000+17/750*x^3-42/125*x^2};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:incdec4}
+	\end{subfigure}
+	\caption{Graphs for \cref{poly:prob:incdec}.}
+	\label{poly:fig:incdec}
+	\end{widepage}
+\end{figure}
+\begin{subproblem}
+	Match each of the formulas with one of the given graphs.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item $p$ is graphed in \vref{poly:fig:incdec1};
+			\item $q$ is graphed in \vref{poly:fig:incdec2};
+			\item $r$ is graphed in \vref{poly:fig:incdec3};
+			\item $s$ is graphed in \vref{poly:fig:incdec4}.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Approximate the zeros of each function using the appropriate graph.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item $p$ has simple zeros at about $-3.8$, $0$, and $5$.
+			\item $q$ has simple zeros at about $-5.9$, $-1$, $1$, and $4$.
+			\item $r$ has simple zeros at about $-5$, $-2.9$, and $4.1$.
+			\item $s$ has simple zeros at about $-9$, $-6$, $4.2$, $8.1$, and a zero of multiplicity $2$ at $0$.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Approximate the local maximums and minimums of each of the functions.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item $p$ has a local maximum of approximately $3.9$ at $-2$, and a local minimum of approximately $-6.5$ at $3$.
+			\item $q$ has a local minimum of approximately $-10$ at $-4$, and $-4$ at $3$; $q$ has a local maximum of approximately $1$ at $0$.
+			\item $r$ has a local minimum of approximately $-5.5$ at $-4$, and a local maximum of approximately $10$ at $3$.
+			\item $s$ has a local maximum of approximately $5$ at $-8$, $0$ at $0$, and $5$ at  $7$; $s$ has local minimums 
+			of approximately $-3$ at $-4$, and $-1$ at $3$.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Approximate the global maximums and minimums of each of the functions.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item $p$ does not have a global maximum, nor a global minimum.
+			\item $q$ has a global minimum of approximately $-10$; it does not have a global maximum.
+			\item $r$ does not have a global maximum, nor a global minimum.
+			\item $s$ has a global maximum of approximately $5$; it does not have a global minimum.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Approximate the intervals on which each function is increasing and decreasing.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item $p$ is increasing on $(-\infty,-2)\cup (3,\infty)$, and decreasing on $(-2,3)$.
+			\item $q$ is increasing on $(-4,0)\cup (3,\infty)$, and decreasing on $(-\infty,-4)\cup (0,3)$.
+			\item $r$ is increasing on $(-4,3)$, and decreasing on $(-\infty,-4)\cup (3,\infty)$.
+			\item $s$ is increasing on $(-\infty,-8)\cup (-4,0)\cup (3,5)$, and decreasing on $(-8,-4)\cup (0,3)\cup (5,\infty)$.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Approximate the intervals on which each function is concave up and concave down.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item $p$ is concave up on  $(1,\infty)$, and concave down on  $(-\infty,1)$.
+			\item $q$ is concave up on $(-\infty,-1)\cup (1,\infty)$, and concave down on $(-1,1)$.
+			\item $r$ is concave up on $(-\infty,-3)\cup (0,2)$, and concave down on $(-3,0)\cup (2,\infty)$.
+			\item $s$ is concave up on $(-6,-2)\cup (2,5)$, and concave down on $(-\infty,-6)\cup (-2,2)\cup (5,\infty)$.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	The degree of $q$ is $5$. Assuming that all of the real zeros of $q$ are
+	shown in its graph, how many complex zeros does $q$ have?
+	\begin{shortsolution}
+		\Vref{poly:fig:incdec2} shows that $q$ has $3$ real zeros 
+		since the curve of $q$ cuts the horizontal axis $3$ times. 
+		Since $q$ has degree $5$, $q$ must have $2$ complex zeros.
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Long-run behaviour of polynomials]
+Describe the long-run behavior of each of polynomial functions in 
+\crefrange{poly:prob:degree5}{poly:prob:degree2}.
+\begin{shortsolution}
+	$\dd\lim_{x\rightarrow-\infty}p(x)=-\infty$,
+	$\dd\lim_{x\rightarrow\infty}p(x)=\infty$,
+	$\dd\lim_{x\rightarrow-\infty}q(x)=\infty$,
+	$\dd\lim_{x\rightarrow\infty}q(x)=-\infty$,
+	$\dd\lim_{x\rightarrow-\infty}r(x)=-\infty$,
+	$\dd\lim_{x\rightarrow\infty}r(x)=\infty$,
+	$\dd\lim_{x\rightarrow-\infty}s(x)=\infty$,
+	$\dd\lim_{x\rightarrow\infty}s(x)=\infty$,
+\end{shortsolution}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[True of false?]
+Let $p$ be a polynomial function. 
+Label each of the following statements as true (T) or false (F); if they are false, 
+provide an example that supports your answer.
+\begin{subproblem}
+	If $p$ has degree $3$, then $p$ has $3$ distinct zeros.
+	\begin{shortsolution}
+		False. Consider $p(x)=x^2(x+1)$ which has only 2 distinct zeros.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	If $p$ has degree $4$, then $\dd\lim_{x\rightarrow-\infty}p(x)=\infty$ and $\dd\lim_{x\rightarrow\infty}p(x)=\infty$.
+	\begin{shortsolution}
+		False. Consider $p(x)=-x^4$.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	If $p$ has even degree, then it is possible that $p$ can have no real zeros.
+	\begin{shortsolution}
+		True.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	If $p$ has odd degree, then it is possible that $p$ can have no real zeros.
+	\begin{shortsolution}
+		False. All odd degree polynomials will cut the horizontal axis at least once.
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a description]
+In each of the following problems, give a possible formula for a polynomial 
+function that has the specified properties.
+\begin{subproblem}
+	 Degree 2 and has zeros at $4$ and $5$.
+	\begin{shortsolution}
+		Possible option: $p(x)=(x-4)(x-5)$. Note we could multiply $p$ by any real number, and still meet the requirements.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	 Degree 3 and has zeros at $4$,$5$ and $-3$.
+	\begin{shortsolution}
+		Possible option: $p(x)=(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	 Degree 4 and has zeros at $0$, $4$, $5$, $-3$.
+	\begin{shortsolution}
+		Possible option: $p(x)=x(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Degree 4, with zeros that make the graph cut at $2$, $-5$, and a zero that makes the graph touch at $-2$;
+	\begin{shortsolution}
+		Possible option: $p(x)=(x-2)(x+5)(x+2)^2$. Note we could multiply $p$ by any real number, and still meet the requirements.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Degree 3, with only one zero at $-1$.
+	\begin{shortsolution}
+		Possible option: $p(x)=(x+1)^3$. Note we could multiply $p$ by any real number, and still meet the requirements.
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{problem}[\Cref{poly:step:last}]
+\pccname{Saheed} is graphing a polynomial function, $p$. 
+He is following \crefrange{poly:step:first}{poly:step:last} and has so far
+marked the zeros of $p$ on \cref{poly:fig:optionsp1}. Saheed tells you that 
+$p$ has degree $3$, but does \emph{not} say if the leading coefficient 
+of $p$ is positive or negative.
+\begin{figure}[!htbp]
+	\begin{widepage}
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-15},
+			   width=\textwidth,
+			   height=.5\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:optionsp1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-15},
+			   width=\textwidth,
+			   height=.5\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-5,0)(6,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{poly:fig:optionsp2}
+	\end{subfigure}%
+	\caption{}
+	\end{widepage}
+\end{figure}
+\begin{subproblem}
+	Use the information in \cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient
+	is positive.
+	\begin{shortsolution}
+		Assuming that $a_3>0$:
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-15},
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot expression[domain=-6.78179:8.35598,samples=50]{1/20*(x+5)*(x-2)*(x-6)};
+			   \addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};
+			\end{axis}
+		\end{tikzpicture}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Use the information in \cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient
+	is negative.
+	\begin{shortsolution}
+		Assuming that $a_3<0$:
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-15},
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot expression[domain=-6.78179:8.35598,samples=50]{-1/20*(x+5)*(x-2)*(x-6)};
+			   \addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};
+			\end{axis}
+		\end{tikzpicture}
+	\end{shortsolution}
+\end{subproblem}
+Saheed now turns his attention to another polynomial function, $q$. He finds
+the zeros of $q$ (there are only $2$) and marks them on \cref{poly:fig:optionsp2}. 
+Saheed knows that $q$ has degree $3$, but doesn't know if the leading 
+coefficient is positive or negative.
+\begin{subproblem}
+	Use the information in \cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading 
+	coefficient of $q$ is positive. Hint: only one of the zeros is simple.
+	\begin{shortsolution}
+		Assuming that $a_4>0$ there are $2$ different options:
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-15},
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot expression[domain=-8.68983:7.31809,samples=50]{1/20*(x+5)^2*(x-6)};
+			   \addplot expression[domain=-6.31809:9.68893,samples=50]{1/20*(x+5)*(x-6)^2};
+			   \addplot[soldot] coordinates{(-5,0)(6,0)};
+			\end{axis}
+		\end{tikzpicture}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Use the information in \cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading 
+	coefficient of $q$ is negative.
+	\begin{shortsolution}
+		Assuming that $a_4<0$ there are $2$ different options:
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-15},
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot expression[domain=-8.68983:7.31809,samples=50]{-1/20*(x+5)^2*(x-6)};
+			   \addplot expression[domain=-6.31809:9.68893,samples=50]{-1/20*(x+5)*(x-6)^2};
+			   \addplot[soldot] coordinates{(-5,0)(6,0)};
+			\end{axis}
+		\end{tikzpicture}
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{problem}[Zeros]
+Find all zeros of each of the following polynomial functions, making 
+sure to detail their multiplicity. Note that
+you may need to use factoring, or the quadratic formula, or both! Also note 
+that some zeros may be repeated, and some may be complex.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$p(x)=x^2+1$ 
+		\begin{shortsolution}
+			$\pm i$ (simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$q(y)=(y^2-9)(y^2-7)$ 
+		\begin{shortsolution}
+			$\pm 3$, $\pm \sqrt{7}$ (all are simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$r(z)=-4z^3(z^2+3)(z^2+64)$ 
+		\begin{shortsolution}
+			$0$ (multiplicity $3$), $\pm\sqrt{3}$ (simple), $\pm\sqrt{8}$ (simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$a(x)=x^4-81$ 
+		\begin{shortsolution}
+			$\pm 3$, $\pm 3i$ (all are simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$b(y)=y^3-8$ 
+		\begin{shortsolution}
+			$2$, $-1\pm i\sqrt{3}$ (all are simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$c(m)=m^3-m^2$
+		\begin{shortsolution}
+			$0$ (multiplicity $2$), $1$ (simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$h(n)=(n+1)(n^2+4)$ 
+		\begin{shortsolution}
+			$-1$, $\pm 2i$ (all are simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$f(\alpha)=(\alpha^2-16)(\alpha^2-5\alpha+4)$ 
+		\begin{shortsolution}
+			$-4$ (simple), $4$ (multiplicity $2$), $1$ (simple).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$g(\beta)=(\beta^2-25)(\beta^2-5\beta-4)$ 
+		\begin{shortsolution}
+			$\pm 5$, $\dfrac{5\pm\sqrt{41}}{2}$ (all are simple).
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{problem}[Given zeros, find a formula]
+In each of the following problems you are given the zeros of a polynomial. 
+Write a possible formula for each polynomial| you may leave your 
+answer in factored form, but it may not contain complex numbers. Unless
+otherwise stated, assume that the zeros are simple.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$1$, $2$ 
+		\begin{shortsolution}
+			$p(x)=(x-1)(x-2)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$0$, $5$, $13$ 
+		\begin{shortsolution}
+			$p(x)=x(x-5)(x-13)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$-7$, $2$ (multiplicity $3$), $5$ 
+		\begin{shortsolution}
+			$p(x)=(x+7)(x-2)^3(x-5)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$0$, $\pm i$ 
+		\begin{shortsolution}
+			$p(x)=x(x^2+1)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$\pm 2i$, $\pm 7$ 
+		\begin{shortsolution}
+			$p(x)=(x^2+4)(x^2-49)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$-2\pm i\sqrt{6}$ 
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{problem}[Composition of polynomials]
+Let $p$ and $q$ be polynomial functions that have formulas
+\[
+	p(x)=(x+1)(x+2)(x+5), \qquad q(x)=3-x^4
+\]
+Evaluate each of the following.
+\begin{multicols}{4}
+	\begin{subproblem}
+		$(p\circ q)(0)$ 
+		\begin{shortsolution}
+			$160$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(q\circ p)(0)$ 
+		\begin{shortsolution}
+			$-9997$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(p\circ q)(1)$ 
+		\begin{shortsolution}
+			$84$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(p\circ p)(0)$ 
+		\begin{shortsolution}
+			$1980$
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{problem}[Piecewise polynomial functions]
+Let $P$ be the piecewise-defined function with formula
+\[
+	P(x)=\begin{cases}
+	(1-x)(2x+5)(x^2+1), &   x\leq -3\\
+	4-x^2,              &   -3<x < 4\\
+	x^3                 &   x\geq 4
+\end{cases}
+\]
+Evaluate each of the following 
+\begin{multicols}{5}
+	\begin{subproblem}
+		$P(-4)$ 
+		\begin{shortsolution}
+			$-255$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$P(0)$ 
+		\begin{shortsolution}
+			$4$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$P(4)$ 
+		\begin{shortsolution}
+			$64$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$P(-3)$ 
+		\begin{shortsolution}
+			$-40$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(P\circ P)(0)$ 
+		\begin{shortsolution}
+			$64$
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   July 2012
+%===================================
+\begin{problem}[Function algebra]
+Let $p$ and $q$ be the polynomial functions that have formulas
+\[
+	p(x)=x(x+1)(x-3)^2, \qquad q(x)=7-x^2
+\]
+Evaluate each of the following (if possible).
+\begin{multicols}{4}
+	\begin{subproblem}
+		$(p+q)(1)$ 
+		\begin{shortsolution}
+			$14$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(p-q)(0)$ 
+		\begin{shortsolution}
+			$7$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(p\cdot q)(\sqrt{7})$ 
+		\begin{shortsolution}
+			$0$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$\left( \frac{q}{p} \right)(1)$ 
+		\begin{shortsolution}
+			$\frac{3}{4}$ 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\begin{subproblem}
+	What is the domain of the function $\frac{q}{p}$?
+	\begin{shortsolution}
+		$(-\infty,-1)\cup (-1,0)\cup (0,3)\cup (3,\infty)$ 
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   July 2012
+%===================================
+\begin{problem}[Transformations: given the transformation, find the formula]
+Let $p$ be the polynomial function that has formula.
+\[
+	p(x)=4x(x^2-1)(x+3)
+\]
+In each of the following 
+problems apply the given transformation to the function $p$ and 
+write a formula for the transformed version of $p$.
+\begin{multicols}{2}
+	\begin{subproblem}
+		Shift $p$ to the right by $5$ units. 
+		\begin{shortsolution}
+			$p(x-5)=4(x-5)(x-2)(x^2-10x+24)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $p$ to the left by $6$ units. 
+		\begin{shortsolution}
+			$p(x+6)=4(x+6)(x+9)(x^2+12x+35)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $p$ up by $12$ units. 
+		\begin{shortsolution}
+			$p(x)+12=4x(x^2-1)(x+3)+12$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $p$ down by $2$ units. 
+		\begin{shortsolution}
+			$p(x)-2=4x(x^2-1)(x+3)-2$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Reflect $p$ over the horizontal axis.
+		\begin{shortsolution}
+			$-p(x)=-4x(x^2-1)(x+3)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Reflect $p$ over the vertical axis.
+		\begin{shortsolution}
+			$p(-x)=-4x(x^2-1)(3-x)$ 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a table]\label{poly:prob:findformula}
+\Crefrange{poly:tab:findformulap}{poly:tab:findformulas} show values of polynomial functions, $p$, $q$, 
+$r$, and $s$.
+
+\begin{table}[!htb]
+	\centering
+	\begin{widepage}
+	\caption{Tables for \cref{poly:prob:findformula}}
+	\label{poly:tab:findformula}
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=p(x)$}
+		\label{poly:tab:findformulap}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ 
+			\afterheading
+			$-4$          & $-56$         \\\normalline
+			$-3$          & $-18$         \\\normalline
+			$-2$          & $0$           \\\normalline
+			$-1$          & $4$           \\\normalline
+			$0$           & $0$           \\\normalline
+			$1$           & $-6$          \\\normalline
+			$2$           & $-8$          \\\normalline
+			$3$           & $0$           \\\normalline
+			$4$           & $24$          \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=q(x)$}
+		\label{poly:tab:findformulaq}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ \afterheading
+			$-4$          & $-16$         \\\normalline
+			$-3$          & $-3$          \\\normalline
+			$-2$          & $0$           \\\normalline
+			$-1$          & $-1$          \\\normalline
+			$0$           & $0$           \\\normalline
+			$1$           & $9$           \\\normalline
+			$2$           & $32$          \\\normalline
+			$3$           & $75$          \\\normalline
+			$4$           & $144$         \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=r(x)$}
+		\label{poly:tab:findformular}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ \afterheading
+			$-4$          & $105$         \\\normalline
+			$-3$          & $0$           \\\normalline
+			$-2$          & $-15$         \\\normalline
+			$-1$          & $0$           \\\normalline
+			$0$           & $9$           \\\normalline
+			$1$           & $0$           \\\normalline
+			$2$           & $-15$         \\\normalline
+			$3$           & $0$           \\\normalline
+			$4$           & $105$         \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=s(x)$}
+		\label{poly:tab:findformulas}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ \afterheading
+			$-4$          & $75$          \\\normalline
+			$-3$          & $0$           \\\normalline
+			$-2$          & $-9$          \\\normalline
+			$-1$          & $0$           \\\normalline
+			$0$           & $3$           \\\normalline
+			$1$           & $0$           \\\normalline
+			$2$           & $15$          \\\normalline
+			$3$           & $96$          \\\normalline
+			$4$           & $760$         \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\end{widepage}
+\end{table}
+
+\begin{subproblem}
+	Assuming that all of the zeros of $p$ are shown (in \cref{poly:tab:findformulap}), how many zeros does $p$ have? 
+	\begin{shortsolution}
+		$p$ has 3 zeros.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	What is the degree of $p$?
+	\begin{shortsolution}
+		$p$ is degree 3.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Write a formula for $p(x)$.
+	\begin{shortsolution}
+		$p(x)=x(x+2)(x-3)$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Assuming that all of the zeros of $q$ are shown (in \cref{poly:tab:findformulaq}), how many zeros does $q$ have?
+	\begin{shortsolution}
+		$q$ has 2 zeros.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Describe the difference in behavior of $p$ and $q$ at $-2$.
+	\begin{shortsolution}
+		$p$ changes sign at $-2$, and $q$ does not change sign at $-2$.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Given that $q$ is a degree-$3$ polynomial, write a formula for $q(x)$.
+	\begin{shortsolution}
+		$q(x)=x(x+2)^2$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Assuming that all of the zeros of $r$ are shown (in \cref{poly:tab:findformular}), find a formula for $r(x)$.
+	\begin{shortsolution}
+		$r(x)=(x+3)(x+1)(x-1)(x-3)$ 
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Assuming that all of the zeros of $s$ are shown (in \cref{poly:tab:findformulas}), find a formula for $s(x)$.
+	\begin{shortsolution}
+		$s(x)=(x+3)(x+1)(x-1)^2$ 
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+\end{exercises}
+
+\section{Rational functions}
+\subsection*{Power functions with negative exponents}
+The study of rational functions will rely upon a good knowledge 
+of power functions with negative exponents. \Cref{rat:ex:oddpow,rat:ex:evenpow} are 
+simple but fundamental to understanding the behavior of rational functions.
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Power functions with odd negative exponents]\label{rat:ex:oddpow}
+Graph each of the following functions on your calculator, state their domain in interval notation, and their 
+behavior as $x\rightarrow 0^-$ and $x\rightarrow 0^+$.
+\[
+	f(x)=\frac{1}{x},\qquad g(x)=\dfrac{1}{x^3},\qquad h(x)=\dfrac{1}{x^5}
+\]
+\begin{pccsolution}
+The functions $f$, $g$, and $k$ are plotted in \cref{rat:fig:oddpow}.
+The domain of each of the functions $f$, $g$, and $h$ is $(-\infty,0)\cup (0,\infty)$. Note that 
+the long-run behavior of each of the functions is the same, and in particular
+\begin{align*}
+	f(x)\rightarrow 0                         & \text{ as } x\rightarrow\infty  \\
+	\mathllap{\text{and }}  f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty 
+\end{align*}
+The same results hold for $g$ and $h$. Note also that each of the functions
+has a \emph{vertical asymptote} at $0$. We see that
+\begin{align*}
+	f(x)\rightarrow -\infty                         & \text{ as } x\rightarrow 0^- \\
+	\mathllap{\text{and }}   f(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ 
+\end{align*}
+The same results hold for $g$ and $h$.
+
+The curve of a function that has a vertical asymptote is necessarily separated 
+into \emph{branches}| each of the functions $f$, $g$, and $h$ have $2$ branches.
+\end{pccsolution}
+\end{pccexample}
+
+\begin{figure}[!htb]
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-3,xmax=3,
+			   ymin=-5,ymax=5,
+			   xtick={-2,-1,...,2},
+			   minor ytick={-3,-1,...,3},
+			   grid=both,
+			   width=\textwidth,
+			   legend pos=north west,
+			   ]
+			   \addplot expression[domain=-3:-0.2]{1/x};
+			   \addplot expression[domain=-3:-0.584]{1/x^3};
+			   \addplot expression[domain=-3:-0.724]{1/x^5};
+			   \addplot expression[domain=0.2:3]{1/x};
+			   \addplot expression[domain=0.584:3]{1/x^3};
+			   \addplot expression[domain=0.724:3]{1/x^5};
+			   \addplot[soldot]coordinates{(-1,-1)}node[axisnode,anchor=north east]{$(-1,-1)$};
+			   \addplot[soldot]coordinates{(1,1)}node[axisnode,anchor=south west]{$(1,1)$};
+			   \legend{$f$,$g$,$h$}
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:oddpow}
+	\end{minipage}%
+	\hfill
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-3,xmax=3,
+			   ymin=-5,ymax=5,
+			   xtick={-2,-1,...,2},
+			   minor ytick={-3,-1,...,3},
+			   grid=both,
+			   width=\textwidth,
+			   legend pos=south east,
+			   ]
+			   \addplot expression[domain=-3:-0.447]{1/x^2};
+			   \addplot expression[domain=-3:-0.668]{1/x^4};
+			   \addplot expression[domain=-3:-0.764]{1/x^6};
+			   \addplot expression[domain=0.447:3]{1/x^2};
+			   \addplot expression[domain=0.668:3]{1/x^4};
+			   \addplot expression[domain=0.764:3]{1/x^6};
+			   \addplot[soldot]coordinates{(-1,1)}node[axisnode,anchor=south east]{$(-1,1)$};
+			   \addplot[soldot]coordinates{(1,1)}node[axisnode,anchor=south west]{$(1,1)$};
+			   \legend{$F$,$G$,$H$}
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:evenpow}
+	\end{minipage}%
+\end{figure}
+
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Power functions with even negative exponents]\label{rat:ex:evenpow}%
+Graph each of the following functions, state their domain, and their 
+behavior as $x\rightarrow 0^-$ and $x\rightarrow 0^+$.
+\[
+	f(x)=\frac{1}{x^2},\qquad g(x)=\frac{1}{x^4},\qquad h(x)=\frac{1}{x^6}
+\]
+\begin{pccsolution}
+The functions $F$, $G$, and $H$ are plotted in \cref{rat:fig:evenpow}.
+The domain of each of the functions $F$, $G$, and $H$ is $(-\infty,0)\cup (0,\infty)$. Note that 
+the long-run behavior of each of the functions is the same, and in particular
+\begin{align*}
+	F(x)\rightarrow 0                           & \text{ as } x\rightarrow\infty  \\
+	\mathllap{\text{and }}    f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty 
+\end{align*}
+As in \cref{rat:ex:oddpow}, $F$ has a horizontal asymptote  that 
+has equation $y=0$.
+The same results hold for $G$ and $H$. Note also that each of the functions
+has a \emph{vertical asymptote} at $0$. We see that
+\begin{align*}
+	F(x)\rightarrow \infty                          & \text{ as } x\rightarrow 0^- \\
+	\mathllap{\text{and }}   F(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ 
+\end{align*}
+The same results hold for $G$ and $H$. Each of the functions $F$, $G$, and $H$ 
+have $2$ branches.
+\end{pccsolution}
+\end{pccexample}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{doyouunderstand}
+	\begin{problem}
+	Repeat \cref{rat:ex:oddpow,rat:ex:evenpow} using (respectively)
+	\begin{subproblem}
+		$k(x)=-\dfrac{1}{x}$, $ m(x)=-\dfrac{1}{x^3}$, $ n(x)=-\dfrac{1}{x^5}$
+		\begin{shortsolution}
+			The functions $k$, $m$, and $n$ have domain $(-\infty,0)\cup (0,\infty)$, and 
+			are graphed below.
+			
+			\begin{tikzpicture}
+				\begin{axis}[
+				   framed,
+				   xmin=-3,xmax=3,
+				   ymin=-5,ymax=5,
+				   xtick={-2,-1,...,2},
+				   minor ytick={-3,-1,...,3},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   legend pos=north east,
+				   ]
+				   \addplot expression[domain=-3:-0.2]{-1/x};
+				   \addplot expression[domain=-3:-0.584]{-1/x^3};
+				   \addplot expression[domain=-3:-0.724]{-1/x^5};
+				   \addplot expression[domain=0.2:3]{-1/x};
+				   \addplot expression[domain=0.584:3]{-1/x^3};
+				   \addplot expression[domain=0.724:3]{-1/x^5};
+				   \legend{$k$,$m$,$n$}
+				\end{axis}
+			\end{tikzpicture}
+			
+			Note that
+			\begin{align*}
+				k(x)\rightarrow 0                                & \text{ as } x\rightarrow\infty  \\
+				\mathllap{\text{and }}    k(x)\rightarrow 0      & \text{ as } x\rightarrow-\infty \\
+				\intertext{and also}
+				k(x)\rightarrow \infty                           & \text{ as } x\rightarrow 0^-    \\
+				\mathllap{\text{and }}   k(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+    
+			\end{align*}
+			The same are true for $m$ and $n$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$ K(x)=-\dfrac{1}{x^2}$, $ M(x)=-\dfrac{1}{x^4}$, $ N(x)=-\dfrac{1}{x^6}$
+		\begin{shortsolution}
+			The functions $K$, $M$, and $N$ have domain $(-\infty,0)\cup (0,\infty)$, and 
+			are graphed below.
+			
+			\begin{tikzpicture}
+				\begin{axis}[
+				   framed,
+				   xmin=-3,xmax=3,
+				   ymin=-5,ymax=5,
+				   xtick={-2,-1,...,2},
+				   minor ytick={-3,-1,...,3},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   legend pos=north east,
+				   ]
+				   \addplot expression[domain=-3:-0.447]{-1/x^2};
+				   \addplot expression[domain=-3:-0.668]{-1/x^4};
+				   \addplot expression[domain=-3:-0.764]{-1/x^6};
+				   \addplot expression[domain=0.447:3]{-1/x^2};
+				   \addplot expression[domain=0.668:3]{-1/x^4};
+				   \addplot expression[domain=0.764:3]{-1/x^6};
+				   \legend{$K$,$M$,$N$}
+				\end{axis}
+			\end{tikzpicture}
+			
+			Note that
+			\begin{align*}
+				K(x)\rightarrow 0                                & \text{ as } x\rightarrow\infty  \\
+				\mathllap{\text{and }}    K(x)\rightarrow 0      & \text{ as } x\rightarrow-\infty \\
+				\intertext{and also}
+				K(x)\rightarrow -\infty                          & \text{ as } x\rightarrow 0^-    \\
+				\mathllap{\text{and }}   K(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+    
+			\end{align*}
+			The same are true for $M$ and $N$.
+		\end{shortsolution}
+	\end{subproblem}
+	\end{problem}
+\end{doyouunderstand}
+
+\subsection*{Rational functions}
+\begin{pccdefinition}[Rational functions]\label{rat:def:function}
+Rational functions have the form
+\[
+	r(x) = \frac{p(x)}{q(x)}
+\]
+where both $p$ and $q$ are polynomials. 
+
+Note that
+\begin{itemize}
+	\item the domain or $r$ will be all real numbers, except those that
+	make the \emph{denominator}, $q(x)$, equal to $0$;
+	\item the zeros of $r$ are the zeros of $p$, i.e the real numbers
+	that make the \emph{numerator}, $p(x)$, equal to $0$.
+\end{itemize}
+
+\Cref{rat:ex:oddpow,rat:ex:evenpow} are particularly important because $r$ 
+will behave like $\frac{1}{x}$, or $\frac{1}{x^2}$ around its vertical asymptotes, 
+depending on the power that the relevant term is raised to| we will demonstrate 
+this in what follows.
+\end{pccdefinition}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Rational or not]
+Identify whether each of the following functions is a rational or not. If 
+the function is rational, state the domain.
+\begin{multicols}{3}
+	\begin{enumerate}
+		\item $r(x)=\dfrac{1}{x}$
+		\item $f(x)=2^x+3$
+		\item $g(x)=19$
+		\item $h(x)=\dfrac{3+x}{4-x}$
+		\item $k(x)=\dfrac{x^3+2x}{x-15}$
+		\item $l(x)=9-4x$
+		\item $m(x)=\dfrac{x+5}{(x-7)(x+9)}$
+		\item $n(x)=x^2+6x+7$
+		\item $q(x)=1-\dfrac{3}{x+1}$
+	\end{enumerate}
+\end{multicols}
+\begin{pccsolution}
+\begin{enumerate}
+	\item $r$ is rational; the domain of $r$ is $(-\infty,0)\cup(0,\infty)$.
+	\item $f$ is not rational.
+	\item $g$ is not rational; $g$ is constant.
+	\item $h$ is rational; the domain of $h$ is $(-\infty,4)\cup(4,\infty)$.
+	\item $k$ is rational; the domain of $k$ is $(-\infty,15)\cup(15,\infty)$.
+	\item $l$ is not rational; $l$ is linear.
+	\item $m$ is rational; the domain of $m$ is $(-\infty,-9)\cup(-9,7)\cup(7,\infty)$.
+	\item $n$ is not rational; $n$ is quadratic (or you might describe $n$ as a polynomial).
+	\item $q$ is rational; the domain of $q$ is $(-\infty,-1)\cup (-1,\infty)$.
+\end{enumerate}
+\end{pccsolution}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Match formula to graph]
+Each of the following functions is graphed in \cref{rat:fig:whichiswhich}. 
+Which is which?
+\[
+	r(x)=\frac{1}{x-3}, \qquad q(x)=\frac{x-2}{x+5}, \qquad k(x)=\frac{1}{(x+2)(x-3)}
+\]
+\begin{figure}[!htb]
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/(x+5);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-6,ymax=6,
+			   xtick={-8,-6,...,8},
+			   minor ytick={-4,-3,...,4},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-6.37]{f};
+			   \addplot[pccplot] expression[domain=-3.97:10]{f};
+			   \addplot[soldot] coordinates{(2,0)};
+			   \addplot[asymptote,domain=-6:6]({-5},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:which1}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=1/(x-3);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-5,ymax=6,
+			   xtick={-8,-6,...,8},
+			   ytick={-4,4},
+			   minor ytick={-3,...,5},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:2.8]{f};
+			   \addplot[pccplot] expression[domain=3.17:10]{f};
+			   \addplot[asymptote,domain=-6:6]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:which2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=1/((x-3)*(x+2));}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-5,ymax=5,
+			   xtick={-8,-6,...,8},
+			   ytick={-4,4},
+			   minor ytick={-3,...,3},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-2.03969]{f};
+			   \addplot[pccplot] expression[domain=-1.95967:2.95967]{f};
+			   \addplot[pccplot] expression[domain=3.03969:10]{f};
+			   \addplot[asymptote,domain=-5:5]({-2},{x});
+			   \addplot[asymptote,domain=-5:5]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:which3}
+	\end{subfigure}
+	\caption{}
+	\label{rat:fig:whichiswhich}
+\end{figure}
+
+\begin{pccsolution}
+Let's start with the function $r$. Note that domain of $r$ is $(-\infty,3)\cup(0,3)$, so 
+we search for a function that has a vertical asymptote at $3$. There 
+are two possible choices: the functions graphed in \cref{rat:fig:which2,rat:fig:which3}, 
+but note that the function in \cref{rat:fig:which3} also has a vertical asymptote at $-2$ 
+which is not consistent with the formula for $r(x)$. Therefore, $y=r(x)$
+is graphed in \cref{rat:fig:which2}. 
+
+The function $q$ has domain $(-\infty,-5)\cup(-5,\infty)$, so we search 
+for a function that has a vertical asymptote at $-5$. The only candidate 
+is the curve shown in \cref{rat:fig:which1}; note that the curve also goes through $(2,0)$, 
+which is consistent with the formula for $q(x)$, since $q(2)=0$, i.e $q$
+has a zero at $2$.
+
+The function $k$ has domain $(-\infty,-2)\cup(-2,3)\cup(3,\infty)$, and 
+has vertical asymptotes at $-2$ and $3$. This is consistent with 
+the graph in \cref{rat:fig:which3} (and is the only curve that 
+has $3$ branches).
+
+We note that each function behaves like $\frac{1}{x}$ around its vertical asymptotes, 
+because each linear factor in each denominator is raised to the power $1$; if (for example) 
+the definition of $r$ was instead
+\[
+	r(x)=\frac{1}{(x-3)^2}
+\]
+then we would see that $r$ behaves like $\frac{1}{x^2}$ around its vertical asymptote, and 
+the graph of $r$ would be very different. We will deal with these cases in the examples that follow.
+\end{pccsolution}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Repeated factors in the denominator]
+Consider the functions $f$, $g$, and $h$ that have formulas
+\[
+	f(x)=\frac{x-2}{(x-3)(x+2)}, \qquad g(x)=\frac{x-2}{(x-3)^2(x+2)}, \qquad h(x)=\frac{x-2}{(x-3)(x+2)^2}
+\]
+which are graphed in \cref{rat:fig:repfactd}. Note that each function has $2$ 
+vertical asymptotes, and the domain of each function is 
+\[
+	(-\infty,-2)\cup(-2,3)\cup(3,\infty)
+\]
+so we are not surprised to see that each curve has $3$ branches. We also note that 
+the numerator of each function is the same, which tells us that each function has 
+only $1$ zero at $2$.
+
+The functions $g$ and $h$ are different from those that we have considered previously, 
+because they have a repeated factor in the denominator. Notice in particular 
+the way that the functions behave around their asymptotes:
+\begin{itemize}
+	\item $f$ behaves like $\frac{1}{x}$ around both of its asymptotes;
+	\item $g$ behaves like $\frac{1}{x}$ around $-2$, and like $\frac{1}{x^2}$ around $3$;
+	\item $h$ behaves like $\frac{1}{x^2}$ around $-2$, and like $\frac{1}{x}$ around $3$.
+\end{itemize}
+\end{pccexample}
+\begin{figure}[!htb]
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)*(x-3));}]
+			\begin{axis}[
+			   %                    framed,
+			   xmin=-5,xmax=5,
+			   ymin=-4,ymax=4,
+			   xtick={-4,-2,...,4},
+			   ytick={-2,2},
+			   %                    grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-5:-2.201]{f};
+			   \addplot[pccplot] expression[domain=-1.802:2.951]{f};
+			   \addplot[pccplot] expression[domain=3.052:5]{f};
+			   \addplot[soldot] coordinates{(2,0)};
+			   %                 \addplot[asymptote,domain=-6:6]({-2},{x});
+			   %                 \addplot[asymptote,domain=-6:6]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=\dfrac{x-2}{(x+2)(x-3)}$}
+		\label{rat:fig:repfactd1}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)*(x-3)^2);}]
+			\begin{axis}[
+			   %                    framed,
+			   xmin=-5,xmax=5,
+			   ymin=-4,ymax=4,
+			   xtick={-4,-2,...,4},
+			   ytick={-2,2},
+			   %                    grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-5:-2.039]{f};
+			   \addplot[pccplot] expression[domain=-1.959:2.796]{f};
+			   \addplot[pccplot] expression[domain=3.243:5]{f};
+			   \addplot[soldot] coordinates{(2,0)};
+			   %                 \addplot[asymptote,domain=-4:4]({-2},{x});
+			   %                 \addplot[asymptote,domain=-4:4]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=\dfrac{x-2}{(x+2)(x-3)^2}$}
+		\label{rat:fig:repfactd2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)^2*(x-3));}]
+			\begin{axis}[
+			   %                    framed,
+			   xmin=-5,xmax=5,
+			   ymin=-4,ymax=4,
+			   xtick={-4,-2,...,2},
+			   ytick={-2,2},
+			   %                    grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-5:-2.451]{f};
+			   \addplot[pccplot] expression[domain=-1.558:2.990]{f};
+			   \addplot[pccplot] expression[domain=3.010:6]{f};
+			   \addplot[soldot] coordinates{(2,0)};
+			   %                 \addplot[asymptote,domain=-4:4]({-2},{x});
+			   %                 \addplot[asymptote,domain=-4:4]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=\dfrac{x-2}{(x+2)^2(x-3)}$}
+		\label{rat:fig:repfactd3}
+	\end{subfigure}
+	\caption{}
+	\label{rat:fig:repfactd}
+\end{figure}
+
+\Cref{rat:def:function} says that the zeros of 
+the rational function $r$ that has formula $r(x)=\frac{p(x)}{q(x)}$ are 
+the zeros of $p$. Let's explore this a little more.
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}[Zeros] Find the zeros of each of the following functions
+\[
+	\alpha(x)=\frac{x+5}{3x-7}, \qquad \beta(x)=\frac{9-x}{x+1}, \qquad \gamma(x)=\frac{17x^2-10}{2x+1}
+\]
+\begin{pccsolution}
+We find the zeros of each function in turn by setting the numerator equal to $0$. The zeros of 
+$\alpha$ are found by solving 
+\[
+	x+5=0
+\]
+The zero of $\alpha$ is $-5$.
+
+Similarly, we may solve $9-x=0$ to find the zero of $\beta$, which is clearly $9$.
+
+The zeros of $\gamma$ satisfy the equation
+\[
+	17x^2-10=0
+\]
+which we can solve using the square root property to obtain
+\[
+	x=\pm\frac{10}{17}
+\]
+The zeros of $\gamma$ are $\pm\frac{10}{17}$.
+\end{pccsolution}
+\end{pccexample}
+
+\subsection*{Long-run behavior}
+Our focus so far has been on the behavior of rational functions around 
+their \emph{vertical} asymptotes. In fact, rational functions also 
+have interesting long-run behavior around their \emph{horizontal} or 
+\emph{oblique} asymptotes. A rational function will always have either 
+a horizontal or an oblique asymptote| the case is determined by the degree
+of the numerator and the degree of the denominator.
+\begin{pccdefinition}[Long-run behavior]\label{rat:def:longrun}
+Let $r$ be the rational function that has formula
+\[
+	r(x) = \frac{a_n x^n + a_{n-1}x^{n-1}+\ldots + a_0}{b_m x^m + b_{m-1}x^{m-1}+\ldots+b_0}
+\]
+We can classify the long-run behavior of the rational function $r$ 
+according to the following criteria:
+\begin{itemize}
+	\item if $n<m$ then  $r$ has a horizontal asymptote with equation $y=0$;
+	\item if $n=m$ then $r$ has a horizontal asymptote with equation $y=\dfrac{a_n}{b_m}$;
+	\item if $n>m$ then $r$ will have an oblique asymptote as $x\rightarrow\pm\infty$ (more on this in \cref{rat:sec:oblique})
+\end{itemize} 
+\end{pccdefinition}
+We will concentrate on functions that have horizontal asymptotes until 
+we reach \cref{rat:sec:oblique}.
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}[Long-run behavior graphically]\label{rat:ex:horizasymp}
+\pccname{Kebede} has graphed the following functions in his graphing calculator
+\[
+	r(x)=\frac{x+1}{x-3}, \qquad s(x)=\frac{2(x+1)}{x-3}, \qquad t(x)=\frac{3(x+1)}{x-3}
+\]
+and obtained the curves shown in \cref{rat:fig:horizasymp}. Kebede decides 
+to test his knowledgeable friend \pccname{Oscar}, and asks him 
+to match the formulas to the graphs.
+
+\begin{figure}[!htb]
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=2*(x+1)/(x-3);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-15,xmax=15,
+			   ymin=-6,ymax=6,
+			   xtick={-12,-8,...,12},
+			   minor ytick={-4,-3,...,4},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-15:2]{f};
+			   \addplot[pccplot] expression[domain=5:15]{f};
+			   \addplot[soldot] coordinates{(-1,0)};
+			   \addplot[asymptote,domain=-6:6]({3},{x});
+			   \addplot[asymptote,domain=-15:15]({x},{2});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:horizasymp1}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x+1)/(x-3);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-15,xmax=15,
+			   ymin=-6,ymax=6,
+			   xtick={-12,-8,...,12},
+			   minor ytick={-4,-3,...,4},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-15:2.42857,samples=50]{f};
+			   \addplot[pccplot] expression[domain=3.8:15,samples=50]{f};
+			   \addplot[soldot] coordinates{(-1,0)};
+			   \addplot[asymptote,domain=-6:6]({3},{x});
+			   \addplot[asymptote,domain=-15:15]({x},{1});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:horizasymp2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=3*(x+1)/(x-3);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-15,xmax=15,
+			   ymin=-6,ymax=6,
+			   xtick={-12,-8,...,12},
+			   minor ytick={-4,-3,...,4},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-15:1.6666,samples=50]{f};
+			   \addplot[pccplot] expression[domain=7:15]{f};
+			   \addplot[soldot] coordinates{(-1,0)};
+			   \addplot[asymptote,domain=-6:6]({3},{x});
+			   \addplot[asymptote,domain=-15:15]({x},{3});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:horizasymp3}
+	\end{subfigure}
+	\caption{Horizontal asymptotes}
+	\label{rat:fig:horizasymp}
+\end{figure}
+
+Oscar notices that each function has a vertical asymptote at $3$ and a zero at $-1$. 
+The main thing that catches Oscar's eye is that each function has a different 
+coefficient in the numerator, and that each curve has a different horizontal asymptote. 
+In particular, Oscar notes that
+\begin{itemize}
+	\item the curve shown in \cref{rat:fig:horizasymp1} has a horizontal asymptote with equation $y=2$;
+	\item the curve shown in \cref{rat:fig:horizasymp2} has a horizontal asymptote with equation $y=1$;
+	\item the curve shown in \cref{rat:fig:horizasymp3} has a horizontal asymptote with equation $y=3$.
+\end{itemize}
+Oscar is able to tie it all together for Kebede by referencing \cref{rat:def:longrun}. He says 
+that since the degree of the numerator and the degree of the denominator is the same 
+for each of the functions $r$, $s$, and $t$, the horizontal asymptote will be determined 
+by evaluating the ratio of their leading coefficients. 
+
+Oscar therefore says that $r$ should have a horizontal asymptote $y=\frac{1}{1}=1$, $s$ should 
+have a horizontal asymptote $y=\frac{2}{1}=2$, and $t$ should have a horizontal asymptote 
+$y=\frac{3}{1}=3$. Kebede is able to finish the problem from here, and says that $r$ is 
+shown in \cref{rat:fig:horizasymp2}, $s$ is shown in \cref{rat:fig:horizasymp1}, and 
+$t$ is shown in \cref{rat:fig:horizasymp3}.
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}[Long-run behavior numerically]
+\pccname{Xiao} and \pccname{Dwayne} saw \cref{rat:ex:horizasymp} but are a little confused 
+about horizontal asymptotes. What does it mean to say that a function $r$ has a horizontal
+asymptote? 
+
+They decide to explore the concept by 
+constructing a table of values for the rational functions $R$ and  $S$ that have formulas
+\[
+	R(x)=\frac{-5(x+1)}{x-3}, \qquad S(x)=\frac{7(x-5)}{2(x+1)}
+\]
+In \cref{rat:tab:plusinfty} they model the behavior of $R$ and $S$ as $x\rightarrow\infty$, 
+and in \cref{rat:tab:minusinfty} they model the behavior of $R$ and $S$ as $x\rightarrow-\infty$
+by substituting very large values of $|x|$ into each function.
+\begin{table}[!htb]
+	\begin{minipage}{.5\textwidth}
+		\centering
+		\caption{$R$ and $S$ as $x\rightarrow\infty$}
+		\label{rat:tab:plusinfty}
+		\begin{tabular}{crr}
+			\beforeheading
+			$x$            & $R(x)$     & $S(x)$    \\ \afterheading
+			$1\times 10^2$ & $-5.20619$ & $3.29208$ \\\normalline
+			$1\times 10^3$ & $-5.02006$ & $3.47902$ \\\normalline
+			$1\times 10^4$ & $-5.00200$ & $3.49790$ \\\normalline
+			$1\times 10^5$ & $-5.00020$ & $3.49979$ \\\normalline
+			$1\times 10^6$ & $-5.00002$ & $3.49998$ \\\lastline
+		\end{tabular}
+	\end{minipage}%
+	\begin{minipage}{.5\textwidth}
+		\centering
+		\caption{$R$ and $S$ as $x\rightarrow-\infty$}
+		\label{rat:tab:minusinfty}
+		\begin{tabular}{crr}
+			\beforeheading
+			$x$             & $R(x)$     & $S(x)$    \\ \afterheading
+			$-1\times 10^2$ & $-4.80583$ & $3.71212$ \\\normalline
+			$-1\times 10^3$ & $-4.98006$ & $3.52102$ \\\normalline
+			$-1\times 10^4$ & $-4.99800$ & $3.50210$ \\\normalline
+			$-1\times 10^5$ & $-4.99980$ & $3.50021$ \\\normalline
+			$-1\times 10^6$ & $-4.99998$ & $3.50002$ \\\lastline
+		\end{tabular}
+	\end{minipage}
+\end{table}
+
+Xiao and Dwayne study \cref{rat:tab:plusinfty,rat:tab:minusinfty} and decide that 
+the functions $R$ and $S$ never actually touch their horizontal asymptotes, but they 
+do get infinitely close. They also feel as if they have a better understanding of 
+what it means to study the behavior of a function as $x\rightarrow\pm\infty$.
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}[Repeated factors in the numerator]
+Consider the functions $f$, $g$, and $h$ that have formulas
+\[
+	f(x)=\frac{(x-2)^2}{(x-3)(x+1)}, \qquad g(x)=\frac{x-2}{(x-3)(x+1)}, \qquad h(x)=\frac{(x-2)^3}{(x-3)(x+1)}
+\]
+which are graphed in \cref{rat:fig:repfactn}. We note that each function has vertical
+asymptotes at $-1$ and $3$, and so the domain of each function is
+\[
+	(-\infty,-1)\cup(-1,3)\cup(3,\infty)
+\]
+We also notice that the numerators of each function are quite similar| indeed, each 
+function has a zero at $2$, but how does each function behave around their zero?
+
+Using \cref{rat:fig:repfactn} to guide us, we note that
+\begin{itemize}
+	\item $f$ has a horizontal intercept $(2,0)$, but the curve of 
+	$f$ does not cut the horizontal axis| it bounces off it;
+	\item $g$ also has a horizontal intercept $(2,0)$, and the curve 
+	of $g$ \emph{does} cut the horizontal axis;
+	\item $h$ has a horizontal intercept $(2,0)$, and the curve of $h$ 
+	also cuts the axis, but appears flattened as it does so.
+\end{itemize}
+
+We can further enrich our study by discussing the long-run behavior of each function. 
+Using the tools of \cref{rat:def:longrun}, we can deduce that 
+\begin{itemize}
+	\item $f$ has a horizontal asymptote with equation $y=1$;
+	\item $g$ has a horizontal asymptote with equation $y=0$;
+	\item $h$ does \emph{not} have a horizontal asymptote| it has an oblique asymptote (we'll 
+	study this more in \cref{rat:sec:oblique}).
+\end{itemize}
+\end{pccexample}
+
+\begin{figure}[!htb]
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-2)^2/((x+1)*(x-3));}]
+			\begin{axis}[
+			   %                    framed,
+			   xmin=-5,xmax=5,
+			   ymin=-10,ymax=10,
+			   xtick={-4,-2,...,4},
+			   ytick={-8,-4,...,8},
+			   %                    grid=both,
+			   width=\figurewidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-5:-1.248,samples=50]{f};
+			   \addplot[pccplot] expression[domain=-0.794:2.976,samples=50]{f};
+			   \addplot[pccplot] expression[domain=3.026:5,samples=50]{f};
+			   \addplot[soldot] coordinates{(2,0)};
+			   %                 \addplot[asymptote,domain=-6:6]({-1},{x});
+			   %                 \addplot[asymptote,domain=-6:6]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=\dfrac{(x-2)^2}{(x+1)(x-3)}$}
+		\label{rat:fig:repfactn1}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+1)*(x-3));}]
+			\begin{axis}[
+			   %                    framed,
+			   xmin=-5,xmax=5,
+			   ymin=-10,ymax=10,
+			   xtick={-4,-2,...,4},
+			   ytick={-8,-4,...,8},
+			   %                    grid=both,
+			   width=\figurewidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-5:-1.075]{f};
+			   \addplot[pccplot] expression[domain=-0.925:2.975]{f};
+			   \addplot[pccplot] expression[domain=3.025:5]{f};
+			   \addplot[soldot] coordinates{(2,0)};
+			   %                 \addplot[asymptote,domain=-6:6]({-1},{x});
+			   %                 \addplot[asymptote,domain=-6:6]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=\dfrac{x-2}{(x+1)(x-3)}$}
+		\label{rat:fig:repfactn2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-2)^3/((x+1)*(x-3));}]
+			\begin{axis}[
+			   %                    framed,
+			   xmin=-5,xmax=5,
+			   xtick={-8,-6,...,8},
+			   %                    grid=both,
+			   ymin=-30,ymax=30,
+			   width=\figurewidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-5:-1.27]{f};
+			   \addplot[pccplot] expression[domain=-0.806:2.99185]{f};
+			   \addplot[pccplot] expression[domain=3.0085:5]{f};
+			   \addplot[soldot] coordinates{(2,0)};
+			   %                 \addplot[asymptote,domain=-30:30]({-1},{x});
+			   %                 \addplot[asymptote,domain=-30:30]({3},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=\dfrac{(x-2)^3}{(x+1)(x-3)}$}
+		\label{rat:fig:repfactn3}
+	\end{subfigure}
+	\caption{}
+	\label{rat:fig:repfactn}
+\end{figure}
+
+\subsection*{Holes}
+Rational functions have a vertical asymptote at $a$ if the denominator is $0$ at $a$. 
+What happens if the numerator is $0$ at the same place? In this case, we say that the rational 
+function has a \emph{hole} at $a$.
+\begin{pccdefinition}[Holes]
+The rational function
+\[
+	r(x)=\frac{p(x)}{q(x)}
+\]
+has a hole at $a$ if $p(a)=q(a)=0$. Note that holes are different from 
+a vertical asymptotes. We represent that $r$ has a hole at the point 
+$(a,r(a))$ on the curve $y=r(x)$ by 
+using a hollow circle, $\circ$.
+\end{pccdefinition}
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{pccexample}
+\pccname{Mohammed} and \pccname{Sue} have graphed the function $r$ that has formula
+\[
+	r(x)=\frac{x^2+x-6}{(x-2)}
+\]
+in their calculators, and can not decide if the correct graph 
+is \cref{rat:fig:hole} or \cref{rat:fig:hole1}.
+
+Luckily for them, Oscar is nearby, and can help them settle the debate. 
+Oscar demonstrates that
+\begin{align*}
+	r(x) & =\frac{(x+3)(x-2)}{(x-2)} \\
+	     & = x+3                     
+\end{align*}
+but only when $x\ne 2$, because the function is undefined at $2$. Oscar 
+says that this necessarily means that the domain or $r$ is 
+\[
+	(-\infty,2)\cup(2,\infty)
+\]
+and that $r$ must have a hole at $2$. 
+
+Mohammed and Sue are very grateful for the clarification, and conclude that 
+the graph of $r$ is shown in \cref{rat:fig:hole1}.
+\begin{figure}[!htb]
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-4,...,8},
+			   ytick={-8,-4,...,8},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-10:7]{x+3};
+			   \addplot[soldot] coordinates{(-3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:hole}
+	\end{minipage}%
+	\hfill
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-4,...,8},
+			   ytick={-8,-4,...,8},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot expression[domain=-10:7]{x+3};
+			   \addplot[holdot] coordinates{(2,5)};
+			   \addplot[soldot] coordinates{(-3,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:hole1}
+	\end{minipage}%
+\end{figure}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{pccexample}
+Consider the function $f$ that has formula
+\[
+	f(x)=\frac{x(x+3)}{x^2-4x}
+\]
+The domain of $f$ is $(-\infty,0)\cup(0,4)\cup(4,\infty)$ because both $0$ and $4$ 
+make the denominator equal to $0$. Notice that
+\begin{align*}
+	f(x) & = \frac{x(x+3)}{x(x-4)} \\
+	     & = \frac{x+3}{x-4}       
+\end{align*}
+provided that $x\ne 0$. Since $0$ makes the numerator 
+and the denominator 0 at the same time, we say that $f$ has a hole at $(0,-\nicefrac{3}{4})$. 
+Note that this necessarily means that $f$ does not have a vertical intercept.
+
+We also note $f$ has a vertical asymptote at $4$; the function is graphed in \cref{rat:fig:holeex}.
+\begin{figure}[!htb]
+	\centering
+	\begin{tikzpicture}[/pgf/declare function={f=(x+3)/(x-4);}]
+		\begin{axis}[
+		   framed,
+		   xmin=-10,xmax=10,
+		   ymin=-10,ymax=10,
+		   xtick={-8,-6,...,8},
+		   ytick={-8,-6,...,8},
+		   grid=both,
+		   ]
+		   \addplot[pccplot] expression[domain=-10:3.36364,samples=50]{f};
+		   \addplot[pccplot] expression[domain=4.77:10]{f};
+		   \addplot[asymptote,domain=-10:10]({4},{x});
+		   \addplot[holdot]coordinates{(0,-0.75)};
+		   \addplot[soldot] coordinates{(-3,0)};
+		\end{axis}
+	\end{tikzpicture}
+	\caption{$y=\dfrac{x(x+3)}{x^2-4x}$}
+	\label{rat:fig:holeex}
+\end{figure}
+\end{pccexample}
+
+
+
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{pccexample}[Minimums and maximums]
+\pccname{Seamus} and \pccname{Trang} are discussing rational functions. Seamus says that 
+if a rational function has a vertical asymptote, then it can 
+not possibly have local minimums and maximums, nor can it have 
+global minimums and maximums.
+
+Trang says this statement is not always true. She plots the functions 
+$f$ and $g$ that have formulas
+\[
+	f(x)=-\frac{32(x-1)(x+1)}{(x-2)^2(x+2)^2}, \qquad g(x)=\frac{32(x-1)(x+1)}{(x-2)^2(x+2)^2}
+\]
+in \cref{rat:fig:minmax1,rat:fig:minmax2} and shows them to Seamus. On seeing the graphs, 
+Seamus quickly corrects himself, and says  that $f$ has a local (and global) 
+maximum of $2$ at $0$, and that $g$ has a local (and global) minimum of $-2$ at $0$.
+
+\begin{figure}[!htb]
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=-32*(x-1)*(x+1)/(( x-2)^2*(x+2)^2);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-8,-6,...,8},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-3.01]{f};
+			   \addplot[pccplot] expression[domain=-1.45:1.45]{f};
+			   \addplot[pccplot] expression[domain=3.01:10]{f};
+			   \addplot[soldot] coordinates{(-1,0)(1,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=f(x)$}
+		\label{rat:fig:minmax1}
+	\end{minipage}%
+	\hfill
+	\begin{minipage}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=32*(x-1)*(x+1)/(( x-2)^2*(x+2)^2);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-8,-6,...,8},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-3.01]{f};
+			   \addplot[pccplot] expression[domain=-1.45:1.45]{f};
+			   \addplot[pccplot] expression[domain=3.01:10]{f};
+			   \addplot[soldot] coordinates{(-1,0)(1,0)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{$y=g(x)$}
+		\label{rat:fig:minmax2}
+	\end{minipage}%
+\end{figure}
+
+Seamus also notes that (in its domain) the function $f$ is always concave down, and 
+that (in its domain) the function $g$ is always concave up. Furthermore, Trang
+observes that each function behaves like $\frac{1}{x^2}$ around each of its vertical 
+asymptotes, because each linear factor in the denominator is raised to the power $2$.
+
+\pccname{Oscar} stops by and reminds both students about the long-run behavior; according 
+to \cref{rat:def:longrun} since the degree of the denominator is greater than the
+degree of the numerator (in both functions), each function has a horizontal asymptote
+at $y=0$.
+\end{pccexample}
+
+
+\investigation*{}
+%===================================
+%   Author: Pettit/Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[The spaghetti incident]
+The same Queen from \vref{exp:prob:queenschessboard} has recovered from 
+the rice experiments, and has called her loyal jester for another challenge.
+
+The jester has an $11-$inch piece of uncooked spaghetti that he puts on a table; 
+he uses a book to cover $\unit[1]{inch}$ of it so that 
+$\unit[10]{inches}$ hang over the edge. The jester then produces a box of $\unit{mg}$
+weights that can be hung from the spaghetti.
+
+The jester says it will take $\unit[y]{mg}$ to break the spaghetti when hung
+$\unit[x]{inches}$ from the edge, according to the rule $y=\frac{100}{x}$.
+\begin{margintable}
+	\centering
+	\captionof{table}{}
+	\label{rat:tab:spaghetti}
+	\begin{tabular}{cc}
+		\beforeheading
+		\heading{$x$} & \heading{$y$} \\
+		\afterheading
+		$1$           &               \\\normalline
+		$2$           &               \\\normalline
+		$3$           &               \\\normalline
+		$4$           &               \\\normalline
+		$5$           &               \\\normalline
+		$6$           &               \\\normalline
+		$7$           &               \\\normalline
+		$8$           &               \\\normalline
+		$9$           &               \\\normalline
+		$10$          &               \\\lastline
+	\end{tabular}
+\end{margintable}
+\begin{subproblem}\label{rat:prob:spaggt1}
+	Help the Queen complete \cref{rat:tab:spaghetti}, and use $2$ digits after the decimal
+	where appropriate.
+	\begin{shortsolution}
+		\begin{tabular}[t]{ld{2}}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\
+			\afterheading
+			$1$           & 100           \\\normalline
+			$2$           & 50            \\\normalline
+			$3$           & 33.33         \\\normalline
+			$4$           & 25            \\\normalline
+			$5$           & 20            \\\normalline
+			$6$           & 16.67         \\\normalline
+			$7$           & 14.29         \\\normalline
+			$8$           & 12.50         \\\normalline
+			$9$           & 11.11         \\\normalline
+			$10$          & 10            \\\lastline
+		\end{tabular}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	What do you notice about the number of $\unit{mg}$ that it takes to break 
+	the spaghetti as $x$ increases?
+	\begin{shortsolution}
+		It seems that the number of $\unit{mg}$ that it takes to break the spaghetti decreases
+		as $x$ increases.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}\label{rat:prob:spaglt1}
+	The Queen wonders what happens when $x$ gets very small| help the Queen construct 
+	a table of values for $x$ and $y$ when $x=0.0001, 0.001, 0.01, 0.1, 0.5, 1$.
+	\begin{shortsolution}
+		\begin{tabular}[t]{d{2}l}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\
+			\afterheading
+			0.0001        & $1000000$     \\\normalline
+			0.001         & $100000$      \\\normalline
+			0.01          & $10000$       \\\normalline
+			0.1           & $1000$        \\\normalline
+			0.5           & $200$         \\\normalline
+			1             & $100$         \\\lastline
+		\end{tabular}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	What do you notice about the number of $\unit{mg}$ that it takes to break the spaghetti
+	as $x\rightarrow 0$? Would it ever make sense to let $x=0$?
+	\begin{shortsolution}
+		The number of $\unit{mg}$ required to break the spaghetti increases as $x\rightarrow 0$.
+		We can not allow $x$ to be $0$, as we can not divide by $0$, and we can not 
+		be $0$ inches from the edge of the table.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Plot your results from \cref{rat:prob:spaggt1,rat:prob:spaglt1} on the same graph, 
+	and join the points using a smooth curve| set the maximum value of $y$ as $200$, and 
+	note that this necessarily means that you will not be able to plot all of the points.
+	\begin{shortsolution}
+		The graph of $y=\frac{100}{x}$ is shown below.
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-2,xmax=11,
+			   ymin=-20,ymax=200,
+			   xtick={2,4,...,10},
+			   ytick={20,40,...,180},
+			   grid=major,
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot+[-] expression[domain=0.5:10]{100/x};
+			   \addplot[soldot] coordinates{(0.5,200)(1,100)(2,50)(3,33.33)
+			   (4,25)(5,20)(16.67)(7,14.29)(8,12.50)(9,11.11)(10,10)};
+			\end{axis}
+		\end{tikzpicture}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Using your graph, observe what happens to $y$ as $x$ increases. If we could somehow
+	construct a piece of uncooked spaghetti that was $\unit[101]{inches}$ long, how many 
+	$\unit{mg}$ would it take to break the spaghetti?
+	\begin{shortsolution}
+		As $x$ increases, $y\rightarrow 0$. If we could construct a piece of spaghetti 
+		$\unit[101]{inches}$ long, it would only take $\unit[1]{mg}$ to break it $\left(\frac{100}{100}=1\right)$. Of course, 
+		the weight of spaghetti would probably cause it to break without the weight.
+	\end{shortsolution}
+\end{subproblem}
+The Queen looks forward to more food-related investigations from her jester.
+\end{problem}
+
+
+
+%===================================
+%   Author: Adams (Hughes)
+%   Date:   March 2012
+%===================================
+\begin{problem}[Debt Amortization]
+To amortize a debt means to pay it off in a given length of time using 
+equal periodic payments. The payments include interest on the unpaid 
+balance. The following formula gives the monthly payment, $M$, in dollars
+that is necessary to amortize a debt of $P$ dollars in $n$ months 
+at a monthly interest rate of $i$
+\[
+	M=\frac{P\cdot i}{1-(1+i)^{-n}}
+\]
+Use this formula in each of the following problems.
+\begin{subproblem}
+	What monthly payments are necessary on a credit card debt of \$2000 at 
+	$\unit[1.5]{\%}$ monthly if you want to pay off the debt in $2$ years?
+	In one year? How much money will you save by paying off the debt in the
+	shorter amount of time?
+	\begin{shortsolution}
+		Paying off the debt in $2$ years, we use
+		\begin{align*}
+			M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-24}} \\
+			  & \approx 99.85                               
+		\end{align*}
+		The monthly payments are \$99.85.
+		
+		Paying off the debt in $1$ year, we use
+		\begin{align*}
+			M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-12}} \\
+			  & \approx 183.36                              
+		\end{align*}
+		The monthly payments are \$183.36
+		
+		In the $2$-year model we would pay a total of $\$99.85\cdot 12=\$2396.40$. In the
+		$1$-year model we would pay a total of $\$183.36\cdot 12=\$2200.32$. We would therefore
+		save $\$196.08$ if we went with the $1$-year model instead of the $2$-year model.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	To purchase a home, a family needs a loan of \$300,000 at $\unit[5.2]{\%}$ 
+	annual interest.  Compare a $20$ year loan to a $30$ year loan and make 
+	a recommendation for the family.
+	(Note: when given an annual interest rate, it is a common business practice to divide by
+	$12$ to get a monthly rate.)
+	\begin{shortsolution}
+		For the $20$-year loan we use
+		\begin{align*}
+			M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 20}} \\
+			  & \approx 2013.16                                                                          
+		\end{align*}
+		The monthly payments are \$2013.16.
+		
+		For the $30$-year loan we use
+		\begin{align*}
+			M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 30}} \\
+			  & \approx 1647.33                                                                          
+		\end{align*}
+		The monthly payments are \$1647.33.
+		
+		The total amount paid during the $20$-year loan is $\$2013.16\cdot 12\cdot 20=\$483,158.40$. 
+		The total amount paid during the $30$-year loan is $\$1647.33\cdot 12\cdot 30=\$593,038.80$.
+		
+		Recommendation: if you can afford the payments, choose the $20$-year loan.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	\pccname{Ellen} wants to make monthly payments of \$100 to pay off a debt of \$3000 
+	at \unit[12]{\%} annual interest. How long will it take her to pay off the 
+	debt?
+	\begin{shortsolution}
+		We are given $M=100$, $P=3000$, $i=0.01$, and we need to find $n$
+		in the equation
+		\[
+			100 = \frac{3000\cdot 0.01}{1-(1+0.01)^{-n}}
+		\]
+		Using logarithms, we find that $n\approx 36$. It will take 
+		Ellen about $3$ years to pay off the debt.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	\pccname{Jake} is going to buy a new car. He puts \$2000 down and wants to finance the
+	remaining \$14,000. The dealer will offer him \unit[4]{\%} annual interest for 
+	$5$ years, or a \$2000
+	rebate which he can use to reduce the amount of the loan and \unit[8]{\%} 
+	annual interest for 5 years. Which should he choose?
+	\begin{shortsolution}
+		\begin{description}
+			\item[Option 1:] $\unit[4]{\%}$ annual interest for $5$ years on \$14,000.
+			This means that the monthly payments will be calculated using
+			\begin{align*}
+				M & = \frac{14000\cdot \frac{0.04}{12}}{1-\left( 1+\frac{0.04}{12} \right)^{-12\cdot 5}} \\
+				  & \approx 257.83                                                                       
+			\end{align*}
+			The monthly payments will be $\$257.83$. The total amount paid will be
+			$\$257.83\cdot 5\cdot 12=\$15,469.80$, of which $\$1469.80$ is interest.
+			\item[Option 2:] $\unit[8]{\%}$ annual interest for $5$ years on \$12,000.
+			This means that the monthly payments will be calculated using
+			\begin{align*}
+				M & = \frac{12000\cdot \frac{0.08}{12}}{1-\left( 1+\frac{0.08}{12} \right)^{-12\cdot 5}} \\
+				  & \approx 243.32                                                                       
+			\end{align*}
+			The monthly payments will be $\$243.32$. The total amount paid 
+			will be $\$243.32\cdot 5\cdot 12 =\$14,599.20$, of which $\$2599.2$ is 
+			interest.
+		\end{description}
+		Jake should choose option 1 to minimize the amount of interest 
+		he has to pay.
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+\begin{exercises}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Rational or not]
+Decide if each of the following functions are rational or not. If 
+they are rational, state their domain.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$r(x)=\dfrac{3}{x}$    
+		\begin{shortsolution}
+			$r$ is rational; the domain of $r$ is $(-\infty,0)\cup (0,\infty)$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$s(y)=\dfrac{y}{6}$    
+		\begin{shortsolution}
+			$s$ is not rational ($s$ is linear). 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$t(z)=\dfrac{4-x}{7-8z}$    
+		\begin{shortsolution}
+			$t$ is rational; the domain of $t$ is $\left( -\infty,\dfrac{7}{8} \right)\cup \left( \dfrac{7}{8},\infty \right)$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$u(w)=\dfrac{w^2}{(w-3)(w+4)}$ 
+		\begin{shortsolution}
+			$u$ is rational; the domain of $w$ is $(-\infty,-4)\cup(-4,3)\cup(3,\infty)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$v(x)=\dfrac{4}{(x-2)^2}$ 
+		\begin{shortsolution}
+			$v$ is rational; the domain of $v$ is $(-\infty,2)\cup(2,\infty)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$w(x)=\dfrac{9-x}{x+17}$ 
+		\begin{shortsolution}
+			$w$ is rational; the domain of $w$ is $(-\infty,-17)\cup(-17,\infty)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$a(x)=x^2+4$ 
+		\begin{shortsolution}
+			$a$ is not rational ($a$ is quadratic, or a polynomial of degree $2$).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$b(y)=3^y$
+		\begin{shortsolution}
+			$b$ is not rational ($b$ is exponential).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$c(z)=\dfrac{z^2}{z^3}$ 
+		\begin{shortsolution}
+			$c$ is rational; the domain of $c$ is $(-\infty,0)\cup (0,\infty)$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$d(x)=x^2(x+3)(5x-7)$ 
+		\begin{shortsolution}
+			$d$ is not rational ($d$ is a polynomial).
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$e(\alpha)=\dfrac{\alpha^2}{\alpha^2-1}$ 
+		\begin{shortsolution}
+			$e$ is rational; the domain of $e$ is $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$f(\beta)=\dfrac{3}{4}$ 
+		\begin{shortsolution}
+			$f$ is not rational ($f$ is constant).
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Function evaluation]
+Let $r$ be the function that has formula 
+\[
+	r(x)=\frac{(x-2)(x+3)}{(x+5)(x-7)}
+\]
+Evaluate each of the following (if possible); if the value is undefined, 
+then state so.
+\begin{multicols}{4}
+	\begin{subproblem}
+		$r(0)$ 
+		\begin{shortsolution}
+			$\begin{aligned}[t]
+			r(0)&=\frac{(0-2)(0+3)}{(0+5)(0-7)}\\
+			&=\frac{-6}{-35}\\
+			&=\frac{6}{35}
+		\end{aligned}$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r(1)$ 
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r(1)&=\frac{(1-2)(1+3)}{(1+5)(1-7)}\\
+		&=\frac{-4}{-36}\\
+		&=\frac{1}{9}
+	\end{aligned}$
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r(2)$ 
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r(2)&=\frac{(2-2)(2+3)}{(2+5)(2-7)}\\
+		& = \frac{0}{-50}\\
+		&=0
+	\end{aligned}$
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r(4)$ 
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r(4)&=\frac{(4-2)(4+3)}{(4+5)(4-7)}\\
+		&=\frac{14}{-27}\\
+		&=-\frac{14}{27}
+	\end{aligned}$
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r(7)$ 
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r(7)&=\frac{(7-2)(7+3)}{(7+5)(7-7)}\\
+		& =\frac{50}{0}
+	\end{aligned}$
+	
+	$r(7)$ is undefined.
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r(-3)$ 
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r(-3)&=\frac{(-3-2)(-3+3)}{(-3+5)(-3-7)}\\
+		&=\frac{0}{-20}\\
+		&=0
+	\end{aligned}$
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r(-5)$ 
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r(-5)&=\frac{(-5-2)(-5+3)}{(-5+5)(-5-7)}\\
+		&=\frac{14}{0}
+	\end{aligned}$
+	
+	$r(-5)$ is undefined.
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r\left( \frac{1}{2} \right)$
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r\left( \frac{1}{2} \right)& = \frac{\left( \frac{1}{2}-2 \right)\left( \frac{1}{2}+3 \right)}{\left( \frac{1}{2}+5 \right)\left( \frac{1}{2}-7 \right)}\\
+		&=\frac{-\frac{3}{2}\cdot\frac{7}{2}}{\frac{11}{2}\left( -\frac{13}{2} \right)}\\
+		&=\frac{-\frac{21}{4}}{-\frac{143}{4}}\\
+		&=\frac{37}{143}
+	\end{aligned}$
+\end{shortsolution}
+\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Holes or asymptotes?]
+State the domain of each of the following rational functions. Identify
+any holes or asymptotes.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$f(x)=\dfrac{12}{x-2}$ 
+		\begin{shortsolution}
+			$f$ has a vertical asymptote at $2$; the domain of $f$ is $(-\infty,2)\cup (2,\infty)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$g(x)=\dfrac{x^2+x}{(x+1)(x-2)}$ 
+		\begin{shortsolution}
+			$g$ has a vertical asymptote at $2$, and a hole at $-1$; the domain of $g$ is $(-\infty,-1)\cup(-1,2)\cup(2,\infty)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$h(x)=\dfrac{x^2+5x+4}{x^2+x-12}$ 
+		\begin{shortsolution}
+			$h$ has a vertical asymptote at $3$, and a whole at $-4$; the domain of $h$ is $(-\infty,-4)\cup(-4,3)\cup(3,\infty)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$k(z)=\dfrac{z+2}{2z-3}$ 
+		\begin{shortsolution}
+			$k$ has a vertical asymptote at $\dfrac{3}{2}$; the domain of $k$ is $\left( -\infty,\dfrac{3}{2} \right)\cup\left( \dfrac{3}{2},\infty \right)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$l(w)=\dfrac{w}{w^2+1}$ 
+		\begin{shortsolution}
+			$l$ does not have any vertical asymptotes nor holes; the domain of $w$ is $(-\infty,\infty)$. 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$m(t)=\dfrac{14}{13-t^2}$ 
+		\begin{shortsolution}
+			$m$ has vertical asymptotes at $\pm\sqrt{13}$; the domain of $m$ is $(-\infty,\sqrt{13})\cup(-\sqrt{13},\sqrt{13})\cup(\sqrt{13},\infty)$.
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a graph]
+Consider the rational functions graphed in \cref{rat:fig:findformula}. Find 
+the vertical asymptotes for each function, together with any zeros, and 
+give a possible formula for each. 
+\begin{shortsolution}
+	\begin{itemize}
+		\item \Vref{rat:fig:formula1}: possible formula is $r(x)=\dfrac{1}{x+5}$ 
+		\item \Vref{rat:fig:formula2}: possible formula is $r(x)=\dfrac{(x+3)}{(x-5)}$
+		\item \Vref{rat:fig:formula3}: possible formula is $r(x)=\dfrac{1}{(x-4)(x+3)}$.
+	\end{itemize}
+\end{shortsolution}
+\end{problem}
+
+\begin{figure}[!htb]
+	\begin{widepage}
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=1/(x+4);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-6,ymax=6,
+			   xtick={-8,-6,...,8},
+			   minor ytick={-4,-3,...,4},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-4.16667,samples=50]{f};
+			   \addplot[pccplot] expression[domain=-3.83333:10,samples=50]{f};
+			   \addplot[asymptote,domain=-6:6]({-4},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:formula1}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x+3)/(x-5);}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-6,ymax=6,
+			   xtick={-8,-6,...,8},
+			   minor ytick={-4,-3,...,4},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:3.85714]{f};
+			   \addplot[pccplot] expression[domain=6.6:10]{f};
+			   \addplot[soldot] coordinates{(-3,0)};
+			   \addplot[asymptote,domain=-6:6]({5},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{1});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:formula2}
+	\end{subfigure}
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}[/pgf/declare function={f=1/((x-4)*(x+3));}]
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-3,ymax=3,
+			   xtick={-8,-6,...,8},
+			   minor ytick={-4,-3,...,4},
+			   grid=both,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-3.0473]{f};
+			   \addplot[pccplot] expression[domain=-2.95205:3.95205]{f};
+			   \addplot[pccplot] expression[domain=4.0473:10]{f};
+			   \addplot[asymptote,domain=-3:3]({-3},{x});
+			   \addplot[asymptote,domain=-3:3]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{0});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:formula3}
+	\end{subfigure}
+	\caption{}
+	\label{rat:fig:findformula}
+	\end{widepage}
+\end{figure}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a description]
+In each of the following problems, give a formula of a rational 
+function that has the listed properties.
+\begin{subproblem}
+	Vertical asymptote at $2$.
+	\begin{shortsolution}
+		Possible option: $r(x)=\dfrac{1}{x-2}$. Note that we could multiply the 
+		numerator or denominator by any real number and still have the desired properties.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Vertical asymptote at $5$.
+	\begin{shortsolution}
+		Possible option: $r(x)=\dfrac{1}{x-5}$. Note that we could multiply the 
+		numerator or denominator by any real number and still have the desired properties.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Vertical asymptote at $-2$, and zero at $6$.
+	\begin{shortsolution}
+		Possible option: $r(x)=\dfrac{x-6}{x+2}$. Note that we could multiply the 
+		numerator or denominator by any real number and still have the desired properties.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Zeros at $2$ and $-5$ and vertical asymptotes at $1$ and $-7$.
+	\begin{shortsolution}
+		Possible option: $r(x)=\dfrac{(x-2)(x+5)}{(x-1)(x+7)}$. Note that we could multiply the 
+		numerator or denominator by any real number and still have the desired properties.
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Given formula, find horizontal asymptotes]
+Each of the following functions has a horizontal asymptote. Write the equation
+of the horizontal asymptote for each function.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$f(x) = \dfrac{1}{x}$
+		\begin{shortsolution}
+			$y=0$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$g(x) = \dfrac{2x+3}{x}$
+		\begin{shortsolution}
+			$y=2$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$h(x) = \dfrac{x^2+2x}{x^2+3}$
+		\begin{shortsolution}
+			$y=1$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$k(x) = \dfrac{x^2+7}{x}$
+		\begin{shortsolution}
+			$y=1$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$l(x)=\dfrac{3x-2}{5x+8}$ 
+		\begin{shortsolution}
+			$y=\dfrac{3}{5}$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$m(x)=\dfrac{3x-2}{5x^2+8}$ 
+		\begin{shortsolution}
+			$y=0$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$n(x)=\dfrac{(6x+1)(x-7)}{(11x-8)(x-5)}$ 
+		\begin{shortsolution}
+			$y=\dfrac{6}{11}$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$p(x)=\dfrac{19x^3}{5-x^4}$ 
+		\begin{shortsolution}
+			$y=0$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$q(x)=\dfrac{14x^2+x}{1-7x^2}$ 
+		\begin{shortsolution}
+			$y=-2$
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{problem}[Given horizontal asymptotes, find formula]
+In each of the following problems, give a formula for a function that 
+has the given horizontal asymptote. Note that there may be more than one option.
+\begin{multicols}{4}
+	\begin{subproblem}
+		$y=7$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{7(x-2)}{x+1}$. Note that there
+			are other options, provided that the degree of the numerator is the same as the degree
+			of the denominator, and that the ratio of the leading 
+			coefficients is $7$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=-1$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{5-x^2}{x^2+10}$. Note that there
+			are other options, provided that the degree of the numerator is the same as the degree
+			of the denominator, and that the ratio of the leading 
+			coefficients is $10$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=53$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{53x^3}{x^3+4x^2-7}$. Note that there
+			are other options, provided that the degree of the numerator is the same as the degree
+			of the denominator, and that the ratio of the leading 
+			coefficients is $53$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=-17$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{34(x+2)}{7-2x}$. Note that there
+			are other options, provided that the degree of the numerator is the same as the degree
+			of the denominator, and that the ratio of the leading 
+			coefficients is $-17$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=\dfrac{3}{2}$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{3x+4}{2(x+1)}$. Note that there
+			are other options, provided that the degree of the numerator is the same as the degree
+			of the denominator, and that the ratio of the leading 
+			coefficients is $\dfrac{3}{2}$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=0$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{4}{x}$. Note that there
+			are other options, provided that the degree of the numerator is less than the degree
+			of the denominator.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=-1$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{10x}{5-10x}$. Note that there
+			are other options, provided that the degree of the numerator is the same as the degree
+			of the denominator, and that the ratio of the leading 
+			coefficients is $-1$.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=2$ 
+		\begin{shortsolution}
+			Possible option: $f(x)=\dfrac{8x-3}{4x+1}$. Note that there
+			are other options, provided that the degree of the numerator is the same as the degree
+			of the denominator, and that the ratio of the leading 
+			coefficients is $2$.
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a description]
+In each of the following problems, give a formula for a function that 
+has the prescribed properties. Note that there may be more than one option.
+\begin{subproblem}
+	$f(x)\rightarrow 3$ as $x\rightarrow\pm\infty$.
+	\begin{shortsolution}
+		Possible option: $f(x)=\dfrac{3(x-2)}{x+7}$. Note that 
+		the zero and asymptote of $f$ could be changed, and $f$ would still have the desired properties.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$r(x)\rightarrow -4$ as $x\rightarrow\pm\infty$.
+	\begin{shortsolution}
+		Possible option: $r(x)=\dfrac{-4(x-2)}{x+7}$. Note that 
+		the zero and asymptote of $r$ could be changed, and $r$ would still have the desired properties.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$k(x)\rightarrow 2$ as $x\rightarrow\pm\infty$, and $k$ has vertical asymptotes at $-3$ and $5$.
+	\begin{shortsolution}
+		Possible option: $k(x)=\dfrac{2x^2}{(x+3)(x-5)}$. Note that the denominator 
+		must have the given factors; the numerator could be any degree $2$ polynomial, provided the 
+		leading coefficient is $2$.
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   Feb 2011
+%===================================
+\begin{problem}
+Let $r$ be the rational function that has
+\[
+	r(x) = \frac{(x+2)(x-1)}{(x+3)(x-4)}
+\]  
+Each of the following questions are in relation to this function.
+\begin{subproblem}
+	What is the vertical intercept of this function? State your answer as an 
+	ordered pair. \index{rational functions!vertical intercept}
+	\begin{shortsolution}
+		$\left(0,\frac{1}{6}\right)$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}\label{rat:prob:rational}
+	What values of $x$ make the denominator equal to $0$?
+	\begin{shortsolution}
+		$-3,4$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Use your answer to \cref{rat:prob:rational} to write the domain of the function in 
+	both interval, and set builder notation. %\index{rational functions!domain}\index{domain!rational functions}
+	\begin{shortsolution}
+		Interval notation: $(-\infty,-3)\cup (-3,4)\cup (4,\infty)$. 
+		Set builder: $\{x|x\ne -3, \mathrm{and}\, x\ne 4\}$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	What are the vertical asymptotes of the function? State your answers in 
+	the form $x=$
+	\begin{shortsolution}
+		$x=-3$ and $x=4$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}\label{rat:prob:zeroes}
+	What values of $x$ make the numerator equal to $0$?
+	\begin{shortsolution}
+		$-2,1$ 
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Use your answer to \cref{rat:prob:zeroes} to write the horizontal intercepts of 
+	$r$ as ordered pairs.
+	\begin{shortsolution}
+		$(-2,0)$ and $(1,0)$
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Holes]
+\pccname{Josh} and \pccname{Pedro} are discussing the function
+\[
+	r(x)=\frac{x^2-1}{(x+3)(x-1)}
+\]
+\begin{subproblem}
+	What is the domain of $r$?
+	\begin{shortsolution}
+		The domain of $r$ is $(-\infty,-3)\cup(-3,1)\cup(1,\infty)$.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Josh notices that the numerator can be factored- can you see how?
+	\begin{shortsolution}
+		$(x^2-1)=(x-1)(x+1)$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Pedro asks, `Doesn't that just mean that 
+	\[
+		r(x)=\frac{x+1}{x+3}
+	\]
+	for all values of $x$?' Josh says, `Nearly\ldots but not for all values of $x$'. 
+	What does Josh mean?
+	\begin{shortsolution}
+		$r(x)=\dfrac{x+1}{x+3}$ provided that $x\ne -1$. 
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Where does $r$ have vertical asymptotes, and where does it have holes?
+	\begin{shortsolution}
+		The function $r$ has a vertical asymptote at $-3$, and a hole at $1$.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Sketch a graph of $r$.
+	\begin{shortsolution}
+		A graph of $r$ is shown below.
+		
+		\begin{tikzpicture}
+			\begin{axis}[
+			   framed,
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-8,-6,...,8},
+			   grid=both,
+			   width=\solutionfigurewidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-3.25]{(x+1)/(x+3)};
+			   \addplot[pccplot] expression[domain=-2.75:10]{(x+1)/(x+3)};
+			   \addplot[asymptote,domain=-10:10]({-3},{x});
+			   \addplot[holdot]coordinates{(1,0.5)};
+			\end{axis}
+		\end{tikzpicture}
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+
+%===================================
+%   Author: Hughes
+%   Date:   July 2012
+%===================================
+\begin{problem}[Function algebra]
+Let $r$ and $s$ be the rational functions that have formulas
+\[
+	r(x)=\frac{2-x}{x+3}, \qquad s(x)=\frac{x^2}{x-4}
+\]
+Evaluate each of the following (if possible).
+\begin{multicols}{4}
+	\begin{subproblem}
+		$(r+s)(5)$ 
+		\begin{shortsolution}
+			$\frac{197}{8}$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(r-s)(3)$
+		\begin{shortsolution}
+			$\frac{53}{6}$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(r\cdot s)(4)$ 
+		\begin{shortsolution}
+			Undefined.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$\left( \frac{r}{s} \right)(1)$ 
+		\begin{shortsolution}
+			$-\frac{3}{4}$ 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+
+
+%===================================
+%   Author: Hughes
+%   Date:   July 2012
+%===================================
+\begin{problem}[Transformations: given the transformation, find the formula]
+Let $r$ be the rational function that has formula.
+\[
+	r(x)=\frac{x+5}{2x-3}
+\]
+In each of the following problems apply the given transformation to the function $r$ and 
+write a formula for the transformed version of $r$.
+\begin{multicols}{2}
+	\begin{subproblem}
+		Shift $r$ to the right by $3$ units. 
+		\begin{shortsolution}
+			$r(x-3)=\frac{x+2}{2x-9}$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $r$ to the left by $4$ units. 
+		\begin{shortsolution}
+			$r(x+4)=\frac{x+9}{2x+5}$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $r$ up by $\pi$ units. 
+		\begin{shortsolution}
+			$r(x)+\pi=\frac{x+5}{2x-3}+\pi$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $r$ down by $17$ units. 
+		\begin{shortsolution}
+			$r(x)-17=\frac{x+5}{2x-3}-17$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Reflect $r$ over the horizontal axis.
+		\begin{shortsolution}
+			$-r(x)=-\frac{x+5}{2x-3}$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Reflect $r$ over the vertical axis.
+		\begin{shortsolution}
+			$r(-x)=\frac{x-5}{2x+3}$
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a table]\label{rat:prob:findformula}
+\Crefrange{rat:tab:findformular}{rat:tab:findformulau} show values of rational functions $r$, $q$, $s$, 
+and $t$. Assume that any values marked with an X are undefined.
+
+\begin{table}[!htb]
+	\begin{widepage}
+	\centering
+	\caption{Tables for \cref{rat:prob:findformula}}
+	\label{rat:tab:findformula}
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=r(x)$}
+		\label{rat:tab:findformular}
+		\begin{tabular}{rr}
+			\beforeheading
+			$x$  & $y$                \\ \afterheading
+			$-4$ & $\nicefrac{7}{2}$  \\\normalline
+			$-3$ & $-18$              \\\normalline
+			$-2$ & X                  \\\normalline
+			$-1$ & $-4$               \\\normalline
+			$0$  & $\nicefrac{-3}{2}$ \\\normalline
+			$1$  & $\nicefrac{-2}{3}$ \\\normalline
+			$2$  & $\nicefrac{-1}{4}$ \\\normalline
+			$3$  & $0$                \\\normalline
+			$4$  & $\nicefrac{1}{6}$  \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=s(x)$}
+		\label{rat:tab:findformulas}
+		\begin{tabular}{rr}
+			\beforeheading
+			$x$  & $y$                 \\ \afterheading
+			$-4$ & $\nicefrac{-2}{21}$ \\\normalline
+			$-3$ & $\nicefrac{-1}{12}$ \\\normalline
+			$-2$ & $0$                 \\\normalline
+			$-1$ & X                   \\\normalline
+			$0$  & $\nicefrac{-2}{3}$  \\\normalline
+			$1$  & $\nicefrac{-3}{4}$  \\\normalline
+			$2$  & $\nicefrac{-4}{3}$  \\\normalline
+			$3$  & X                   \\\normalline
+			$4$  & $\nicefrac{6}{5}$   \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=t(x)$}
+		\label{rat:tab:findformulat}
+		\begin{tabular}{rr}
+			\beforeheading
+			$x$  & $y$               \\ \afterheading
+			$-4$ & $\nicefrac{3}{5}$ \\\normalline
+			$-3$ & $0$               \\\normalline
+			$-2$ & X                 \\\normalline
+			$-1$ & $3$               \\\normalline
+			$0$  & $3$               \\\normalline
+			$1$  & X                 \\\normalline
+			$2$  & $0$               \\\normalline
+			$3$  & $\nicefrac{3}{5}$ \\\normalline
+			$4$  & $\nicefrac{7}{9}$ \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=u(x)$}
+		\label{rat:tab:findformulau}
+		\begin{tabular}{rr}
+			\beforeheading
+			$x$  & $y$                \\ \afterheading
+			$-4$ & $\nicefrac{16}{7}$ \\\normalline
+			$-3$ & X                  \\\normalline
+			$-2$ & $-\nicefrac{4}{5}$ \\\normalline
+			$-1$ & $-\nicefrac{1}{8}$ \\\normalline
+			$0$  & $0$                \\\normalline
+			$1$  & $-\nicefrac{1}{8}$ \\\normalline
+			$2$  & $-\nicefrac{4}{5}$ \\\normalline
+			$3$  & X                  \\\normalline
+			$4$  & $\nicefrac{16}{7}$ \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\end{widepage}
+\end{table}
+\begin{subproblem}
+	Given that the formula for $r(x)$ has the form $r(x)=\dfrac{x-A}{x-B}$, use \cref{rat:tab:findformular}
+	to find values of $A$ and $B$.
+	\begin{shortsolution}
+		$A=3$ and $B=-2$, so $r(x)=\dfrac{x-3}{x+2}$.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Check your formula by computing $r(x)$ at the values specified in the table.
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		r(-4)&= \frac{-4-3}{-4+2}\\
+		&= \frac{7}{2}\\
+	\end{aligned}$
+	
+	$r(-3)=\ldots$ etc
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	The function $s$ in \cref{rat:tab:findformulas} has two vertical asymptotes and one zero. 
+	Can you find a formula for $s(x)$?
+	\begin{shortsolution}
+		$s(x)=\dfrac{x+2}{(x-3)(x+1)}$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Check your formula by computing $s(x)$ at the values specified in the table.
+	\begin{shortsolution}
+		$\begin{aligned}[t]
+		s(-4)&=\frac{-4+2}{(-4-3)(-4+1)}\\
+		&=-\frac{2}{21}
+	\end{aligned}$
+	
+	$s(-3)=\ldots$ etc
+\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Given that the formula for $t(x)$ has  the form $t(x)=\dfrac{(x-A)(x-B)}{(x-C)(x-D)}$, use \cref{rat:tab:findformulat} to find the 
+	values of $A$, $B$, $C$, and $D$; hence write a formula for $t(x)$.
+	\begin{shortsolution}
+		$t(x)=\dfrac{(x+3)(x-2)}{(x+2)(x+1)}$ 
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Given that the formula for $u(x)$ has  the form $u(x)=\dfrac{(x-A)^2}{(x-B)(x-C)}$, use \cref{rat:tab:findformulau} to find the 
+	values of $A$, $B$, and $C$; hence write a formula for $u(x)$.
+	\begin{shortsolution}
+		$u(x)=\dfrac{x^2}{(x+3)(x-3)}$ 
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+\end{exercises}
+
+\section{Graphing rational functions (horizontal asymptotes)}
+\reformatstepslist{R} % the steps list should be R1, R2, \ldots
+We studied rational functions in the previous section, but were 
+not asked to graph them; in this section we will demonstrate the 
+steps to be followed in order to sketch graphs of the functions. 
+
+Remember from \vref{rat:def:function} that rational functions have 
+the form
+\[
+	r(x)=\frac{p(x)}{q(x)}
+\]
+In this section we will restrict attention to the case when 
+\[
+	\text{degree of }p\leq \text{degree of }q
+\]
+Note that this necessarily means that each function that we consider
+in this section \emph{will have a horizontal asymptote} (see \vref{rat:def:longrun}).
+The cases in which the degree of $p$ is greater than the degree of $q$ 
+is covered in the next section.
+
+Before we begin, it is important to remember the following:
+\begin{itemize}
+	\item Our sketches will give a good representation of the overall 
+	shape of the graph, but until we have the tools of calculus (from MTH 251)
+	we can not find local minimums, local maximums, and inflection points algebraically. This
+	means that we will make our best guess as to where these points are.
+	\item We will not concern ourselves too much with the vertical scale (because of 
+	our previous point)| we will, however, mark the vertical intercept (assuming there is one), 
+	and any horizontal asymptotes.
+\end{itemize}
+\begin{pccspecialcomment}[Steps to follow when sketching rational functions]\label{rat:def:stepsforsketch}
+	\begin{steps}
+		\item \label{rat:step:first} Find all vertical asymptotes and holes, and mark them on the 
+		graph using dashed vertical lines and open circles $\circ$ respectively.
+		\item Find any intercepts, and mark them using solid circles $\bullet$;
+		determine if the curve cuts the axis, or bounces off it at each zero.
+		\item Determine the behavior of the function around each asymptote| does
+		it behave like $\frac{1}{x}$ or $\frac{1}{x^2}$?
+		\item \label{rat:step:penultimate} Determine the long-run behavior of the function, and mark the horizontal 
+		asymptote using a dashed horizontal line.
+		\item \label{rat:step:last}  Deduce the overall shape of the curve, and sketch it. If there isn't
+		enough information from the previous steps, then construct a table of values
+		including sample points from each branch.
+	\end{steps}
+	Remember that until we have the tools of calculus, we won't be able to 
+	find the exact coordinates of local minimums, local maximums, and points
+	of inflection.
+\end{pccspecialcomment}
+
+The examples that follow show how \crefrange{rat:step:first}{rat:step:last} can be
+applied to a variety of different rational functions.
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{rat:ex:1overxminus2p2}
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $r$ 
+that has formula
+\[
+	r(x)=\frac{1}{x-2}
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $r$ has a vertical asymptote at $2$; $r$ does not have any holes. The curve of 
+	$r$ will have $2$ branches.
+	\item $r$ does not have any zeros since the numerator is never equal to $0$. The
+	vertical intercept of $r$ is $\left( 0,-\frac{1}{2} \right)$.
+	\item $r$ behaves like $\frac{1}{x}$ around its vertical asymptote since $(x-2)$
+	is raised to the power $1$.
+	\item Since the degree of the numerator is less than the degree of the denominator, 
+	according to \vref{rat:def:longrun} the horizontal asymptote of $r$ has equation $y=0$.
+	\item We put the details we have obtained so far on \cref{rat:fig:1overxminus2p1}. Notice 
+	that there is only one way to complete the graph, which we have done in \cref{rat:fig:1overxminus2p2}.
+\end{steps}
+\end{pccsolution}
+\end{pccexample}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-5,ymax=5,
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-5:5]({2},{x});
+			   \addplot[asymptote,domain=-5:5]({x},{0});
+			   \addplot[soldot] coordinates{(0,-0.5)}node[axisnode,anchor=north east]{$\left( 0,-\frac{1}{2} \right)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:1overxminus2p1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=1/(x-2);}]
+			\begin{axis}[
+			   xmin=-5,xmax=5,
+			   ymin=-5,ymax=5,
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-5:1.8,samples=50]{f};
+			   \addplot[pccplot] expression[domain=2.2:5]{f};
+			   \addplot[asymptote,domain=-5:5]({2},{x});
+			   \addplot[asymptote,domain=-5:5]({x},{0});
+			   \addplot[soldot] coordinates{(0,-0.5)}node[axisnode,anchor=north east]{$\left( 0,-\frac{1}{2} \right)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:1overxminus2p2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{1}{x-2}$}
+\end{figure}
+
+The function $r$ in \cref{rat:ex:1overxminus2p2} has a horizontal asymptote which has equation $y=0$.
+This asymptote lies on the horizontal axis, and you might (understandably) find it hard 
+to distinguish between the two lines (\cref{rat:fig:1overxminus2p2}). When faced
+with such a situation, it is perfectly acceptable to draw the horizontal axis 
+as a dashed line| just make sure to label it correctly. We will demonstrate this 
+in the next example.
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{rat:ex:1overxp1}
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $v$ 
+that has formula
+\[
+	v(x)=\frac{10}{x}
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $v$ has a vertical asymptote at $0$. $v$ does not have
+	any holes. The curve of $v$ will have $2$ branches.
+	\item $v$ does not have any zeros (since $10\ne 0$). Furthermore, $v$ 
+	does not have a vertical intercept since $v(0)$ is undefined.
+	\item $v$ behaves like $\frac{1}{x}$ around its vertical asymptote.
+	\item $v$ has a horizontal asymptote with equation $y=0$.
+	\item We put the details we have obtained so far in \cref{rat:fig:1overxp1}. 
+	We do not have enough information to sketch $v$ yet (because $v$ does 
+	not have any intercepts), so let's pick a sample 
+	point in either of the $2$ branches| it doesn't matter where our sample point 
+	is, because we know what the overall shape will be. Let's compute $v(2)$
+	\begin{align*}
+		v(2) & =\dfrac{10}{2} \\
+		     & = 5            
+	\end{align*}
+	We therefore mark the point $(2,5)$ on \cref{rat:fig:1overxp2}, and then complete the sketch using 
+	the details we found in the previous steps.
+\end{steps}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-5,5},
+			   ytick={-5,5},
+			   axis line style={color=white},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,<->,domain=-10:10]({0},{x});
+			   \addplot[asymptote,<->,domain=-10:10]({x},{0});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:1overxp1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=10/x;}]
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-5,5},
+			   ytick={-5,5},
+			   axis line style={color=white},
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:-1]{f};
+			   \addplot[pccplot] expression[domain=1:10]{f};
+			   \addplot[soldot] coordinates{(2,5)}node[axisnode,anchor=south west]{$(2,5)$};
+			   \addplot[asymptote,<->,domain=-10:10]({0},{x});
+			   \addplot[asymptote,<->,domain=-10:10]({x},{0});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:1overxp2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{10}{x}$}
+\end{figure}
+\end{pccsolution}
+\end{pccexample}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{rat:ex:asympandholep1}
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $u$ 
+that has formula
+\[
+	u(x)=\frac{-4(x^2-9)}{x^2-8x+15}
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item We begin by factoring both the numerator and denominator of $u$ to help
+	us find any vertical asymptotes or holes
+	\begin{align*}
+		u(x) & =\frac{-4(x^2-9)}{x^2-8x+15}     \\
+		     & =\frac{-4(x+3)(x-3)}{(x-5)(x-3)} \\
+		     & =\frac{-4(x+3)}{x-5}             
+	\end{align*}
+	provided that $x\ne 3$. Therefore $u$ has a vertical asymptote at $5$ and 
+	a hole at $3$. The curve of $u$ has $2$ branches.
+	\item $u$ has a simple zero at $-3$. The vertical intercept of $u$ is $\left( 0,\frac{12}{5} \right)$.
+	\item $u$ behaves like $\frac{1}{x}$ around its vertical asymptote at $4$.
+	\item Using \vref{rat:def:longrun} the equation of the horizontal asymptote of $u$ is $y=-4$.
+	\item We put the details we have obtained so far on \cref{rat:fig:1overxminus2p1}. Notice 
+	that there is only one way to complete the graph, which we have done in \cref{rat:fig:1overxminus2p2}.
+\end{steps}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-20,ymax=20,
+			   xtick={-8,-6,...,8},
+			   ytick={-10,10},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-20:20]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{-4});
+			   \addplot[soldot] coordinates{(-3,0)(0,2.4)}node[axisnode,anchor=south east]{$\left( 0,\frac{12}{5} \right)$};
+			   \addplot[holdot] coordinates{(3,12)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:asympandholep1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=-4*(x+3)/(x-5);}]
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-20,ymax=20,
+			   xtick={-8,-6,...,8},
+			   ytick={-10,10},
+			   width=\textwidth,
+			   ]
+			   \addplot[pccplot] expression[domain=-10:3.6666,samples=50]{f};
+			   \addplot[pccplot] expression[domain=7:10]{f};
+			   \addplot[asymptote,domain=-20:20]({5},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{-4});
+			   \addplot[soldot] coordinates{(-3,0)(0,2.4)}node[axisnode,anchor=south east]{$\left( 0,\frac{12}{5} \right)$};
+			   \addplot[holdot] coordinates{(3,12)};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:asympandholep2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{-4(x+3)}{x-5}$}
+\end{figure}
+\end{pccsolution}
+\end{pccexample}
+
+\Cref{rat:ex:1overxminus2p2,rat:ex:1overxp1,rat:ex:asympandholep1} have focused on functions
+that only have one vertical asymptote; the remaining examples in this section
+concern functions that have more than one vertical asymptote. We will demonstrate
+that \crefrange{rat:step:first}{rat:step:last} still apply.
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{rat:ex:sketchtwoasymp}
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $w$ 
+that has formula
+\[
+	w(x)=\frac{2(x+3)(x-5)}{(x+5)(x-4)}
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $w$ has vertical asymptotes at $-5$ and $4$. $w$ does not have 
+	any holes. The curve of $w$ will have $3$ branches.
+	\item $w$ has simple zeros at $-3$ and $5$. The vertical intercept of $w$ 
+	is $\left( 0,\frac{3}{2} \right)$.
+	\item $w$ behaves like $\frac{1}{x}$ around both of its vertical 
+	asymptotes.
+	\item The degree of the numerator of $w$ is $2$ and the degree of the 
+	denominator of $w$ is also $2$. Using the ratio of the leading coefficients
+	of the numerator and denominator, we say that $w$ has a horizontal 
+	asymptote with equation $y=\frac{2}{1}=2$.
+	\item We put the details we have obtained so far on \cref{rat:fig:sketchtwoasymptp1}. 
+	
+	The function $w$ is a little more complicated than the functions that 
+	we have considered in the previous examples because the curve has $3$
+	branches. When graphing such functions, it is generally a good idea to start with the branch
+	for which you have the most information| in this case, that is the \emph{middle} branch
+	on the interval $(-5,4)$.
+	
+	Once we have drawn the middle branch, there is only one way to complete the graph 
+	(because of our observations about the behavior of $w$ around its vertical asymptotes), 
+	which we have done in \cref{rat:fig:sketchtwoasymptp2}.
+\end{steps}
+\end{pccsolution}
+\end{pccexample}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-5},{x});
+			   \addplot[asymptote,domain=-10:10]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{2});
+			   \addplot[soldot] coordinates{(-3,0)(5,0)};
+			   \addplot[soldot] coordinates{(0,1.5)}node[axisnode,anchor=north west]{$\left( 0,\frac{3}{2} \right)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:sketchtwoasymptp1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=2*(x+3)*(x-5)/( (x+5)*(x-4));}]
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-5},{x});
+			   \addplot[asymptote,domain=-10:10]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{2});
+			   \addplot[soldot] coordinates{(-3,0)(5,0)};
+			   \addplot[soldot] coordinates{(0,1.5)}node[axisnode,anchor=north west]{$\left( 0,\frac{3}{2} \right)$};
+			   \addplot[pccplot] expression[domain=-10:-5.56708]{f};
+			   \addplot[pccplot] expression[domain=-4.63511:3.81708]{f};
+			   \addplot[pccplot] expression[domain=4.13511:10]{f};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:sketchtwoasymptp2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{2(x+3)(x-5)}{(x+5)(x-4)}$}
+\end{figure}
+
+The rational functions that we have considered so far have had simple
+factors in the denominator; each function has behaved like $\frac{1}{x}$ 
+around each of its vertical asymptotes. \Cref{rat:ex:2asympnozeros,rat:ex:2squaredasymp}
+consider functions that have a repeated factor in the denominator.
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{rat:ex:2asympnozeros}
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $f$ 
+that has formula
+\[
+	f(x)=\frac{100}{(x+5)(x-4)^2}
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $f$ has vertical asymptotes at $-5$ and $4$. $f$ does not have 
+	any holes. The curve of $f$ will have $3$ branches.
+	\item $f$ does not have any zeros (since $100\ne 0$). The vertical intercept of $f$ 
+	is $\left( 0,\frac{5}{4} \right)$.
+	\item $f$ behaves like $\frac{1}{x}$ around $-5$ and behaves like $\frac{1}{x^2}$
+	around $4$.
+	\item The degree of the numerator of $f$ is $0$ and the degree of the 
+	denominator of $f$ is $2$. $f$ has a horizontal asymptote with 
+	equation $y=0$.
+	\item We put the details we have obtained so far on \cref{rat:fig:2asympnozerosp1}. 
+	
+	The function $f$ is similar to the function $w$ that we considered in \cref{rat:ex:sketchtwoasymp}|
+	it has two vertical asymptotes and $3$ branches, but in contrast to $w$ it does not have any zeros.
+	
+	We sketch $f$ in \cref{rat:fig:2asympnozerosp2}, using the middle branch as our guide
+	because we have the most information about the function on the interval $(-5,4)$.
+	
+	Once we have drawn the middle branch, there is only one way to complete the graph 
+	because of our observations about the behavior of $f$ around its vertical asymptotes (it behaves like $\frac{1}{x}$), 
+	which we have done in \cref{rat:fig:2asympnozerosp2}.
+	
+	Note that we are not yet able to find the local minimum of $f$ algebraically on the interval $(-5,4)$, 
+	so we make a reasonable guess as to where it is| we can be confident that it is above the horizontal axis
+	since $f$ has no zeros. You may think that this is unsatisfactory, but once we have the tools of calculus, we will 
+	be able to find local minimums more precisely.
+\end{steps}
+\end{pccsolution}
+\end{pccexample}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-5},{x});
+			   \addplot[asymptote,domain=-10:10]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{0});
+			   \addplot[soldot] coordinates{(0,1.25)}node[axisnode,anchor=south east]{$\left( 0,\frac{5}{4} \right)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:2asympnozerosp1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=100/( (x+5)*(x-4)^2);}]
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-5},{x});
+			   \addplot[asymptote,domain=-10:10]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{0});
+			   \addplot[soldot] coordinates{(0,1.25)}node[axisnode,anchor=south east]{$\left( 0,\frac{5}{4} \right)$};
+			   \addplot[pccplot] expression[domain=-10:-5.12022]{f};
+			   \addplot[pccplot] expression[domain=-4.87298:2.87298,samples=50]{f};
+			   \addplot[pccplot] expression[domain=5:10]{f};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:2asympnozerosp2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{100}{(x+5)(x-4)^2}$}
+\end{figure}
+
+%===================================
+%   Author: Hughes
+%   Date:   May 2012
+%===================================
+\begin{pccexample}\label{rat:ex:2squaredasymp}
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $g$ 
+that has formula
+\[
+	g(x)=\frac{50(2-x)}{(x+3)^2(x-5)^2}
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $g$ has vertical asymptotes at $-3$ and $5$. $g$ does 
+	not have any holes. The curve of $g$ will have $3$ branches.
+	\item $g$ has a simple zero at $2$. The vertical intercept of $g$ is
+	$\left( 0,\frac{4}{9} \right)$.
+	\item $g$ behaves like $\frac{1}{x^2}$ around both of its 
+	vertical asymptotes.
+	\item The degree of the numerator of $g$ is $1$ and the degree of the denominator
+	of $g$ is $4$. Using \vref{rat:def:longrun}, we calculate that 
+	the horizontal asymptote of $g$ has equation $y=0$.
+	\item The details that we have found so far have been drawn in 
+	\cref{rat:fig:2squaredasymp1}. The function $g$ is similar to the functions 
+	we considered in \cref{rat:ex:sketchtwoasymp,rat:ex:2asympnozeros} because 
+	it has $2$ vertical asymptotes and $3$ branches. 
+	
+	We sketch $g$ using the middle branch as our guide because we have the most information
+	about $g$ on the interval $(-3,5)$. Note that there is no other way to draw this branch
+	without introducing other zeros which $g$ does not have.
+	
+	Once we have drawn the middle branch, there is only one way to complete the graph 
+	because of our observations about the behavior of $g$ around its vertical asymptotes| it 
+	behaves like $\frac{1}{x^2}$.
+	
+\end{steps}
+\end{pccsolution}
+\end{pccexample}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-3},{x});
+			   \addplot[asymptote,domain=-10:10]({5},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{0});
+			   \addplot[soldot] coordinates{(2,0)(0,4/9)}node[axisnode,anchor=south west]{$\left( 0,\frac{4}{9} \right)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:2squaredasymp1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=50*(2-x)/( (x+3)^2*(x-5)^2);}]
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   ytick={-5,5},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-3},{x});
+			   \addplot[asymptote,domain=-10:10]({5},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{0});
+			   \addplot[soldot] coordinates{(2,0)(0,4/9)}node[axisnode,anchor=south west]{$\left( 0,\frac{4}{9} \right)$};
+			   \addplot[pccplot] expression[domain=-10:-3.61504]{f};
+			   \addplot[pccplot] expression[domain=-2.3657:4.52773]{f};
+			   \addplot[pccplot] expression[domain=5.49205:10]{f};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:2squaredasymp2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{50(2-x)}{(x+3)^2(x-5)^2}$}
+\end{figure}
+
+Each of the rational functions that we have considered so far has had either 
+a \emph{simple} zero, or no zeros at all. Remember from our work on polynomial 
+functions, and particularly \vref{poly:def:multzero}, that a \emph{repeated} zero
+corresponds to the curve of the function behaving differently at the zero
+when compared to how the curve behaves at a simple zero. \Cref{rat:ex:doublezero} details a 
+function that has a non-simple zero.
+
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{pccexample}\label{rat:ex:doublezero}
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $g$ 
+that has formula
+\[
+	h(x)=\frac{(x-3)^2}{(x+4)(x-6)}
+\]
+\begin{pccsolution}
+\begin{steps}
+	\item $h$ has vertical asymptotes at $-4$ and $6$. $h$ does 
+	not have any holes. The curve of $h$ will have $3$ branches.
+	\item $h$ has a zero at $3$ that has \emph{multiplicity $2$}. 
+	The vertical intercept of $h$ is
+	$\left( 0,-\frac{3}{8} \right)$.
+	\item $h$ behaves like $\frac{1}{x}$ around both of its 
+	vertical asymptotes.
+	\item The degree of the numerator of $h$ is $2$ and the degree of the denominator
+	of $h$ is $2$. Using \vref{rat:def:longrun}, we calculate that 
+	the horizontal asymptote of $h$ has equation $y=1$.
+	\item The details that we have found so far have been drawn in 
+	\cref{rat:fig:doublezerop1}. The function $h$ is different 
+	from the functions that we have considered in previous examples because 
+	of the multiplicity of the zero at $3$.
+	
+	We sketch $h$ using the middle branch as our guide because we have the most information
+	about $h$ on the interval $(-4,6)$. Note that there is no other way to draw this branch
+	without introducing other zeros which $h$ does not have| also note how 
+	the curve bounces off the horizontal axis at $3$.
+	
+	Once we have drawn the middle branch, there is only one way to complete the graph 
+	because of our observations about the behavior of $h$ around its vertical asymptotes| it 
+	behaves like $\frac{1}{x}$.
+	
+\end{steps}
+\end{pccsolution}
+\end{pccexample}
+
+\begin{figure}[!htbp]
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-5,ymax=5,
+			   xtick={-8,-6,...,8},
+			   ytick={-3,3},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-4},{x});
+			   \addplot[asymptote,domain=-10:10]({6},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{1});
+			   \addplot[soldot] coordinates{(3,0)(0,-3/8)}node[axisnode,anchor=north west]{$\left( 0,-\frac{3}{8} \right)$};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:doublezerop1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{.45\textwidth}
+		\begin{tikzpicture}[/pgf/declare function={f=(x-3)^2/((x+4)*(x-6));}]
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-5,ymax=5,
+			   xtick={-8,-6,...,8},
+			   ytick={-3,3},
+			   width=\textwidth,
+			   ]
+			   \addplot[asymptote,domain=-10:10]({-4},{x});
+			   \addplot[asymptote,domain=-10:10]({6},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{1});
+			   \addplot[soldot] coordinates{(3,0)(0,-3/8)}node[axisnode,anchor=north west]{$\left( 0,-\frac{3}{8} \right)$};
+			   \addplot[pccplot] expression[domain=-10:-5.20088]{f};
+			   \addplot[pccplot] expression[domain=-3.16975:5.83642,samples=50]{f};
+			   \addplot[pccplot] expression[domain=6.20088:10]{f};
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:doublezerop2}
+	\end{subfigure}%
+	\caption{$y=\dfrac{(x-3)^2}{(x+4)(x-6)}$}
+\end{figure}
+\begin{exercises}
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{problem}[\Cref{rat:step:last}]\label{rat:prob:deduce}
+\pccname{Katie} is working on graphing rational functions. She 
+has been concentrating on functions that have the form
+\begin{equation}\label{rat:eq:deducecurve}
+	f(x)=\frac{a(x-b)}{x-c}
+\end{equation}
+Katie notes that functions with this type of formula have a zero
+at $b$, and a vertical asymptote at $c$. Furthermore, these functions
+behave like $\frac{1}{x}$ around their vertical asymptote, and the 
+curve of each function will have $2$ branches.
+
+Katie has been working with $3$ functions that have the form given 
+in \cref{rat:eq:deducecurve}, and has followed \crefrange{rat:step:first}{rat:step:penultimate};
+her results are shown in \cref{rat:fig:deducecurve}. There is just one
+more thing to do to complete the graphs| follow \cref{rat:step:last}.
+Help Katie finish each graph by deducing the curve of each function.
+\begin{shortsolution}
+	\Vref{rat:fig:deducecurve1}
+	
+	\begin{tikzpicture}[/pgf/declare function={f=3*(x+4)/(x+5);}]
+		\begin{axis}[
+		   xmin=-10,xmax=10,
+		   ymin=-10,ymax=10,
+		   xtick={-8,-6,...,8},
+		   width=\solutionfigurewidth,
+		   ]
+		   \addplot[soldot] coordinates{(-4,0)(0,12/5)};
+		   \addplot[asymptote,domain=-10:10]({-5},{x});
+		   \addplot[asymptote,domain=-10:10]({x},{3});
+		   \addplot[pccplot] expression[domain=-10:-5.42857]{f};
+		   \addplot[pccplot] expression[domain=-4.76923:10,samples=50]{f};
+		\end{axis}
+	\end{tikzpicture}
+	
+	\Vref{rat:fig:deducecurve2}
+	
+	\begin{tikzpicture}[/pgf/declare function={f=-3*(x-2)/(x-4);}]
+		\begin{axis}[
+		   xmin=-10,xmax=10,
+		   ymin=-10,ymax=10,
+		   xtick={-8,-6,...,8},
+		   width=\solutionfigurewidth,
+		   ]
+		   \addplot[soldot] coordinates{(2,0)(0,-3/2)};
+		   \addplot[asymptote,domain=-10:10]({4},{x});
+		   \addplot[asymptote,domain=-10:10]({x},{-3});
+		   \addplot[pccplot] expression[domain=-10:3.53846,samples=50]{f};
+		   \addplot[pccplot] expression[domain=4.85714:10]{f};
+		\end{axis}
+	\end{tikzpicture}
+	
+	\Vref{rat:fig:deducecurve4}
+	
+	\begin{tikzpicture}[/pgf/declare function={f=2*(x-6)/(x-4);}]
+		\begin{axis}[
+		   xmin=-10,xmax=10,
+		   ymin=-10,ymax=10,
+		   xtick={-8,-6,...,8},
+		   width=\solutionfigurewidth,
+		   ]
+		   \addplot[soldot] coordinates{(6,0)(0,3)};
+		   \addplot[asymptote,domain=-10:10]({x},{2});
+		   \addplot[asymptote,domain=-10:10]({4},{x});
+		   \addplot[pccplot] expression[domain=-10:3.5,samples=50]{f};
+		   \addplot[pccplot] expression[domain=4.3333:10]{f};
+		\end{axis}
+	\end{tikzpicture}
+\end{shortsolution}
+\end{problem}
+
+\begin{figure}[!htb]
+	\begin{widepage}
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-4,0)(0,12/5)};
+			   \addplot[asymptote,domain=-10:10]({-5},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{3});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:deducecurve1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(2,0)(0,-3/2)};
+			   \addplot[asymptote,domain=-10:10]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{-3});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:deducecurve2}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(6,0)(0,3)};
+			   \addplot[asymptote,domain=-10:10]({x},{2});
+			   \addplot[asymptote,domain=-10:10]({4},{x});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:deducecurve4}
+	\end{subfigure}
+	\caption{Graphs for \cref{rat:prob:deduce}}
+	\label{rat:fig:deducecurve}
+	\end{widepage}
+\end{figure}
+
+%===================================
+%   Author: Hughes
+%   Date:   June 2012
+%===================================
+\begin{problem}[\Cref{rat:step:last} for  more complicated rational functions]\label{rat:prob:deducehard}
+\pccname{David} is also working on graphing rational functions, and 
+has been concentrating on functions that have the form
+\[
+	r(x)=\frac{a(x-b)(x-c)}{(x-d)(x-e)}
+\]
+David notices that functions with this type of formula have simple zeros 
+at $b$ and $c$, and vertical asymptotes at $d$ and $e$. Furthermore, 
+these functions behave like $\frac{1}{x}$ around both vertical asymptotes,
+and the curve of the function will have $3$ branches. 
+
+David has followed \crefrange{rat:step:first}{rat:step:penultimate} for 
+$3$ separate functions, and drawn the results in \cref{rat:fig:deducehard}.
+Help David finish each graph by deducing the curve of each function.
+\begin{shortsolution}
+	\Vref{rat:fig:deducehard1}
+	
+	\begin{tikzpicture}[/pgf/declare function={f=(x-6)*(x+3)/( (x-4)*(x+1));}]
+		\begin{axis}[
+		   xmin=-10,xmax=10,
+		   ymin=-10,ymax=10,
+		   xtick={-8,-6,...,8},
+		   width=\solutionfigurewidth,
+		   ]
+		   \addplot[soldot] coordinates{(-3,0)(6,0)(0,9/2)};
+		   \addplot[asymptote,domain=-10:10]({-1},{x});
+		   \addplot[asymptote,domain=-10:10]({4},{x});
+		   \addplot[asymptote,domain=-10:10]({x},{2});
+		   \addplot[pccplot] expression[domain=-10:-1.24276]{f};
+		   \addplot[pccplot] expression[domain=-0.6666:3.66667]{f};
+		   \addplot[pccplot] expression[domain=4.24276:10]{f};
+		\end{axis}
+	\end{tikzpicture}
+	
+	\Vref{rat:fig:deducehard2}
+	
+	\begin{tikzpicture}[/pgf/declare function={f=3*(x-2)*(x+3)/( (x-6)*(x+5));}]
+		\begin{axis}[
+		   xmin=-10,xmax=10,
+		   ymin=-10,ymax=10,
+		   xtick={-8,-6,...,8},
+		   width=\solutionfigurewidth,
+		   ]
+		   \addplot[soldot] coordinates{(-3,0)(2,0)(0,3/5)};
+		   \addplot[asymptote,domain=-10:10]({-5},{x});
+		   \addplot[asymptote,domain=-10:10]({6},{x});
+		   \addplot[asymptote,domain=-10:10]({x},{3});
+		   \addplot[pccplot] expression[domain=-10:-5.4861]{f};
+		   \addplot[pccplot] expression[domain=-4.68395:5.22241]{f};
+		   \addplot[pccplot] expression[domain=7.34324:10]{f};
+		\end{axis}
+	\end{tikzpicture}
+	
+	\Vref{rat:fig:deducehard3}
+	
+	\begin{tikzpicture}[/pgf/declare function={f=2*(x-7)*(x+3)/( (x+6)*(x-5));}]
+		\begin{axis}[
+		   xmin=-10,xmax=10,
+		   ymin=-10,ymax=10,
+		   xtick={-8,-6,...,8},
+		   width=\solutionfigurewidth,
+		   ]
+		   \addplot[soldot] coordinates{(-3,0)(7,0)(0,1.4)};
+		   \addplot[asymptote,domain=-10:10]({-6},{x});
+		   \addplot[asymptote,domain=-10:10]({5},{x});
+		   \addplot[asymptote,domain=-10:10]({x},{2});
+		   \addplot[pccplot] expression[domain=-10:-6.91427]{f};
+		   \addplot[pccplot] expression[domain=-5.42252:4.66427]{f};
+		   \addplot[pccplot] expression[domain=5.25586:10]{f};
+		\end{axis}
+	\end{tikzpicture}
+	
+\end{shortsolution}
+\end{problem}
+
+\begin{figure}[!htb]
+	\begin{widepage}
+	\setlength{\figurewidth}{0.3\textwidth}
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-3,0)(6,0)(0,9/2)};
+			   \addplot[asymptote,domain=-10:10]({-1},{x});
+			   \addplot[asymptote,domain=-10:10]({4},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{2});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:deducehard1}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-3,0)(2,0)(0,3/5)};
+			   \addplot[asymptote,domain=-10:10]({-5},{x});
+			   \addplot[asymptote,domain=-10:10]({6},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{3});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:deducehard2}
+	\end{subfigure}%
+	\hfill
+	\begin{subfigure}{\figurewidth}
+		\begin{tikzpicture}
+			\begin{axis}[
+			   xmin=-10,xmax=10,
+			   ymin=-10,ymax=10,
+			   xtick={-8,-6,...,8},
+			   width=\textwidth,
+			   ]
+			   \addplot[soldot] coordinates{(-3,0)(7,0)(0,1.4)};
+			   \addplot[asymptote,domain=-10:10]({-6},{x});
+			   \addplot[asymptote,domain=-10:10]({5},{x});
+			   \addplot[asymptote,domain=-10:10]({x},{2});
+			\end{axis}
+		\end{tikzpicture}
+		\caption{}
+		\label{rat:fig:deducehard3}
+	\end{subfigure}%
+	\hfill
+	\caption{Graphs for \cref{rat:prob:deducehard}}
+	\label{rat:fig:deducehard}
+	\end{widepage}
+\end{figure}
+%===================================
+%   Author: Adams (Hughes)
+%   Date:   March 2012
+%===================================
+\begin{problem}[\Crefrange{rat:step:first}{rat:step:last}]
+Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of 
+each of the following functions
+\fixthis{need 2 more subproblems here}
+\begin{multicols}{4}
+	\begin{subproblem}
+		$y=\dfrac{4}{x+2}$ 
+		\begin{shortsolution}
+			Vertical intercept: $(0,2)$; vertical asymptote: $x=-2$, horizontal asymptote: $y=0$.
+			
+			\begin{tikzpicture}
+				\begin{axis}[
+				   framed,
+				   xmin=-5,xmax=5,
+				   ymin=-5,ymax=5,
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot] expression[domain=-5:-2.8]{4/(x+2)};
+				   \addplot[pccplot] expression[domain=-1.2:5]{4/(x+2)};
+				   \addplot[soldot]coordinates{(0,2)};
+				   \addplot[asymptote,domain=-5:5]({-2},{x});
+				   \addplot[asymptote,domain=-5:5]({x},{0});
+				\end{axis}
+			\end{tikzpicture}
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=\dfrac{2x-1}{x^2-9}$ 
+		\begin{shortsolution}
+			Vertical intercept:$\left( 0,\frac{1}{9} \right)$; 
+			horizontal intercept: $\left( \frac{1}{2},0 \right)$;
+			vertical asymptotes: $x=-3$, $x=3$, horizontal asymptote: $y=0$.
+			
+			\begin{tikzpicture}
+				\begin{axis}[
+				   framed,
+				   xmin=-5,xmax=5,
+				   ymin=-5,ymax=5,
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot] expression[domain=-5:-3.23974]{(2*x-1)/(x^2-9)};
+				   \addplot[pccplot,samples=50] expression[domain=-2.77321:2.83974]{(2*x-1)/(x^2-9)};
+				   \addplot[pccplot] expression[domain=3.17321:5]{(2*x-1)/(x^2-9)};
+				   \addplot[soldot]coordinates{(0,1/9)(1/2,0)};
+				   \addplot[asymptote,domain=-5:5]({-3},{x});
+				   \addplot[asymptote,domain=-5:5]({3},{x});
+				   \addplot[asymptote,domain=-5:5]({x},{0});
+				\end{axis}
+			\end{tikzpicture}
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=\dfrac{x+3}{x-5}$ 
+		\begin{shortsolution}
+			Vertical intercept $\left( 0,-\frac{3}{5} \right)$; horizontal
+			intercept: $(-3,0)$; vertical asymptote: $x=5$; horizontal asymptote: $y=1$.
+			
+			\begin{tikzpicture}
+				\begin{axis}[
+				   framed,
+				   xmin=-10,xmax=10,
+				   ymin=-5,ymax=5,
+				   xtick={-8,-6,...,8},
+				   minor ytick={-3,-1,...,3},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot] expression[domain=-10:3.666]{(x+3)/(x-5)};
+				   \addplot[pccplot] expression[domain=7:10]{(x+3)/(x-5)};
+				   \addplot[asymptote,domain=-5:5]({5},{x});
+				   \addplot[asymptote,domain=-10:10]({x},{1});
+				   \addplot[soldot]coordinates{(0,-3/5)(-3,0)};
+				\end{axis}
+			\end{tikzpicture}
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=\dfrac{2x+3}{3x-1}$ 
+		\begin{shortsolution}
+			Vertical intercept: $(0,-3)$; horizontal intercept: $\left( -\frac{3}{2},0 \right)$;
+			vertical asymptote: $x=\frac{1}{3}$, horizontal asymptote: $y=\frac{2}{3}$.
+			
+			\begin{tikzpicture}[/pgf/declare function={f=(2*x+3)/(3*x-1);}]
+				\begin{axis}[
+				   framed,
+				   xmin=-5,xmax=5,
+				   ymin=-5,ymax=5,
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot] expression[domain=-5:0.1176]{f};
+				   \addplot[pccplot] expression[domain=0.6153:5]{f};
+				   \addplot[asymptote,domain=-5:5]({1/3},{x});
+				   \addplot[asymptote,domain=-5:5]({x},{2/3});
+				   \addplot[soldot]coordinates{(0,-3)(-3/2,0)};
+				\end{axis}
+			\end{tikzpicture}
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=\dfrac{4-x^2}{x^2-9}$ 
+		\begin{shortsolution}
+			Vertical intercept: $\left( 0,-\frac{4}{9} \right)$;
+			horizontal intercepts: $(2,0)$, $(-2,0)$; 
+			vertical asymptotes: $x=-3$, $x=3$; horizontal asymptote: $y=-1$.
+			
+			\begin{tikzpicture}[/pgf/declare function={f=(4-x^2)/(x^2-9);}]
+				\begin{axis}[
+				   framed,
+				   xmin=-5,xmax=5,
+				   ymin=-5,ymax=5,
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot] expression[domain=-5:-3.20156]{f};
+				   \addplot[pccplot,samples=50] expression[domain=-2.85774:2.85774]{f};
+				   \addplot[pccplot] expression[domain=3.20156:5]{f};
+				   \addplot[asymptote,domain=-5:5]({-3},{x});
+				   \addplot[asymptote,domain=-5:5]({3},{x});
+				   \addplot[asymptote,domain=-5:5]({x},{-1});
+				   \addplot[soldot] coordinates{(-2,0)(2,0)(0,-4/9)};
+				\end{axis}
+			\end{tikzpicture}
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$y=\dfrac{(4x+5)(3x-4)}{(2x+5)(x-5)}$ 
+		\begin{shortsolution}
+			Vertical intercept: $\left( 0,\frac{4}{5} \right)$; 
+			horizontal intercepts: $\left( -\frac{5}{4},0 \right)$, $\left( \frac{4}{3},0 \right)$; 
+			vertical asymptotes: $x=-\frac{5}{2}$, $x=5$; horizontal asymptote: $y=6$.
+			
+			\begin{tikzpicture}[/pgf/declare function={f=(4*x+5)*(3*x-4)/((2*x+5)*(x-5));}]
+				\begin{axis}[
+				   framed,
+				   xmin=-10,xmax=10,
+				   ymin=-20,ymax=20,
+				   xtick={-8,-6,...,8},
+				   ytick={-10,0,...,10},
+				   minor ytick={-15,-5,...,15},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot] expression[domain=-10:-2.73416]{f};
+				   \addplot[pccplot] expression[domain=-2.33689:4.2792]{f};
+				   \addplot[pccplot] expression[domain=6.26988:10]{f};
+				   \addplot[asymptote,domain=-20:20]({-5/2},{x});
+				   \addplot[asymptote,domain=-20:20]({5},{x});
+				   \addplot[asymptote,domain=-10:10]({x},{6});
+				   \addplot[soldot]coordinates{(0,4/5)(-5/4,0)(4/3,0)};
+				\end{axis}
+			\end{tikzpicture}
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Inverse functions]
+Each of the following rational functions are invertible 
+\[
+	F(x)=\frac{2x+1}{x-3}, \qquad G(x)= \frac{1-4x}{x+3}
+\]
+\begin{subproblem}
+	State the domain of each function. 
+	\begin{shortsolution}
+		\begin{itemize}
+			\item The domain of $F$ is $(-\infty,3)\cup(3,\infty)$.
+			\item The domain of $G$ is $(-\infty,-3)\cup(-3,\infty)$.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Find the inverse of each function, and state its domain.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item $F^{-1}(x)=\frac{3x+1}{x-2}$; the domain of $F^{-1}$ is $(-\infty,2)\cup(2,\infty)$.
+			\item $G^{-1}(x)=\frac{3x+1}{x+4}$; the domain of $G^{-1}$ is $(-\infty,-4)\cup(-4,\infty)$.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Hence state the range of the original functions.
+	\begin{shortsolution}
+		\begin{itemize}
+			\item The range of $F$ is the domain of $F^{-1}$, which is $(-\infty,2)\cup(2,\infty)$.
+			\item The range of $G$ is the domain of $G^{-1}$, which is $(-\infty,-4)\cup(-4,\infty)$.
+		\end{itemize}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	State the range of each inverse function. 
+	\begin{shortsolution}
+		\begin{itemize}
+			\item The range of $F^{-1}$ is the domain of $F$, which is $(-\infty,3)\cup(3,\infty)$.
+			\item The range of $G^{-1}$ is the domain of $G$, which is $(-\infty,-3)\cup(-3,\infty)$.
+		\end{itemize}<++>
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Composition]
+Let $r$ and $s$ be the rational functions that have formulas
+\[
+	r(x)=\frac{3}{x^2},\qquad s(x)=\frac{4-x}{x+5}
+\]
+Evaluate each of the following.
+\begin{multicols}{3}
+	\begin{subproblem}
+		$(r\circ s)(0)$
+		\begin{shortsolution}
+			$\frac{75}{16}$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(s\circ r)(0)$ 
+		\begin{shortsolution}
+			$(s\circ r)(0)$ is undefined.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(r\circ s)(2)$ 
+		\begin{shortsolution}
+			$\frac{147}{4}$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(s\circ r)(3)$ 
+		\begin{shortsolution}
+			$192$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(s\circ r)(4)$ 
+		\begin{shortsolution}
+			$(s\circ r)(4)$ is undefined.
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(s\circ r)(x)$ 
+		\begin{shortsolution}
+			$\dfrac{4x^2-3}{1+5x^2}$
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+%===================================
+%   Author: Hughes
+%   Date:   March 2012
+%===================================
+\begin{problem}[Piecewise rational functions]
+The function $R$ has formula
+\[
+	R(x)=
+	\begin{dcases}
+		\frac{2}{x+3},    & x<-5     \\
+		\frac{x-4}{x-10}, & x\geq -5 
+	\end{dcases}
+\]
+Evaluate each of the following.
+\begin{multicols}{4}
+	\begin{subproblem}
+		$R(-6)$    
+		\begin{shortsolution}
+			$-\frac{2}{3}$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$R(-5)$ 
+		\begin{shortsolution}
+			$\frac{3}{5}$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$R(-3)$ 
+		\begin{shortsolution}
+			$\frac{7}{13}$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$R(5)$ 
+		\begin{shortsolution}
+			$-\frac{1}{5}$ 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\begin{subproblem}
+	What is the domain of $R$? 
+	\begin{shortsolution}
+		$(-\infty,10)\cup(10,\infty)$ 
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+\end{exercises}
+
+\section{Graphing rational functions (oblique asymptotes)}\label{rat:sec:oblique}
+\begin{subproblem}
+	$y=\dfrac{x^2+1}{x-4}$ 
+	\begin{shortsolution}
+		\begin{enumerate}
+			\item $\left( 0,-\frac{1}{4} \right)$
+			\item Vertical asymptote: $x=4$.
+			\item A graph of the function is shown below
+			
+			\begin{tikzpicture}[/pgf/declare function={f=(x^2+1)/(x-4);}]
+				\begin{axis}[
+				   framed,
+				   xmin=-20,xmax=20,
+				   ymin=-30,ymax=30,
+				   xtick={-10,10},
+				   minor xtick={-15,-5,...,15},
+				   minor ytick={-10,10},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot,samples=50] expression[domain=-20:3.54724]{f};
+				   \addplot[pccplot,samples=50] expression[domain=4.80196:20]{f};
+				   \addplot[asymptote,domain=-30:30]({4},{x});
+				\end{axis}
+			\end{tikzpicture}
+		\end{enumerate}
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	$y=\dfrac{x^3(x+3)}{x-5}$ 
+	\begin{shortsolution}
+		\begin{enumerate}
+			\item $(0,0)$, $(-3,0)$
+			\item Vertical asymptote: $x=5$, horizontal asymptote: none.
+			\item A graph of the function is shown below
+			
+			\begin{tikzpicture}[/pgf/declare function={f=x^3*(x+3)/(x-5);}]
+				\begin{axis}[
+				   framed,
+				   xmin=-10,xmax=10,
+				   ymin=-500,ymax=2500,
+				   xtick={-8,-6,...,8},
+				   ytick={500,1000,1500,2000},
+				   grid=both,
+				   width=\solutionfigurewidth,
+				   ]
+				   \addplot[pccplot,samples=50] expression[domain=-10:4]{f};
+				   \addplot[pccplot] expression[domain=5.6068:9.777]{f};
+				   \addplot[asymptote,domain=-500:2500]({5},{x});
+				\end{axis}
+			\end{tikzpicture}
+		\end{enumerate}
+	\end{shortsolution}
+\end{subproblem}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex b/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex
new file mode 100644
index 00000000000..757f65c557e
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex
@@ -0,0 +1,5819 @@
+% arara: indent: {overwrite: true, trace: on}
+% A sample chapter file- it contains a lot of 
+% environments, including tabulars, align, etc
+%   
+% Don't try and compile this file using pdflatex etc, just 
+% compare the *format* of it to the format of the
+% sampleAFTER.tex
+%
+% In particular, compare the tabular and align-type
+% environments before and after running the script
+
+\section{Polynomial functions}
+ \reformatstepslist{P} % the steps list should be P1, P2, \ldots
+ In your previous mathematics classes you have studied \emph{linear} and 
+ \emph{quadratic} functions. The most general forms of these types of 
+ functions can be represented (respectively) by  the functions $f$ 
+ and $g$ that have formulas
+ \begin{equation}\label{poly:eq:linquad}
+ 	f(x)=mx+b, \qquad g(x)=ax^2+bx+c
+ \end{equation}
+ We know that $m$ is the slope of $f$, and that $a$ is the \emph{leading coefficient} 
+ of $g$. We also know that the \emph{signs} of $m$ and $a$ completely 
+ determine the behavior of the functions $f$ and $g$. For example, if $m>0$
+ then $f$ is an \emph{increasing} function, and if $m<0$ then $f$ is 
+ a \emph{decreasing} function.  Similarly, if $a>0$ then $g$ is 
+ \emph{concave up} and if $a<0$ then $g$ is \emph{concave down}. Graphical 
+ representations of these statements are given in \cref{poly:fig:linquad}.
+    
+ \begin{figure}[!htb]
+ 	\setlength{\figurewidth}{.2\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-10:8]{(x+2)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$m>0$}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-10:8]{-(x+2)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$m<0$}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-4:4]{(x^2-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$a>0$}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-4:4]{-(x^2-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$a<0$}
+ 	\end{subfigure}
+ 	\caption{Typical graphs of linear and quadratic functions.}
+ 	\label{poly:fig:linquad}
+ \end{figure}
+    
+ Let's look a little more closely at the formulas for $f$ and $g$ in 
+ \cref{poly:eq:linquad}. Note that the \emph{degree} 
+ of $f$ is $1$ since the highest power of $x$ that is present in the 
+ formula for $f(x)$ is $1$. Similarly, the degree of $g$ is $2$ since
+ the highest power of $x$ that is present in the formula for $g(x)$ 
+ is $2$.
+    
+ In this section we will build upon our knowledge of these elementary
+ functions. In particular, we will generalize the functions $f$ and $g$ to a function $p$ that has 
+ any degree that we wish.
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{essentialskills}
+ 	%===================================
+ 	%   Author: Hughes
+ 	%   Date:   March 2012
+ 	%===================================
+ 	\begin{problem}[Quadratic functions]
+ 	Every quadratic function has the form $y=ax^2+bx+c$; state the value 
+ 	of $a$ for each of the following functions, and hence decide if the 
+ 	parabola that represents the function opens upward or downward.
+ 	\begin{multicols}{2}
+ 		\begin{subproblem}
+ 			$F(x)=x^2+3$ 
+ 			\begin{shortsolution}
+ 				$a=1$; the parabola opens upward. 
+ 			\end{shortsolution}
+ 		\end{subproblem}
+ 		\begin{subproblem}
+ 			$G(t)=4-5t^2$ 
+ 			\begin{shortsolution}
+ 				$a=-5$; the parabola opens downward. 
+ 			\end{shortsolution}
+ 		\end{subproblem}
+ 		\begin{subproblem}
+ 			$H(y)=4y^2-96y+8$ 
+ 			\begin{shortsolution}
+ 				$a=4$; the parabola opens upward. 
+ 			\end{shortsolution}
+ 		\end{subproblem}
+ 		\begin{subproblem}
+ 			$K(z)=-19z^2$ 
+ 			\begin{shortsolution}
+ 				$m=-19$; the parabola opens downward. 
+ 			\end{shortsolution}
+ 		\end{subproblem}
+ 	\end{multicols}
+ 	Now let's generalize our findings for the most general quadratic function $g$
+ 	that has formula $g(x)=a_2x^2+a_1x+a_0$. Complete the following sentences.
+ 	\begin{subproblem}
+ 		When $a_2>0$, the parabola that represents $y=g(x)$ opens $\ldots$ 
+ 		\begin{shortsolution}
+ 			When $a_2>0$, the parabola that represents the function opens upward.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		When $a_2<0$, the parabola that represents $y=g(x)$ opens $\ldots$ 
+ 		\begin{shortsolution}
+ 			When $a_2<0$, the parabola that represents the function opens downward.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\end{problem}
+ \end{essentialskills}
+    
+ \subsection*{Power functions with positive exponents}
+ The study of polynomials will rely upon a good knowledge 
+ of power functions| you may reasonably ask, what is a power function?
+ \begin{pccdefinition}[Power functions]
+ Power functions have the form
+ \[
+ 	f(x) = a_n x^n
+ \]
+ where $n$ can be any real number.
+  	  
+ Note that for this section we will only be concerned with the 
+ case when $n$ is a positive integer.
+ \end{pccdefinition}
+  	  
+ You may find assurance in the fact that you are already very comfortable 
+ with power functions that have $n=1$ (linear) and $n=2$ (quadratic). Let's 
+ explore some power functions that you might not be so familiar with.
+ As you read \cref{poly:ex:oddpow,poly:ex:evenpow}, try and spot 
+ as many patterns and similarities as you can.
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{pccexample}[Power functions with odd positive exponents]
+ \label{poly:ex:oddpow}
+ Graph each of the following functions, state their domain, and their 
+ long-run behavior as $x\rightarrow\pm\infty$
+ \[
+ 	f(x)=x^3,   \qquad  g(x)=x^5, \qquad h(x)=x^7
+ \]
+ \begin{pccsolution}
+ The functions $f$, $g$, and $h$ are plotted in \cref{poly:fig:oddpow}.
+ The domain of each of the functions $f$, $g$, and $h$ is $(-\infty,\infty)$. Note that 
+ the long-run behavior of each of the functions is the same, and in particular
+ \begin{align*}
+ 	f(x)\rightarrow\infty                           & \text{ as } x\rightarrow\infty  \\
+ 	\mathllap{\text{and }}   f(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty 
+ \end{align*}
+ The same results hold for $g$ and $h$.
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ \begin{figure}[!htb]
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-1.5,xmax=1.5,
+ 					ymin=-5,ymax=5,
+ 					xtick={-1.0,-0.5,...,1.0},
+ 					minor ytick={-3,-1,...,3},
+ 					grid=both,
+ 					width=\textwidth,
+ 					legend pos=north west,
+ 				]
+ 				\addplot expression[domain=-1.5:1.5]{x^3};
+ 				\addplot expression[domain=-1.379:1.379]{x^5};
+ 				\addplot expression[domain=-1.258:1.258]{x^7};
+ 				\addplot[soldot]coordinates{(-1,-1)} node[axisnode,anchor=north west]{$(-1,-1)$};
+ 				\addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=south east]{$(1,1)$};
+ 				\legend{$f$,$g$,$h$}
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{Odd power functions}
+ 		\label{poly:fig:oddpow}
+ 	\end{minipage}%
+ 	\hfill
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-2.5,xmax=2.5,
+ 					ymin=-5,ymax=5,
+ 					xtick={-2.0,-1.5,...,2.0},
+ 					minor ytick={-3,-1,...,3},
+ 					grid=both,
+ 					width=\textwidth,
+ 					legend pos=south east,
+ 				]
+ 				\addplot expression[domain=-2.236:2.236]{x^2};
+ 				\addplot expression[domain=-1.495:1.495]{x^4};
+ 				\addplot expression[domain=-1.307:1.307]{x^6};
+ 				\addplot[soldot]coordinates{(-1,1)} node[axisnode,anchor=east]{$(-1,1)$};
+ 				\addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=west]{$(1,1)$};
+ 				\legend{$F$,$G$,$H$}
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{Even power functions}
+ 		\label{poly:fig:evenpow}
+ 	\end{minipage}%
+ \end{figure}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{pccexample}[Power functions with even positive exponents]\label{poly:ex:evenpow}%
+ Graph each of the following functions, state their domain, and their 
+ long-run behavior as $x\rightarrow\pm\infty$
+ \[
+ 	F(x)=x^2, \qquad G(x)=x^4, \qquad H(x)=x^6
+ \]
+ \begin{pccsolution}
+ The functions $F$, $G$, and $H$ are plotted in \cref{poly:fig:evenpow}. The domain
+ of each of the functions is $(-\infty,\infty)$. Note that the long-run behavior 
+ of each of the functions is the same, and in particular
+ \begin{align*}
+ 	F(x)\rightarrow\infty                          & \text{ as } x\rightarrow\infty  \\
+ 	\mathllap{\text{and }}   F(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty 
+ \end{align*}
+ The same result holds for $G$ and $H$.
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ \begin{doyouunderstand}
+ 	\begin{problem}
+ 	Repeat \cref{poly:ex:oddpow,poly:ex:evenpow} using (respectively)
+ 	\begin{subproblem}
+ 		$f(x)=-x^3,   \qquad  g(x)=-x^5, \qquad h(x)=-x^7$
+ 		\begin{shortsolution}
+ 			The functions $f$, $g$, and $h$ have domain $(-\infty,\infty)$ and 
+ 			are graphed below.
+ 			 				 			 			
+ 			\begin{tikzpicture}
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-1.5,xmax=1.5,
+ 						ymin=-5,ymax=5,
+ 						xtick={-1.0,-0.5,...,0.5},
+ 						minor ytick={-3,-1,...,3},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 						legend pos=north east,
+ 					]
+ 					\addplot expression[domain=-1.5:1.5]{-x^3};
+ 					\addplot expression[domain=-1.379:1.379]{-x^5};
+ 					\addplot expression[domain=-1.258:1.258]{-x^7};
+ 					\legend{$f$,$g$,$h$}
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 			 				 			 			
+ 			Note that
+ 			\begin{align*}
+ 				f(x)\rightarrow-\infty                         & \text{ as } x\rightarrow\infty  \\
+ 				\mathllap{\text{and }}   f(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty 
+ 			\end{align*}
+ 			The same is true for $g$ and $h$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$F(x)=-x^2,   \qquad  G(x)=-x^4, \qquad H(x)=-x^6$
+ 		\begin{shortsolution}
+ 			The functions $F$, $G$, and $H$ have domain $(-\infty,\infty)$ and 
+ 			are graphed below.
+ 			 				 			 			
+ 			\begin{tikzpicture}
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-2.5,xmax=2.5,
+ 						ymin=-5,ymax=5,
+ 						xtick={-1.0,-0.5,...,0.5},
+ 						minor ytick={-3,-1,...,3},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 						legend pos=north east,
+ 					]
+ 					\addplot expression[domain=-2.236:2.236]{-x^2};
+ 					\addplot expression[domain=-1.495:1.495]{-x^4};
+ 					\addplot expression[domain=-1.307:1.307]{-x^6};
+ 					\legend{$F$,$G$,$H$}
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 			 				 			 			
+ 			Note that
+ 			\begin{align*}
+ 				F(x)\rightarrow-\infty                          & \text{ as } x\rightarrow\infty  \\
+ 				\mathllap{\text{and }}   F(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty 
+ 			\end{align*}
+ 			The same is true for $G$ and $H$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\end{problem}
+ \end{doyouunderstand}
+  	  
+ \subsection*{Polynomial functions}
+ Now that we have a little more familiarity with power functions, 
+ we can define polynomial functions. Provided that you were comfortable
+ with our opening discussion about linear and quadratic functions (see 
+ $f$ and $g$ in \cref{poly:eq:linquad}) then there is every chance 
+ that you'll be able to master polynomial functions as well; just remember
+ that polynomial functions are a natural generalization of linear
+ and quadratic functions. Once you've studied the examples and problems
+ in this section, you'll hopefully agree that polynomial functions
+ are remarkably predictable.
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccdefinition}[Polynomial functions]
+ Polynomial functions have the form
+ \[
+ 	p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0
+ \]
+ where $a_n$, $a_{n-1}$, $a_{n-2}$, \ldots, $a_0$ are real numbers.
+ \begin{itemize}
+ 	\item We call $n$ the degree of the polynomial, and require that $n$
+ 	is a non-negative integer;
+ 	\item $a_n$, $a_{n-1}$, $a_{n-2}$, \ldots, $a_0$ are called the coefficients;
+ 	\item We typically write polynomial functions in descending powers of $x$.
+ \end{itemize}
+ In particular, we call $a_n$ the \emph{leading} coefficient, and $a_nx^n$ the 
+ \emph{leading term}.
+  	  
+ Note that if a polynomial is given in factored form, then the degree can be found 
+ by counting the number of linear factors.
+ \end{pccdefinition}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{pccexample}[Polynomial or not]
+ Identify the following functions as polynomial or not; if the function 
+ is a polynomial, state its degree.
+ \begin{multicols}{3}
+ 	\begin{enumerate}
+ 		\item $p(x)=x^2-3$     
+ 		\item $q(x)=-4x^{\nicefrac{1}{2}}+10$     
+ 		\item $r(x)=10x^5$
+ 		\item $s(x)=x^{-2}+x^{23}$
+ 		\item $f(x)=-8$
+ 		\item $g(x)=3^x$
+ 		\item $h(x)=\sqrt[3]{x^7}-x^2+x$
+ 		\item $k(x)=4x(x+2)(x-3)$
+ 		\item $j(x)=x^2(x-4)(5-x)$
+ 	\end{enumerate}
+ \end{multicols}
+ \begin{pccsolution}
+ \begin{enumerate}
+ 	\item $p$ is a polynomial, and its degree is $2$.
+ 	\item $q$ is \emph{not} a polynomial, because $\frac{1}{2}$ is not an integer.
+ 	\item $r$ is a polynomial, and its degree is $5$.
+ 	\item $s$ is \emph{not} a polynomial, because $-2$ is not a positive integer.
+ 	\item $f$ is a polynomial, and its degree is $0$.
+ 	\item $g$ is \emph{not} a polynomial, because the independent 
+ 	variable, $x$, is in the exponent.
+ 	\item $h$ is \emph{not} a polynomial, because $\frac{7}{3}$ is not an integer.
+ 	\item $k$ is a polynomial, and its degree is $3$.
+ 	\item $j$ is a polynomial, and its degree is $4$.
+ \end{enumerate}
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{pccexample}[Typical graphs]\label{poly:ex:typical}
+ \Cref{poly:fig:typical} shows graphs of some polynomial functions;
+ the ticks have deliberately been left off the axis to allow us to concentrate
+ on the features of each graph. Note in particular that:
+ \begin{itemize}
+ 	\item \cref{poly:fig:typical1} shows a degree-$1$ polynomial (you might also 
+ 	classify the function as linear) whose leading coefficient, $a_1$, is positive.
+ 	\item \cref{poly:fig:typical2} shows a degree-$2$ polynomial (you might also
+ 	classify the function as quadratic) whose leading coefficient, $a_2$, is positive.
+ 	\item \cref{poly:fig:typical3} shows a degree-$3$ polynomial whose leading coefficient, $a_3$,
+ 	is positive| compare its overall
+ 	shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}.
+ 	\item \cref{poly:fig:typical4} shows a degree-$4$ polynomial whose leading coefficient, $a_4$,
+ 	is positive|compare its overall shape and long-run behavior to the functions described in \cref{poly:ex:evenpow}.
+ 	\item \cref{poly:fig:typical5} shows a degree-$5$ polynomial whose leading coefficient, $a_5$,
+ 	is positive| compare its overall
+ 	shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}.
+ \end{itemize}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{figure}[!htb]
+ 	\begin{widepage}
+ 	\setlength{\figurewidth}{\textwidth/6}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-10:8]{(x+2)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$a_1>0$}
+ 		\label{poly:fig:typical1}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-4:4]{(x^2-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$a_2>0$}
+ 		\label{poly:fig:typical2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-7.5:7.5]{0.05*(x+6)*x*(x-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$a_3>0$}
+ 		\label{poly:fig:typical3}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-2.35:5.35,samples=100]{0.2*(x-5)*x*(x-3)*(x+2)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$a_4>0$}
+ 		\label{poly:fig:typical4}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\textwidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-5.5:6.3,samples=100]{0.01*(x+2)*x*(x-3)*(x+5)*(x-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$a_5>0$}
+ 		\label{poly:fig:typical5}
+ 	\end{subfigure}
+ 	\end{widepage}
+ 	\caption{Graphs to illustrate typical curves of polynomial functions.}
+ 	\label{poly:fig:typical}
+ \end{figure}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{doyouunderstand}
+ 	\begin{problem}
+ 	Use \cref{poly:ex:typical} and \cref{poly:fig:typical} to help you sketch 
+ 	the graphs of polynomial functions that have negative leading coefficients| note
+ 	that there are many ways to do this! The intention with this problem
+ 	is to use your knowledge of transformations- in particular, \emph{reflections}- 
+ 	to guide you.
+ 	\begin{shortsolution}
+ 		$a_1<0$:
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\solutionfigurewidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-10:8]{-(x+2)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		 			 		 		
+ 		$a_2<0$
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\solutionfigurewidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-4:4]{-(x^2-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		 			 		 		
+ 		$a_3<0$
+ 		 			 		 		        
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\solutionfigurewidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-7.5:7.5]{-0.05*(x+6)*x*(x-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		 			 		 		
+ 		$a_4<0$
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\solutionfigurewidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-2.35:5.35,samples=100]{-0.2*(x-5)*x*(x-3)*(x+2)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		 			 		 		
+ 		$a_5<0$
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					width=\solutionfigurewidth,
+ 					xtick={-11},
+ 					ytick={-11},
+ 				]
+ 				\addplot expression[domain=-5.5:6.3,samples=100]{-0.01*(x+2)*x*(x-3)*(x+5)*(x-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 	\end{shortsolution}
+ 	\end{problem}
+ \end{doyouunderstand}
+  	  
+ \fixthis{poly: Need a more basic example here- it can have a similar
+ 	format to the multiple zeros example, but just keep it simple; it should
+ be halfway between the 2 examples surrounding it}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Multiple zeros]
+ Consider the polynomial functions $p$, $q$, and $r$ which are 
+ graphed in \cref{poly:fig:moremultiple}. 
+ The formulas for $p$, $q$, and $r$ are as follows
+ \begin{align*}
+ 	p(x) & =(x-3)^2(x+4)^2       \\
+ 	q(x) & =x(x+2)^2(x-1)^2(x-3) \\
+ 	r(x) & =x(x-3)^3(x+1)^2      
+ \end{align*}
+ Find the degree of $p$, $q$, and $r$, and decide if the functions bounce off or cut 
+ through the horizontal axis at each of their zeros.
+ \begin{pccsolution}
+ The degree of $p$ is 4. Referring to \cref{poly:fig:bouncep}, 
+ the curve bounces off the horizontal axis at both zeros, $3$ and $4$.
+  	  
+ The degree of $q$ is 6. Referring to \cref{poly:fig:bounceq},
+ the curve bounces off the horizontal axis at $-2$ and $1$, and cuts 
+ through the horizontal axis at $0$ and $3$.
+  	  
+ The degree of $r$ is 6. Referring to \cref{poly:fig:bouncer},
+ the curve bounces off the horizontal axis at $-1$, and cuts through 
+ the horizontal axis at $0$ and at $3$, although is flattened immediately to the left and right of $3$.
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ \setlength{\figurewidth}{0.25\textwidth}
+ \begin{figure}[!htb]
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-6,xmax=5,
+ 					ymin=-30,ymax=200,
+ 					xtick={-4,-2,...,4},
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-5.63733:4.63733,samples=50]{(x-3)^2*(x+4)^2};
+ 				\addplot[soldot]coordinates{(3,0)(-4,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=p(x)$}
+ 		\label{poly:fig:bouncep}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-3,xmax=4,
+ 					xtick={-2,...,3},
+ 					ymin=-60,ymax=40,
+ 					width=\textwidth,
+ 				]
+ 				\addplot+[samples=50] expression[domain=-2.49011:3.11054]{x*(x+2)^2*(x-1)^2*(x-3)};
+ 				\addplot[soldot]coordinates{(-2,0)(0,0)(1,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=q(x)$}
+ 		\label{poly:fig:bounceq}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-2,xmax=4,
+ 					xtick={-1,...,3},
+ 					ymin=-40,ymax=40,
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-1.53024:3.77464,samples=50]{x*(x-3)^3*(x+1)^2};
+ 				\addplot[soldot]coordinates{(-1,0)(0,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=r(x)$}
+ 		\label{poly:fig:bouncer}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\label{poly:fig:moremultiple}
+ \end{figure}
+  	  
+ \begin{pccdefinition}[Multiple zeros]\label{poly:def:multzero}
+ Let $p$ be a polynomial that has a repeated linear factor $(x-a)^n$. Then we say
+ that $p$ has a multiple zero at $a$ of multiplicity $n$ and 
+ \begin{itemize}
+ 	\item if the factor $(x-a)$ is repeated an even number of times, the graph of $y=p(x)$ does not
+ 	cross the $x$ axis at $a$, but `bounces' off the horizontal axis at $a$.
+ 	\item if the factor $(x-a)$ is repeated an odd number of times, the graph of $y=p(x)$ crosses the
+ 	horizontal axis at $a$, but it looks `flattened' there
+ \end{itemize}
+ If $n=1$, then we say that $p$ has a \emph{simple} zero at $a$.
+ \end{pccdefinition}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Find a formula]
+ Find formulas for the polynomial functions, $p$ and $q$, graphed in \cref{poly:fig:findformulademoboth}.
+ \begin{figure}[!htb]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[framed,
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					xtick={-4,-2,...,4},
+ 					minor xtick={-3,-1,...,3},
+ 					ytick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				grid=both]
+ 				\addplot expression[domain=-3.25842:2.25842,samples=50]{-x*(x-2)*(x+3)*(x+1)};
+ 				\addplot[soldot]coordinates{(1,8)}node[axisnode,inner sep=.35cm,anchor=west]{$(1,8)$};
+ 				\addplot[soldot]coordinates{(-3,0)(-1,0)(0,0)(2,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$p$}
+ 		\label{poly:fig:findformulademo}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[framed,
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					xtick={-4,-2,...,4},
+ 					minor xtick={-3,-1,...,3},
+ 					ytick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				grid=both]
+ 				\addplot expression[domain=-4.33:4.08152]{-.25*(x+2)^2*(x-3)};
+ 				\addplot[soldot]coordinates{(2,4)}node[axisnode,anchor=south west]{$(2,4)$};
+ 				\addplot[soldot]coordinates{(-2,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$q$}
+ 		\label{poly:fig:findformulademo1}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\label{poly:fig:findformulademoboth}
+ \end{figure}
+ \begin{pccsolution}
+ \begin{enumerate}
+ 	\item We begin by noting that the horizontal intercepts of $p$ are $(-3,0)$, $(-1,0)$, $(0,0)$ and $(2,0)$. 
+ 	We also note that each zero is simple (multiplicity $1$).
+ 	If we assume that $p$ has no other zeros, then we can start by writing
+ 	\begin{align*}
+ 		p(x) & =(x+3)(x+1)(x-0)(x-2) \\
+ 		     & =x(x+3)(x+1)(x-2)     \\
+ 	\end{align*}
+ 	According to \cref{poly:fig:findformulademo}, the point $(1,8)$ lies 
+ 	on the curve $y=p(x)$.
+ 	Let's check if the formula we have written satisfies this requirement
+ 	\begin{align*}
+ 		p(1) & = (1)(4)(2)(-1) \\
+ 		     & = -8            
+ 	\end{align*}
+ 	which is clearly not correct| it is close though. We can correct this by 
+ 	multiplying $p$ by a constant $k$; so let's assume that
+ 	\[
+ 		p(x)=kx(x+3)(x+1)(x-2)
+ 	\]
+ 	Then $p(1)=-8k$, and if this is to equal $8$, then $k=-1$. Therefore
+ 	the formula for $p(x)$ is
+ 	\[
+ 		p(x)=-x(x+3)(x+1)(x-2)
+ 	\]
+ 	\item The function $q$ has a zero at $-2$ of multiplicity $2$, and zero of 
+ 	multiplicity $1$ at $3$ (so $3$ is a simple zero of $q$); we can therefore assume that $q$ has the form
+ 	\[
+ 		q(x)=k(x+2)^2(x-3)
+ 	\]
+ 	where $k$ is some real number. In order to find $k$, we use the given ordered pair, $(2,4)$, and
+ 	evaluate $p(2)$
+ 	\begin{align*}
+ 		p(2) & =k(4)^2(-1) \\
+ 		     & =-16k       
+ 	\end{align*}
+ 	We solve the equation $4=-8k$ and obtain $k=-\frac{1}{4}$ and conclude that the 
+ 	formula for $q(x)$ is
+ 	\[
+ 		q(x)=-\frac{1}{4}(x+2)^2(x-3)
+ 	\]
+ \end{enumerate}
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+  	  
+ \fixthis{Chris: need sketching polynomial problems}
+ \begin{pccspecialcomment}[Steps to follow when sketching polynomial functions]
+ 	\begin{steps}
+ 		\item \label{poly:step:first} Determine the degree of the polynomial, 
+ 		its leading term and leading coefficient, and hence determine
+ 		the long-run behavior of the polynomial| does it behave like $\pm x^2$ or $\pm x^3$ 
+ 		as $x\rightarrow\pm\infty$?
+ 		\item Determine the zeros and their multiplicity. Mark all zeros 
+ 		and the vertical intercept on the graph using solid circles $\bullet$.
+ 		\item \label{poly:step:last}  Deduce the overall shape of the curve, and sketch it. If there isn't
+ 		enough information from the previous steps, then construct a table of values.
+ 	\end{steps}
+ 	Remember that until we have the tools of calculus, we won't be able to 
+ 	find the exact coordinates of local minimums, local maximums, and points
+ 	of inflection.
+ \end{pccspecialcomment}
+ Before we demonstrate some examples, it is important to remember the following:
+ \begin{itemize}
+ 	\item our sketches will give a good representation of the overall 
+ 	shape of the graph, but until we have the tools of calculus (from MTH 251)
+ 	we can not find local minimums, local maximums, and inflection points algebraically. This
+ 	means that we will make our best guess as to where these points are.
+ 	\item we will not concern ourselves too much with the vertical scale (because of 
+ 	our previous point)| we will, however, mark the vertical intercept (assuming there is one), 
+ 	and any horizontal asymptotes.
+ \end{itemize}
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{poly:ex:simplecubic}
+ Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $p$ 
+ that has formula
+ \[
+ 	p(x)=\frac{1}{2}(x-4)(x-1)(x+3)
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $p$ has degree $3$. The leading term of $p$ is $\frac{1}{2}x^3$, so the leading coefficient of $p$ 
+ 	is $\frac{1}{2}$. The long-run behavior of $p$ is therefore similar to that of $x^3$.
+ 	\item The zeros of $p$ are $-3$, $1$, and $4$; each zero is simple (i.e, it has multiplicity $1$).
+ 	This means that the curve of $p$ cuts the horizontal axis at each zero. The vertical 
+ 	intercept of $p$ is $(0,6)$.
+ 	\item We draw the details we have obtained so far on \cref{poly:fig:simplecubicp1}. Given 
+ 	that the curve of $p$ looks like the curve of $x^3$ in the long-run, we are able to complete a sketch of the 
+ 	graph of $p$ in \cref{poly:fig:simplecubicp2}.
+ 	 		 	 	
+ 	Note that we can not find the coordinates of the local minimums, local maximums, and inflection
+ 	points| for the moment we make reasonable guesses as to where these points are (you'll find how
+ 	to do this in calculus).
+ \end{steps}
+  	  
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=15,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:simplecubicp1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=15,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$};
+ 				\addplot[pccplot] expression[domain=-3.57675:4.95392,samples=100]{.5*(x-4)*(x-1)*(x+3)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:simplecubicp2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{1}{2}(x-4)(x-1)(x+3)$}
+ 	\label{poly:fig:simplecubic}
+ \end{figure}
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{poly:ex:degree5}
+ Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $q$ 
+ that has formula
+ \[
+ 	q(x)=\frac{1}{200}(x+7)^2(2-x)(x-6)^2
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $q$ has degree $4$. The leading term of $q$ is 
+ 	\[
+ 		-\frac{1}{200}x^5
+ 	\]
+ 	so the leading coefficient of $q$ is $-\frac{1}{200}$. The long-run behavior of $q$
+ 	is therefore similar to that of $-x^5$.
+ 	\item The zeros of $q$ are $-7$ (multiplicity 2), $2$ (simple), and $6$ (multiplicity $2$).
+ 	The curve of $q$ bounces off the horizontal axis at the zeros with multiplicity $2$ and
+ 	cuts the horizontal axis at the simple zeros. The vertical intercept of $q$ is $\left( 0,\frac{441}{25} \right)$.
+ 	\item We mark the details we have found so far on \cref{poly:fig:degree5p1}. Given that 
+ 	the curve of $q$ looks like the curve of $-x^5$ in the long-run, we can complete \cref{poly:fig:degree5p2}.
+ \end{steps}
+  	  
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=40,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,0,...,35},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\left( 0, \frac{441}{25} \right)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:degree5p1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=40,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,0,...,35},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\left( 0, \frac{441}{25} \right)$};
+ 				\addplot[pccplot] expression[domain=-8.83223:7.34784,samples=50]{1/200*(x+7)^2*(2-x)*(x-6)^2};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:degree5p2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{1}{200}(x+7)^2(2-x)(x-6)^2$}
+ 	\label{poly:fig:degree5}
+ \end{figure}
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}
+ Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $r$ 
+ that has formula
+ \[
+ 	r(x)=\frac{1}{100}x^3(x+4)(x-4)(x-6)
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $r$ has degree $6$. The leading term of $r$ is 
+ 	\[
+ 		\frac{1}{100}x^6
+ 	\]
+ 	so the leading coefficient of $r$ is $\frac{1}{100}$. The long-run behavior of $r$
+ 	is therefore similar to that of $x^6$.
+ 	\item The zeros of $r$ are $-4$ (simple), $0$ (multiplicity $3$), $4$ (simple), 
+ 	and $6$ (simple). The vertical intercept of $r$ is $(0,0)$. The curve of $r$
+ 	cuts the horizontal axis at the simple zeros, and goes through the axis
+ 	at $(0,0)$, but does so in a flattened way.
+ 	\item We mark the zeros and vertical intercept on \cref{poly:fig:degree6p1}. Given that
+ 	the curve of $r$ looks like the curve of $x^6$ in the long-run, we complete the graph
+ 	of $r$ in \cref{poly:fig:degree6p2}.
+ \end{steps}
+  	  
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=10,
+ 					ymin=-20,ymax=10,
+ 					xtick={-4,-2,...,8},
+ 					ytick={-15,-10,...,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:degree6p1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=10,
+ 					ymin=-20,ymax=10,
+ 					xtick={-4,-2,...,8},
+ 					ytick={-15,-10,...,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)};
+ 				\addplot[pccplot] expression[domain=-4.16652:6.18911,samples=100]{1/100*(x+4)*x^3*(x-4)*(x-6)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:degree6p2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{1}{100}(x+4)x^3(x-4)(x-6)$}
+ \end{figure}
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{pccexample}[An open-topped box] 
+ A cardboard company makes open-topped boxes for their clients. The specifications
+ dictate that the box must have a square base, and that it must be open-topped. 
+ The company uses sheets of cardboard that are $\unit[1200]{cm^2}$. Assuming that 
+ the base of each box has side $x$ (measured in cm), it can be shown that the volume of each box, $V(x)$,
+ has formula
+ \[
+ 	V(x)=\frac{x}{4}(1200-x^2)
+ \]
+ Find the dimensions of the box that maximize the volume.
+ \begin{pccsolution}
+ We graph $y=V(x)$ in \cref{poly:fig:opentoppedbox}. Note that because 
+ $x$ represents the length of a side, and $V(x)$ represents the volume
+ of the box, we necessarily require both values to be positive; we illustrate
+ the part of the curve that applies to this problem using a solid line. 
+  	  
+ \begin{figure}[!htb]
+ 	\centering
+ 	\begin{tikzpicture}
+ 		\begin{axis}[framed,
+ 				xmin=-50,xmax=50,
+ 				ymin=-5000,ymax=5000,
+ 				xtick={-40,-30,...,40},
+ 				minor xtick={-45,-35,...,45},
+ 				minor ytick={-3000,-1000,1000,3000},
+ 				width=.75\textwidth,
+ 				height=.5\textwidth,
+ 			grid=both]
+ 			\addplot[pccplot,dashed,<-] expression[domain=-40:0,samples=50]{x/4*(1200-x^2)};
+ 			\addplot[pccplot,-] expression[domain=0:34.64,samples=50]{x/4*(1200-x^2)};
+ 			\addplot[pccplot,dashed,->] expression[domain=34.64:40,samples=50]{x/4*(1200-x^2)};
+ 			\addplot[soldot] coordinates{(20,4000)};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ 	\caption{$y=V(x)$}
+ 	\label{poly:fig:opentoppedbox}
+ \end{figure}
+  	  
+ According to \cref{poly:fig:opentoppedbox}, the maximum volume of such a box is 
+ approximately $\unit[4000]{cm^2}$, and we achieve it using a base of length 
+ approximately $\unit[20]{cm}$. Since the base is square and each sheet of cardboard
+ is $\unit[1200]{cm^2}$, we conclude that the dimensions of each box are $\unit[20]{cm}\times\unit[20]{cm}\times\unit[30]{cm}$.
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ \subsection*{Complex zeros}
+ There has been a pattern to all of the examples that we have seen so far|
+ the degree of the polynomial has dictated the number of \emph{real} zeros that the 
+ polynomial has. For example, the function $p$ in \cref{poly:ex:simplecubic}
+ has degree $3$, and $p$ has $3$ real zeros; the function $q$ in \cref{poly:ex:degree5}
+ has degree $5$ and $q$ has $5$ real zeros.
+  	  
+ You may wonder if this result can be generalized| does every polynomial that
+ has degree $n$ have $n$ real zeros? Before we tackle the general result, 
+ let's consider an example that may help motivate it.
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{pccexample}\label{poly:ex:complx}
+ Consider the polynomial function $c$ that has formula
+ \[
+ 	c(x)=x(x^2+1)
+ \]
+ It is clear that $c$ has degree $3$, and that $c$ has a (simple) zero at $0$. Does
+ $c$ have any other zeros, i.e, can we find any values of $x$ that satisfy the equation
+ \begin{equation}\label{poly:eq:complx}
+ 	x^2+1=0
+ \end{equation}
+ The solutions to \cref{poly:eq:complx} are $\pm i$.
+  	  
+ We conclude that $c$ has $3$ zeros: $0$ and $\pm i$; we note that \emph{not 
+ all of them are real}.
+ \end{pccexample}
+ \Cref{poly:ex:complx} shows that not every degree-$3$ polynomial has $3$
+ \emph{real} zeros; however, if we are prepared to venture into the complex numbers, 
+ then we can state the following theorem.
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{pccspecialcomment}[The fundamental theorem of algebra]
+ 	Every polynomial function of degree $n$ has $n$ roots, some of which may 
+ 	be complex, and some may be repeated.
+ \end{pccspecialcomment}
+ \fixthis{Fundamental theorem of algebra: is this wording ok? do we want
+ it as a theorem?}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{pccexample}
+ Find all the zeros of the polynomial function $p$ that has formula
+ \[
+ 	p(x)=x^4-2x^3+5x^2
+ \]
+ \begin{pccsolution}
+ We begin by factoring $p$
+ \begin{align*}
+ 	p(x) & =x^4-2x^3+5x^2 \\
+ 	     & =x^2(x^2-2x+5) 
+ \end{align*}
+ We note that $0$ is a zero of $p$ with multiplicity $2$. The other zeros of $p$ 
+ can be found by solving the equation
+ \[
+ 	x^2-2x+5=0
+ \]
+ This equation can not be factored, so we use the quadratic formula
+ \begin{align*}
+ 	x & =\frac{2\pm\sqrt{(-2)^2}-20}{2(1)} \\
+ 	  & =\frac{2\pm\sqrt{-16}}{2}          \\
+ 	  & =1\pm 2i                           
+ \end{align*}
+ We conclude that $p$ has $4$ zeros: $0$ (multiplicity $2$), and $1\pm 2i$ (simple).
+ \end{pccsolution}
+ \end{pccexample}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{pccexample}
+ Find a polynomial that has zeros at $2\pm i\sqrt{2}$. 
+ \begin{pccsolution}
+ We know that the zeros of a polynomial can be found by analyzing the linear
+ factors. We are given the zeros, and have to work backwards to find the 
+ linear factors. 
+  	  
+ We begin by assuming that $p$ has the form
+ \begin{align*}
+ 	p(x) & =(x-(2-i\sqrt{2}))(x-(2+i\sqrt{2}))                           \\
+ 	     & =x^2-x(2+i\sqrt{2})-x(2-i\sqrt{2})+(2-i\sqrt{2})(2+i\sqrt{2}) \\
+ 	     & =x^2-4x+(4-2i^2)                                              \\
+ 	     & =x^2-4x+6                                                     
+ \end{align*}
+ We conclude that a possible formula for a polynomial function, $p$, 
+ that has zeros at $2\pm i\sqrt{2}$ is
+ \[
+ 	p(x)=x^2-4x+6
+ \]
+ Note that we could multiply $p$ by any real number and still ensure
+ that $p$ has the same zeros.
+ \end{pccsolution}
+ \end{pccexample}
+ \investigation*{}
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Find a formula from a graph]
+ For each of the polynomials in \cref{poly:fig:findformula}
+ \begin{enumerate}
+ 	\item count the number of times the curve turns round, and cuts/bounces off the $x$ axis;
+ 	\item approximate the degree of the polynomial;
+ 	\item use your information to find the linear factors of each polynomial, and therefore write a possible formula for each;
+ 	\item make sure your polynomial goes through the given ordered pair.
+ \end{enumerate}
+ \begin{shortsolution}
+ 	\Vref{poly:fig:findformdeg2}: 
+ 	\begin{enumerate}
+ 		\item the curve turns round once;
+ 		\item the degree could be 2;
+ 		\item based on the zeros, the linear factors are $(x+5)$ and $(x-3)$; since the 
+ 		graph opens downwards, we will assume the leading coefficient is negative: $p(x)=-k(x+5)(x-3)$;
+ 		\item $p$ goes through $(2,2)$, so we need to solve $2=-k(7)(-1)$ and therefore $k=\nicefrac{2}{7}$, so
+ 		\[
+ 			p(x)=-\frac{2}{7}(x+5)(x-3)
+ 		\]
+ 	\end{enumerate}
+ 	\Vref{poly:fig:findformdeg3}:
+ 	\begin{enumerate}
+ 		\item the curve turns around twice;
+ 		\item the degree could be 3;
+ 		\item based on the zeros, the linear factors are $(x+2)^2$, and $(x-1)$;
+ 		based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+2)^2(x-1)$;
+ 		\item $p$ goes through $(0,-2)$, so we need to solve $-2=k(4)(-1)$ and therefore $k=\nicefrac{1}{2}$, so
+ 		\[
+ 			p(x)=\frac{1}{2}(x+2)^2(x-1)
+ 		\]
+ 	\end{enumerate}
+ 	\Vref{poly:fig:findformdeg5}:
+ 	\begin{enumerate}
+ 		\item the curve turns around 4 times;
+ 		\item the degree could be 5;
+ 		\item based on the zeros, the linear factors are $(x+5)^2$, $(x+1)$, $(x-2)$, $(x-3)$;
+ 		based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+5)^2(x+1)(x-2)(x-3)$;
+ 		\item $p$ goes through $(-3,-50)$, so we need to solve $-50=k(64)(-2)(-5)(-6)$ and therefore $k=\nicefrac{5}{384}$, so
+ 		\[
+ 			p(x)=\frac{5}{384}(x+5)^2(x+1)(x-2)(x-3)
+ 		\]
+ 	\end{enumerate}
+ \end{shortsolution}
+ \end{problem}
+  	  
+  	  
+ \begin{figure}[!htb]
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-2,ymax=5,
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-4.5:3.75]{-1/3*(x+4)*(x-3)};
+ 				\addplot[soldot] coordinates{(-4,0)(3,0)(2,2)} node[axisnode,above right]{$(2,2)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:findformdeg2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-3,xmax=2,
+ 					ymin=-2,ymax=4,
+ 					xtick={-2,...,1},
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-2.95:1.75]{1/3*(x+2)^2*(x-1)};
+ 				\addplot[soldot]coordinates{(-2,0)(1,0)(0,-1.33)}node[axisnode,anchor=north west]{$(0,-2)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:findformdeg3}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-100,ymax=150,
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-4.5:3.4,samples=50]{(x+4)^2*(x+1)*(x-2)*(x-3)};
+ 				\addplot[soldot]coordinates{(-4,0)(-1,0)(2,0)(3,0)(-3,-60)}node[axisnode,anchor=north]{$(-3,-50)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:findformdeg5}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\label{poly:fig:findformula}
+ \end{figure}
+  	  
+  	  
+  	  
+  	  
+ \begin{exercises}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Prerequisite classifacation skills]
+ Decide if each of the following functions are linear or quadratic.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$f(x)=2x+3$ 
+ 		\begin{shortsolution}
+ 			$f$ is linear. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$g(x)=10-7x$ 
+ 		\begin{shortsolution}
+ 			$g$ is linear 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$h(x)=-x^2+3x-9$ 
+ 		\begin{shortsolution}
+ 			$h$ is quadratic. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$k(x)=-17$ 
+ 		\begin{shortsolution}
+ 			$k$ is linear.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$l(x)=-82x^2-4$
+ 		\begin{shortsolution}
+ 			$l$ is quadratic 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$m(x)=6^2x-8$ 
+ 		\begin{shortsolution}
+ 			$m$ is linear. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Prerequisite slope identification]
+ State the slope of each of the following linear functions, and 
+ hence decide if each function is increasing or decreasing.
+ \begin{multicols}{4}
+ 	\begin{subproblem}
+ 		$\alpha(x)=4x+1$ 
+ 		\begin{shortsolution}
+ 			$m=4$; $\alpha$ is increasing. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$\beta(x)=-9x$ 
+ 		\begin{shortsolution}
+ 			$m=-9$; $\beta$ is decreasing. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$\gamma(t)=18t+100$ 
+ 		\begin{shortsolution}
+ 			$m=18$; $\gamma$ is increasing.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$\delta(y)=23-y$ 
+ 		\begin{shortsolution}
+ 			$m=-1$; $\delta$ is decreasing. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ Now let's generalize our findings for the most general linear function $f$
+ that has formula $f(x)=mx+b$. Complete the following sentences.
+ \begin{subproblem}
+ 	When $m>0$, the function $f$ is $\ldots$  
+ 	\begin{shortsolution}
+ 		When $m>0$, the function $f$ is $\ldots$  \emph{increasing}.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	When $m<0$, the function $f$ is $\ldots$  
+ 	\begin{shortsolution}
+ 		When $m<0$, the function $f$ is $\ldots$  \emph{decreasing}.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Polynomial or not?]
+ Identify whether each of the following functions is a polynomial or not. 
+ If the function is a polynomial, state its degree.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$p(x)=2x+1$
+ 		\begin{shortsolution}
+ 			$p$ is a polynomial (you might also describe $p$ as linear). The degree of $p$ is 1.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=7x^2+4x$
+ 		\begin{shortsolution}
+ 			$p$ is a polynomial (you might also describe $p$ as quadratic). The degree of $p$ is 2.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=\sqrt{x}+2x+1$
+ 		\begin{shortsolution}
+ 			$p$ is not a polynomial; we require the powers of $x$ to be integer values.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=2^x-45$
+ 		\begin{shortsolution}
+ 			$p$ is not a polynomial; the $2^x$ term is exponential.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=6x^4-5x^3+9$
+ 		\begin{shortsolution}
+ 			$p$ is a polynomial, and the degree of $p$ is $6$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=-5x^{17}+9x+2$
+ 		\begin{shortsolution}
+ 			$p$ is a polynomial, and the degree of $p$ is 17.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=4x(x+7)^2(x-3)^3$ 
+ 		\begin{shortsolution}
+ 			$p$ is a polynomial, and the degree of $p$ is $6$.     
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=4x^{-5}-x^2+x$ 
+ 		\begin{shortsolution}
+ 			$p$ is not a polynomial because $-5$ is not a positive integer. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=-x^6(x^2+1)(x^3-2)$ 
+ 		\begin{shortsolution}
+ 			$p$ is a polynomial, and the degree of $p$ is $11$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Polynomial graphs]
+ Three polynomial functions $p$, $m$, and $n$ are shown in \crefrange{poly:fig:functionp}{poly:fig:functionn}.
+ The functions have the following formulas
+ \begin{align*}
+ 	p(x) & = (x-1)(x+2)(x-3)           \\ 
+ 	m(x) & = -(x-1)(x+2)(x-3)          \\
+ 	n(x) & = (x-1)(x+2)(x-3)(x+1)(x+4) 
+ \end{align*}
+ Note that for our present purposes we are not concerned with the vertical scale of the graphs.
+ \begin{subproblem}
+ 	Identify both on the graph {\em and} algebraically, the zeros of each polynomial.
+ 	\begin{shortsolution}
+ 		$y=p(x)$ is shown below.
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)};
+ 				\addplot[soldot] coordinates{(-2,0)(1,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		 			 		 		
+ 		$y=m(x)$ is shown below.
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)};
+ 				\addplot[soldot] coordinates{(-2,0)(1,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		 			 		 		
+ 		$y=n(x)$ is shown below.
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-90,ymax=70,
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot expression[domain=-4.15:3.15,samples=50]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)};
+ 				\addplot[soldot] coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		 			 		 		
+ 		The zeros of $p$ are $-2$, $1$, and $3$; the zeros of $m$ are $-2$, $1$, and $3$; the zeros of $n$ are 
+ 		$-4$, $-2$, $-1$, and $3$.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Write down the degree, how many times the curve of each function `turns around', 
+ 	and how many zeros it has
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item The degree of $p$ is 3, and the curve $y=p(x)$ turns around twice. 
+ 			\item The degree of $q$ is also 3, and the curve $y=q(x)$ turns around twice. 
+ 			\item The degree of $n$ is $5$, and the curve $y=n(x)$ turns around 4 times.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+  	  
+ \begin{figure}[!htb]
+ 	\begin{widepage}
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)};
+ 				\addplot[soldot]coordinates{(-2,0)(1,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=p(x)$}
+ 		\label{poly:fig:functionp}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)};
+ 				\addplot[soldot]coordinates{(-2,0)(1,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=m(x)$}
+ 		\label{poly:fig:functionm}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-90,ymax=70,
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-4.15:3.15,samples=100]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)};
+ 				\addplot[soldot]coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=n(x)$}
+ 		\label{poly:fig:functionn}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\end{widepage}
+ \end{figure}
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Horizontal intercepts]\label{poly:prob:matchpolys}%
+ State the horizontal intercepts (as ordered pairs) of the following polynomials.
+ \begin{multicols}{2}
+ 	\begin{subproblem}\label{poly:prob:degree5}
+ 		$p(x)=(x-1)(x+2)(x-3)(x+1)(x+4)$
+ 		\begin{shortsolution}
+ 			$(-4,0)$, $(-2,0)$, $(-1,0)$, $(1,0)$, $(3,0)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$q(x)=-(x-1)(x+2)(x-3)$
+ 		\begin{shortsolution}
+ 			$(-2,0)$, $(1,0)$, $(3,0)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(x)=(x-1)(x+2)(x-3)$
+ 		\begin{shortsolution}
+ 			$(-2,0)$, $(1,0)$, $(3,0)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}\label{poly:prob:degree2}
+ 		$s(x)=(x-2)(x+2)$
+ 		\begin{shortsolution}
+ 			$(-2,0)$, $(2,0)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Minimums, maximums, and concavity]\label{poly:prob:incdec}
+ Four polynomial functions are graphed in \cref{poly:fig:incdec}. The formulas
+ for these functions are (not respectively)
+ \begin{gather*}
+ 	p(x)=\frac{x^3}{6}-\frac{x^2}{4}-3x, \qquad q(x)=\frac{x^4}{20}+\frac{x^3}{15}-\frac{6}{5}x^2+1\\
+ 	r(x)=-\frac{x^5}{50}-\frac{x^4}{40}+\frac{2x^3}{5}+6, \qquad s(x)=-\frac{x^6}{6000}-\frac{x^5}{2500}+\frac{67x^4}{4000}+\frac{17x^3}{750}-\frac{42x^2}{125}
+ \end{gather*}
+ \begin{figure}[!htb]
+ 	\begin{widepage}
+ 	\setlength{\figurewidth}{.23\textwidth}
+ 	\centering
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					width=\textwidth,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-8,-6,...,8},
+ 					grid=major,
+ 				]
+ 				\addplot expression[domain=-5.28:4.68,samples=50]{-x^5/50-x^4/40+2*x^3/5+6};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:incdec3}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					width=\textwidth,
+ 					xmin=-10,xmax=10,ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-8,-6,...,8},
+ 					grid=major,
+ 				]
+ 				\addplot expression[domain=-6.08:4.967,samples=50]{x^4/20+x^3/15-6/5*x^2+1};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:incdec2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					width=\textwidth,
+ 					xmin=-6,xmax=8,ymin=-10,ymax=10,
+ 					xtick={-4,-2,...,6},
+ 					ytick={-8,-4,4,8},
+ 					minor ytick={-6,-2,...,6},
+ 					grid=both,
+ 				]
+ 				\addplot expression[domain=-4.818:6.081,samples=50]{x^3/6-x^2/4-3*x};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:incdec1}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					width=\textwidth,
+ 					xmin=-10,xmax=10,ymin=-10,ymax=10,
+ 					xtick={-8,-4,4,8},
+ 					ytick={-8,-4,4,8},
+ 					minor xtick={-6,-2,...,6},
+ 					minor ytick={-6,-2,...,6},
+ 					grid=both,
+ 				]
+ 				\addplot expression[domain=-9.77:8.866,samples=50]{-x^6/6000-x^5/2500+67*x^4/4000+17/750*x^3-42/125*x^2};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:incdec4}
+ 	\end{subfigure}
+ 	\caption{Graphs for \cref{poly:prob:incdec}.}
+ 	\label{poly:fig:incdec}
+ 	\end{widepage}
+ \end{figure}
+ \begin{subproblem}
+ 	Match each of the formulas with one of the given graphs.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item $p$ is graphed in \vref{poly:fig:incdec1};
+ 			\item $q$ is graphed in \vref{poly:fig:incdec2};
+ 			\item $r$ is graphed in \vref{poly:fig:incdec3};
+ 			\item $s$ is graphed in \vref{poly:fig:incdec4}.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Approximate the zeros of each function using the appropriate graph.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item $p$ has simple zeros at about $-3.8$, $0$, and $5$.
+ 			\item $q$ has simple zeros at about $-5.9$, $-1$, $1$, and $4$.
+ 			\item $r$ has simple zeros at about $-5$, $-2.9$, and $4.1$.
+ 			\item $s$ has simple zeros at about $-9$, $-6$, $4.2$, $8.1$, and a zero of multiplicity $2$ at $0$.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Approximate the local maximums and minimums of each of the functions.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item $p$ has a local maximum of approximately $3.9$ at $-2$, and a local minimum of approximately $-6.5$ at $3$.
+ 			\item $q$ has a local minimum of approximately $-10$ at $-4$, and $-4$ at $3$; $q$ has a local maximum of approximately $1$ at $0$.
+ 			\item $r$ has a local minimum of approximately $-5.5$ at $-4$, and a local maximum of approximately $10$ at $3$.
+ 			\item $s$ has a local maximum of approximately $5$ at $-8$, $0$ at $0$, and $5$ at  $7$; $s$ has local minimums 
+ 			of approximately $-3$ at $-4$, and $-1$ at $3$.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Approximate the global maximums and minimums of each of the functions.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item $p$ does not have a global maximum, nor a global minimum.
+ 			\item $q$ has a global minimum of approximately $-10$; it does not have a global maximum.
+ 			\item $r$ does not have a global maximum, nor a global minimum.
+ 			\item $s$ has a global maximum of approximately $5$; it does not have a global minimum.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Approximate the intervals on which each function is increasing and decreasing.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item $p$ is increasing on $(-\infty,-2)\cup (3,\infty)$, and decreasing on $(-2,3)$.
+ 			\item $q$ is increasing on $(-4,0)\cup (3,\infty)$, and decreasing on $(-\infty,-4)\cup (0,3)$.
+ 			\item $r$ is increasing on $(-4,3)$, and decreasing on $(-\infty,-4)\cup (3,\infty)$.
+ 			\item $s$ is increasing on $(-\infty,-8)\cup (-4,0)\cup (3,5)$, and decreasing on $(-8,-4)\cup (0,3)\cup (5,\infty)$.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Approximate the intervals on which each function is concave up and concave down.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item $p$ is concave up on  $(1,\infty)$, and concave down on  $(-\infty,1)$.
+ 			\item $q$ is concave up on $(-\infty,-1)\cup (1,\infty)$, and concave down on $(-1,1)$.
+ 			\item $r$ is concave up on $(-\infty,-3)\cup (0,2)$, and concave down on $(-3,0)\cup (2,\infty)$.
+ 			\item $s$ is concave up on $(-6,-2)\cup (2,5)$, and concave down on $(-\infty,-6)\cup (-2,2)\cup (5,\infty)$.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	The degree of $q$ is $5$. Assuming that all of the real zeros of $q$ are
+ 	shown in its graph, how many complex zeros does $q$ have?
+ 	\begin{shortsolution}
+ 		\Vref{poly:fig:incdec2} shows that $q$ has $3$ real zeros 
+ 		since the curve of $q$ cuts the horizontal axis $3$ times. 
+ 		Since $q$ has degree $5$, $q$ must have $2$ complex zeros.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Long-run behaviour of polynomials]
+ Describe the long-run behavior of each of polynomial functions in 
+ \crefrange{poly:prob:degree5}{poly:prob:degree2}.
+ \begin{shortsolution}
+ 	$\dd\lim_{x\rightarrow-\infty}p(x)=-\infty$,
+ 	$\dd\lim_{x\rightarrow\infty}p(x)=\infty$,
+ 	$\dd\lim_{x\rightarrow-\infty}q(x)=\infty$,
+ 	$\dd\lim_{x\rightarrow\infty}q(x)=-\infty$,
+ 	$\dd\lim_{x\rightarrow-\infty}r(x)=-\infty$,
+ 	$\dd\lim_{x\rightarrow\infty}r(x)=\infty$,
+ 	$\dd\lim_{x\rightarrow-\infty}s(x)=\infty$,
+ 	$\dd\lim_{x\rightarrow\infty}s(x)=\infty$,
+ \end{shortsolution}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[True of false?]
+ Let $p$ be a polynomial function. 
+ Label each of the following statements as true (T) or false (F); if they are false, 
+ provide an example that supports your answer.
+ \begin{subproblem}
+ 	If $p$ has degree $3$, then $p$ has $3$ distinct zeros.
+ 	\begin{shortsolution}
+ 		False. Consider $p(x)=x^2(x+1)$ which has only 2 distinct zeros.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	If $p$ has degree $4$, then $\dd\lim_{x\rightarrow-\infty}p(x)=\infty$ and $\dd\lim_{x\rightarrow\infty}p(x)=\infty$.
+ 	\begin{shortsolution}
+ 		False. Consider $p(x)=-x^4$.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	If $p$ has even degree, then it is possible that $p$ can have no real zeros.
+ 	\begin{shortsolution}
+ 		True.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	If $p$ has odd degree, then it is possible that $p$ can have no real zeros.
+ 	\begin{shortsolution}
+ 		False. All odd degree polynomials will cut the horizontal axis at least once.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Find a formula from a description]
+ In each of the following problems, give a possible formula for a polynomial 
+ function that has the specified properties.
+ \begin{subproblem}
+ 	Degree 2 and has zeros at $4$ and $5$.
+ 	\begin{shortsolution}
+ 		Possible option: $p(x)=(x-4)(x-5)$. Note we could multiply $p$ by any real number, and still meet the requirements.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Degree 3 and has zeros at $4$,$5$ and $-3$.
+ 	\begin{shortsolution}
+ 		Possible option: $p(x)=(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Degree 4 and has zeros at $0$, $4$, $5$, $-3$.
+ 	\begin{shortsolution}
+ 		Possible option: $p(x)=x(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Degree 4, with zeros that make the graph cut at $2$, $-5$, and a zero that makes the graph touch at $-2$;
+ 	\begin{shortsolution}
+ 		Possible option: $p(x)=(x-2)(x+5)(x+2)^2$. Note we could multiply $p$ by any real number, and still meet the requirements.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Degree 3, with only one zero at $-1$.
+ 	\begin{shortsolution}
+ 		Possible option: $p(x)=(x+1)^3$. Note we could multiply $p$ by any real number, and still meet the requirements.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{problem}[\Cref{poly:step:last}]
+ \pccname{Saheed} is graphing a polynomial function, $p$. 
+ He is following \crefrange{poly:step:first}{poly:step:last} and has so far
+ marked the zeros of $p$ on \cref{poly:fig:optionsp1}. Saheed tells you that 
+ $p$ has degree $3$, but does \emph{not} say if the leading coefficient 
+ of $p$ is positive or negative.
+ \begin{figure}[!htbp]
+ 	\begin{widepage}
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-15},
+ 					width=\textwidth,
+ 					height=.5\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:optionsp1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-15},
+ 					width=\textwidth,
+ 					height=.5\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-5,0)(6,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{poly:fig:optionsp2}
+ 	\end{subfigure}%
+ 	\caption{}
+ 	\end{widepage}
+ \end{figure}
+ \begin{subproblem}
+ 	Use the information in \cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient
+ 	is positive.
+ 	\begin{shortsolution}
+ 		Assuming that $a_3>0$:
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-15},
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot expression[domain=-6.78179:8.35598,samples=50]{1/20*(x+5)*(x-2)*(x-6)};
+ 				\addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Use the information in \cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient
+ 	is negative.
+ 	\begin{shortsolution}
+ 		Assuming that $a_3<0$:
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-15},
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot expression[domain=-6.78179:8.35598,samples=50]{-1/20*(x+5)*(x-2)*(x-6)};
+ 				\addplot[soldot] coordinates{(-5,0)(2,0)(6,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 	\end{shortsolution}
+ \end{subproblem}
+ Saheed now turns his attention to another polynomial function, $q$. He finds
+ the zeros of $q$ (there are only $2$) and marks them on \cref{poly:fig:optionsp2}. 
+ Saheed knows that $q$ has degree $3$, but doesn't know if the leading 
+ coefficient is positive or negative.
+ \begin{subproblem}
+ 	Use the information in \cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading 
+ 	coefficient of $q$ is positive. Hint: only one of the zeros is simple.
+ 	\begin{shortsolution}
+ 		Assuming that $a_4>0$ there are $2$ different options:
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-15},
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot expression[domain=-8.68983:7.31809,samples=50]{1/20*(x+5)^2*(x-6)};
+ 				\addplot expression[domain=-6.31809:9.68893,samples=50]{1/20*(x+5)*(x-6)^2};
+ 				\addplot[soldot] coordinates{(-5,0)(6,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Use the information in \cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading 
+ 	coefficient of $q$ is negative.
+ 	\begin{shortsolution}
+ 		Assuming that $a_4<0$ there are $2$ different options:
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-15},
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot expression[domain=-8.68983:7.31809,samples=50]{-1/20*(x+5)^2*(x-6)};
+ 				\addplot expression[domain=-6.31809:9.68893,samples=50]{-1/20*(x+5)*(x-6)^2};
+ 				\addplot[soldot] coordinates{(-5,0)(6,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{problem}[Zeros]
+ Find all zeros of each of the following polynomial functions, making 
+ sure to detail their multiplicity. Note that
+ you may need to use factoring, or the quadratic formula, or both! Also note 
+ that some zeros may be repeated, and some may be complex.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$p(x)=x^2+1$ 
+ 		\begin{shortsolution}
+ 			$\pm i$ (simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$q(y)=(y^2-9)(y^2-7)$ 
+ 		\begin{shortsolution}
+ 			$\pm 3$, $\pm \sqrt{7}$ (all are simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(z)=-4z^3(z^2+3)(z^2+64)$ 
+ 		\begin{shortsolution}
+ 			$0$ (multiplicity $3$), $\pm\sqrt{3}$ (simple), $\pm\sqrt{8}$ (simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$a(x)=x^4-81$ 
+ 		\begin{shortsolution}
+ 			$\pm 3$, $\pm 3i$ (all are simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$b(y)=y^3-8$ 
+ 		\begin{shortsolution}
+ 			$2$, $-1\pm i\sqrt{3}$ (all are simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$c(m)=m^3-m^2$
+ 		\begin{shortsolution}
+ 			$0$ (multiplicity $2$), $1$ (simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$h(n)=(n+1)(n^2+4)$ 
+ 		\begin{shortsolution}
+ 			$-1$, $\pm 2i$ (all are simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$f(\alpha)=(\alpha^2-16)(\alpha^2-5\alpha+4)$ 
+ 		\begin{shortsolution}
+ 			$-4$ (simple), $4$ (multiplicity $2$), $1$ (simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$g(\beta)=(\beta^2-25)(\beta^2-5\beta-4)$ 
+ 		\begin{shortsolution}
+ 			$\pm 5$, $\dfrac{5\pm\sqrt{41}}{2}$ (all are simple).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{problem}[Given zeros, find a formula]
+ In each of the following problems you are given the zeros of a polynomial. 
+ Write a possible formula for each polynomial| you may leave your 
+ answer in factored form, but it may not contain complex numbers. Unless
+ otherwise stated, assume that the zeros are simple.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$1$, $2$ 
+ 		\begin{shortsolution}
+ 			$p(x)=(x-1)(x-2)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$0$, $5$, $13$ 
+ 		\begin{shortsolution}
+ 			$p(x)=x(x-5)(x-13)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$-7$, $2$ (multiplicity $3$), $5$ 
+ 		\begin{shortsolution}
+ 			$p(x)=(x+7)(x-2)^3(x-5)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$0$, $\pm i$ 
+ 		\begin{shortsolution}
+ 			$p(x)=x(x^2+1)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$\pm 2i$, $\pm 7$ 
+ 		\begin{shortsolution}
+ 			$p(x)=(x^2+4)(x^2-49)$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$-2\pm i\sqrt{6}$ 
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{problem}[Composition of polynomials]
+ Let $p$ and $q$ be polynomial functions that have formulas
+ \[
+ 	p(x)=(x+1)(x+2)(x+5), \qquad q(x)=3-x^4
+ \]
+ Evaluate each of the following.
+ \begin{multicols}{4}
+ 	\begin{subproblem}
+ 		$(p\circ q)(0)$ 
+ 		\begin{shortsolution}
+ 			$160$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(q\circ p)(0)$ 
+ 		\begin{shortsolution}
+ 			$-9997$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(p\circ q)(1)$ 
+ 		\begin{shortsolution}
+ 			$84$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(p\circ p)(0)$ 
+ 		\begin{shortsolution}
+ 			$1980$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{problem}[Piecewise polynomial functions]
+ Let $P$ be the piecewise-defined function with formula
+ \[
+ 	P(x)=\begin{cases}
+ 	(1-x)(2x+5)(x^2+1), &   x\leq -3\\
+ 	4-x^2,              &   -3<x < 4\\
+ 	x^3                 &   x\geq 4
+ \end{cases}
+\]
+Evaluate each of the following 
+\begin{multicols}{5}
+	\begin{subproblem}
+		$P(-4)$ 
+		\begin{shortsolution}
+			$-255$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$P(0)$ 
+		\begin{shortsolution}
+			$4$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$P(4)$ 
+		\begin{shortsolution}
+			$64$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$P(-3)$ 
+		\begin{shortsolution}
+			$-40$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(P\circ P)(0)$ 
+		\begin{shortsolution}
+			$64$
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+ 
+%===================================
+%   Author: Hughes
+%   Date:   July 2012
+%===================================
+\begin{problem}[Function algebra]
+Let $p$ and $q$ be the polynomial functions that have formulas
+\[
+	p(x)=x(x+1)(x-3)^2, \qquad q(x)=7-x^2
+\]
+Evaluate each of the following (if possible).
+\begin{multicols}{4}
+	\begin{subproblem}
+		$(p+q)(1)$ 
+		\begin{shortsolution}
+			$14$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(p-q)(0)$ 
+		\begin{shortsolution}
+			$7$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$(p\cdot q)(\sqrt{7})$ 
+		\begin{shortsolution}
+			$0$ 
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		$\left( \frac{q}{p} \right)(1)$ 
+		\begin{shortsolution}
+			$\frac{3}{4}$ 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\begin{subproblem}
+	What is the domain of the function $\frac{q}{p}$?
+	\begin{shortsolution}
+		$(-\infty,-1)\cup (-1,0)\cup (0,3)\cup (3,\infty)$ 
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+ 
+%===================================
+%   Author: Hughes
+%   Date:   July 2012
+%===================================
+\begin{problem}[Transformations: given the transformation, find the formula]
+Let $p$ be the polynomial function that has formula.
+\[
+	p(x)=4x(x^2-1)(x+3)
+\]
+In each of the following 
+problems apply the given transformation to the function $p$ and 
+write a formula for the transformed version of $p$.
+\begin{multicols}{2}
+	\begin{subproblem}
+		Shift $p$ to the right by $5$ units. 
+		\begin{shortsolution}
+			$p(x-5)=4(x-5)(x-2)(x^2-10x+24)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $p$ to the left by $6$ units. 
+		\begin{shortsolution}
+			$p(x+6)=4(x+6)(x+9)(x^2+12x+35)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $p$ up by $12$ units. 
+		\begin{shortsolution}
+			$p(x)+12=4x(x^2-1)(x+3)+12$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Shift $p$ down by $2$ units. 
+		\begin{shortsolution}
+			$p(x)-2=4x(x^2-1)(x+3)-2$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Reflect $p$ over the horizontal axis.
+		\begin{shortsolution}
+			$-p(x)=-4x(x^2-1)(x+3)$
+		\end{shortsolution}
+	\end{subproblem}
+	\begin{subproblem}
+		Reflect $p$ over the vertical axis.
+		\begin{shortsolution}
+			$p(-x)=-4x(x^2-1)(3-x)$ 
+		\end{shortsolution}
+	\end{subproblem}
+\end{multicols}
+\end{problem}
+ 
+%===================================
+%   Author: Hughes
+%   Date:   May 2011
+%===================================
+\begin{problem}[Find a formula from a table]\label{poly:prob:findformula}
+\Crefrange{poly:tab:findformulap}{poly:tab:findformulas} show values of polynomial functions, $p$, $q$, 
+$r$, and $s$.
+ 
+\begin{table}[!htb]
+	\centering
+	\begin{widepage}
+	\caption{Tables for \cref{poly:prob:findformula}}
+	\label{poly:tab:findformula}
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=p(x)$}
+		\label{poly:tab:findformulap}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ 
+			\afterheading
+			$-4$          & $-56$         \\\normalline
+			$-3$          & $-18$         \\\normalline
+			$-2$          & $0$           \\\normalline
+			$-1$          & $4$           \\\normalline
+			$0$           & $0$           \\\normalline
+			$1$           & $-6$          \\\normalline
+			$2$           & $-8$          \\\normalline
+			$3$           & $0$           \\\normalline
+			$4$           & $24$          \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=q(x)$}
+		\label{poly:tab:findformulaq}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ \afterheading
+			$-4$          & $-16$         \\\normalline
+			$-3$          & $-3$          \\\normalline
+			$-2$          & $0$           \\\normalline
+			$-1$          & $-1$          \\\normalline
+			$0$           & $0$           \\\normalline
+			$1$           & $9$           \\\normalline
+			$2$           & $32$          \\\normalline
+			$3$           & $75$          \\\normalline
+			$4$           & $144$         \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=r(x)$}
+		\label{poly:tab:findformular}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ \afterheading
+			$-4$          & $105$         \\\normalline
+			$-3$          & $0$           \\\normalline
+			$-2$          & $-15$         \\\normalline
+			$-1$          & $0$           \\\normalline
+			$0$           & $9$           \\\normalline
+			$1$           & $0$           \\\normalline
+			$2$           & $-15$         \\\normalline
+			$3$           & $0$           \\\normalline
+			$4$           & $105$         \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\hfill
+	\begin{subtable}{.2\textwidth}
+		\centering
+		\caption{$y=s(x)$}
+		\label{poly:tab:findformulas}
+		\begin{tabular}{rr}
+			\beforeheading
+			\heading{$x$} & \heading{$y$} \\ \afterheading
+			$-4$          & $75$          \\\normalline
+			$-3$          & $0$           \\\normalline
+			$-2$          & $-9$          \\\normalline
+			$-1$          & $0$           \\\normalline
+			$0$           & $3$           \\\normalline
+			$1$           & $0$           \\\normalline
+			$2$           & $15$          \\\normalline
+			$3$           & $96$          \\\normalline
+			$4$           & $760$         \\\lastline
+		\end{tabular}
+	\end{subtable}
+	\end{widepage}
+\end{table}
+ 
+\begin{subproblem}
+	Assuming that all of the zeros of $p$ are shown (in \cref{poly:tab:findformulap}), how many zeros does $p$ have? 
+	\begin{shortsolution}
+		$p$ has 3 zeros.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	What is the degree of $p$?
+	\begin{shortsolution}
+		$p$ is degree 3.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Write a formula for $p(x)$.
+	\begin{shortsolution}
+		$p(x)=x(x+2)(x-3)$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Assuming that all of the zeros of $q$ are shown (in \cref{poly:tab:findformulaq}), how many zeros does $q$ have?
+	\begin{shortsolution}
+		$q$ has 2 zeros.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Describe the difference in behavior of $p$ and $q$ at $-2$.
+	\begin{shortsolution}
+		$p$ changes sign at $-2$, and $q$ does not change sign at $-2$.
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Given that $q$ is a degree-$3$ polynomial, write a formula for $q(x)$.
+	\begin{shortsolution}
+		$q(x)=x(x+2)^2$
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Assuming that all of the zeros of $r$ are shown (in \cref{poly:tab:findformular}), find a formula for $r(x)$.
+	\begin{shortsolution}
+		$r(x)=(x+3)(x+1)(x-1)(x-3)$ 
+	\end{shortsolution}
+\end{subproblem}
+\begin{subproblem}
+	Assuming that all of the zeros of $s$ are shown (in \cref{poly:tab:findformulas}), find a formula for $s(x)$.
+	\begin{shortsolution}
+		$s(x)=(x+3)(x+1)(x-1)^2$ 
+	\end{shortsolution}
+\end{subproblem}
+\end{problem}
+\end{exercises}
+ 
+\section{Rational functions}
+ \subsection*{Power functions with negative exponents}
+ The study of rational functions will rely upon a good knowledge 
+ of power functions with negative exponents. \Cref{rat:ex:oddpow,rat:ex:evenpow} are 
+ simple but fundamental to understanding the behavior of rational functions.
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Power functions with odd negative exponents]\label{rat:ex:oddpow}
+ Graph each of the following functions on your calculator, state their domain in interval notation, and their 
+ behavior as $x\rightarrow 0^-$ and $x\rightarrow 0^+$.
+ \[
+ 	f(x)=\frac{1}{x},\qquad g(x)=\dfrac{1}{x^3},\qquad h(x)=\dfrac{1}{x^5}
+ \]
+ \begin{pccsolution}
+ The functions $f$, $g$, and $k$ are plotted in \cref{rat:fig:oddpow}.
+ The domain of each of the functions $f$, $g$, and $h$ is $(-\infty,0)\cup (0,\infty)$. Note that 
+ the long-run behavior of each of the functions is the same, and in particular
+ \begin{align*}
+ 	f(x)\rightarrow 0                         & \text{ as } x\rightarrow\infty  \\
+ 	\mathllap{\text{and }}  f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty 
+ \end{align*}
+ The same results hold for $g$ and $h$. Note also that each of the functions
+ has a \emph{vertical asymptote} at $0$. We see that
+ \begin{align*}
+ 	f(x)\rightarrow -\infty                         & \text{ as } x\rightarrow 0^- \\
+ 	\mathllap{\text{and }}   f(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ 
+ \end{align*}
+ The same results hold for $g$ and $h$.
+  	  
+ The curve of a function that has a vertical asymptote is necessarily separated 
+ into \emph{branches}| each of the functions $f$, $g$, and $h$ have $2$ branches.
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ \begin{figure}[!htb]
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-3,xmax=3,
+ 					ymin=-5,ymax=5,
+ 					xtick={-2,-1,...,2},
+ 					minor ytick={-3,-1,...,3},
+ 					grid=both,
+ 					width=\textwidth,
+ 					legend pos=north west,
+ 				]
+ 				\addplot expression[domain=-3:-0.2]{1/x};
+ 				\addplot expression[domain=-3:-0.584]{1/x^3};
+ 				\addplot expression[domain=-3:-0.724]{1/x^5};
+ 				\addplot expression[domain=0.2:3]{1/x};
+ 				\addplot expression[domain=0.584:3]{1/x^3};
+ 				\addplot expression[domain=0.724:3]{1/x^5};
+ 				\addplot[soldot]coordinates{(-1,-1)}node[axisnode,anchor=north east]{$(-1,-1)$};
+ 				\addplot[soldot]coordinates{(1,1)}node[axisnode,anchor=south west]{$(1,1)$};
+ 				\legend{$f$,$g$,$h$}
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:oddpow}
+ 	\end{minipage}%
+ 	\hfill
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-3,xmax=3,
+ 					ymin=-5,ymax=5,
+ 					xtick={-2,-1,...,2},
+ 					minor ytick={-3,-1,...,3},
+ 					grid=both,
+ 					width=\textwidth,
+ 					legend pos=south east,
+ 				]
+ 				\addplot expression[domain=-3:-0.447]{1/x^2};
+ 				\addplot expression[domain=-3:-0.668]{1/x^4};
+ 				\addplot expression[domain=-3:-0.764]{1/x^6};
+ 				\addplot expression[domain=0.447:3]{1/x^2};
+ 				\addplot expression[domain=0.668:3]{1/x^4};
+ 				\addplot expression[domain=0.764:3]{1/x^6};
+ 				\addplot[soldot]coordinates{(-1,1)}node[axisnode,anchor=south east]{$(-1,1)$};
+ 				\addplot[soldot]coordinates{(1,1)}node[axisnode,anchor=south west]{$(1,1)$};
+ 				\legend{$F$,$G$,$H$}
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:evenpow}
+ 	\end{minipage}%
+ \end{figure}
+  	  
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Power functions with even negative exponents]\label{rat:ex:evenpow}%
+ Graph each of the following functions, state their domain, and their 
+ behavior as $x\rightarrow 0^-$ and $x\rightarrow 0^+$.
+ \[
+ 	f(x)=\frac{1}{x^2},\qquad g(x)=\frac{1}{x^4},\qquad h(x)=\frac{1}{x^6}
+ \]
+ \begin{pccsolution}
+ The functions $F$, $G$, and $H$ are plotted in \cref{rat:fig:evenpow}.
+ The domain of each of the functions $F$, $G$, and $H$ is $(-\infty,0)\cup (0,\infty)$. Note that 
+ the long-run behavior of each of the functions is the same, and in particular
+ \begin{align*}
+ 	F(x)\rightarrow 0                           & \text{ as } x\rightarrow\infty  \\
+ 	\mathllap{\text{and }}    f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty 
+ \end{align*}
+ As in \cref{rat:ex:oddpow}, $F$ has a horizontal asymptote  that 
+ has equation $y=0$.
+ The same results hold for $G$ and $H$. Note also that each of the functions
+ has a \emph{vertical asymptote} at $0$. We see that
+ \begin{align*}
+ 	F(x)\rightarrow \infty                          & \text{ as } x\rightarrow 0^- \\
+ 	\mathllap{\text{and }}   F(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ 
+ \end{align*}
+ The same results hold for $G$ and $H$. Each of the functions $F$, $G$, and $H$ 
+ have $2$ branches.
+ \end{pccsolution}
+ \end{pccexample}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{doyouunderstand}
+ 	\begin{problem}
+ 	Repeat \cref{rat:ex:oddpow,rat:ex:evenpow} using (respectively)
+ 	\begin{subproblem}
+ 		$k(x)=-\dfrac{1}{x}$, $ m(x)=-\dfrac{1}{x^3}$, $ n(x)=-\dfrac{1}{x^5}$
+ 		\begin{shortsolution}
+ 			The functions $k$, $m$, and $n$ have domain $(-\infty,0)\cup (0,\infty)$, and 
+ 			are graphed below.
+ 			 				 			 			
+ 			\begin{tikzpicture}
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-3,xmax=3,
+ 						ymin=-5,ymax=5,
+ 						xtick={-2,-1,...,2},
+ 						minor ytick={-3,-1,...,3},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 						legend pos=north east,
+ 					]
+ 					\addplot expression[domain=-3:-0.2]{-1/x};
+ 					\addplot expression[domain=-3:-0.584]{-1/x^3};
+ 					\addplot expression[domain=-3:-0.724]{-1/x^5};
+ 					\addplot expression[domain=0.2:3]{-1/x};
+ 					\addplot expression[domain=0.584:3]{-1/x^3};
+ 					\addplot expression[domain=0.724:3]{-1/x^5};
+ 					\legend{$k$,$m$,$n$}
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 			 				 			 			
+ 			Note that
+ 			\begin{align*}
+ 				k(x)\rightarrow 0                                & \text{ as } x\rightarrow\infty  \\
+ 				\mathllap{\text{and }}    k(x)\rightarrow 0      & \text{ as } x\rightarrow-\infty \\
+ 				\intertext{and also}
+ 				k(x)\rightarrow \infty                           & \text{ as } x\rightarrow 0^-    \\
+ 				\mathllap{\text{and }}   k(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+    
+ 			\end{align*}
+ 			The same are true for $m$ and $n$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$ K(x)=-\dfrac{1}{x^2}$, $ M(x)=-\dfrac{1}{x^4}$, $ N(x)=-\dfrac{1}{x^6}$
+ 		\begin{shortsolution}
+ 			The functions $K$, $M$, and $N$ have domain $(-\infty,0)\cup (0,\infty)$, and 
+ 			are graphed below.
+ 			 				 			 			
+ 			\begin{tikzpicture}
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-3,xmax=3,
+ 						ymin=-5,ymax=5,
+ 						xtick={-2,-1,...,2},
+ 						minor ytick={-3,-1,...,3},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 						legend pos=north east,
+ 					]
+ 					\addplot expression[domain=-3:-0.447]{-1/x^2};
+ 					\addplot expression[domain=-3:-0.668]{-1/x^4};
+ 					\addplot expression[domain=-3:-0.764]{-1/x^6};
+ 					\addplot expression[domain=0.447:3]{-1/x^2};
+ 					\addplot expression[domain=0.668:3]{-1/x^4};
+ 					\addplot expression[domain=0.764:3]{-1/x^6};
+ 					\legend{$K$,$M$,$N$}
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 			 				 			 			
+ 			Note that
+ 			\begin{align*}
+ 				K(x)\rightarrow 0                                & \text{ as } x\rightarrow\infty  \\
+ 				\mathllap{\text{and }}    K(x)\rightarrow 0      & \text{ as } x\rightarrow-\infty \\
+ 				\intertext{and also}
+ 				K(x)\rightarrow -\infty                          & \text{ as } x\rightarrow 0^-    \\
+ 				\mathllap{\text{and }}   K(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+    
+ 			\end{align*}
+ 			The same are true for $M$ and $N$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\end{problem}
+ \end{doyouunderstand}
+  	  
+ \subsection*{Rational functions}
+ \begin{pccdefinition}[Rational functions]\label{rat:def:function}
+ Rational functions have the form
+ \[
+ 	r(x) = \frac{p(x)}{q(x)}
+ \]
+ where both $p$ and $q$ are polynomials. 
+  	  
+ Note that
+ \begin{itemize}
+ 	\item the domain or $r$ will be all real numbers, except those that
+ 	make the \emph{denominator}, $q(x)$, equal to $0$;
+ 	\item the zeros of $r$ are the zeros of $p$, i.e the real numbers
+ 	that make the \emph{numerator}, $p(x)$, equal to $0$.
+ \end{itemize}
+  	  
+ \Cref{rat:ex:oddpow,rat:ex:evenpow} are particularly important because $r$ 
+ will behave like $\frac{1}{x}$, or $\frac{1}{x^2}$ around its vertical asymptotes, 
+ depending on the power that the relevant term is raised to| we will demonstrate 
+ this in what follows.
+ \end{pccdefinition}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Rational or not]
+ Identify whether each of the following functions is a rational or not. If 
+ the function is rational, state the domain.
+ \begin{multicols}{3}
+ 	\begin{enumerate}
+ 		\item $r(x)=\dfrac{1}{x}$
+ 		\item $f(x)=2^x+3$
+ 		\item $g(x)=19$
+ 		\item $h(x)=\dfrac{3+x}{4-x}$
+ 		\item $k(x)=\dfrac{x^3+2x}{x-15}$
+ 		\item $l(x)=9-4x$
+ 		\item $m(x)=\dfrac{x+5}{(x-7)(x+9)}$
+ 		\item $n(x)=x^2+6x+7$
+ 		\item $q(x)=1-\dfrac{3}{x+1}$
+ 	\end{enumerate}
+ \end{multicols}
+ \begin{pccsolution}
+ \begin{enumerate}
+ 	\item $r$ is rational; the domain of $r$ is $(-\infty,0)\cup(0,\infty)$.
+ 	\item $f$ is not rational.
+ 	\item $g$ is not rational; $g$ is constant.
+ 	\item $h$ is rational; the domain of $h$ is $(-\infty,4)\cup(4,\infty)$.
+ 	\item $k$ is rational; the domain of $k$ is $(-\infty,15)\cup(15,\infty)$.
+ 	\item $l$ is not rational; $l$ is linear.
+ 	\item $m$ is rational; the domain of $m$ is $(-\infty,-9)\cup(-9,7)\cup(7,\infty)$.
+ 	\item $n$ is not rational; $n$ is quadratic (or you might describe $n$ as a polynomial).
+ 	\item $q$ is rational; the domain of $q$ is $(-\infty,-1)\cup (-1,\infty)$.
+ \end{enumerate}
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Match formula to graph]
+ Each of the following functions is graphed in \cref{rat:fig:whichiswhich}. 
+ Which is which?
+ \[
+ 	r(x)=\frac{1}{x-3}, \qquad q(x)=\frac{x-2}{x+5}, \qquad k(x)=\frac{1}{(x+2)(x-3)}
+ \]
+ \begin{figure}[!htb]
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/(x+5);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-6,ymax=6,
+ 					xtick={-8,-6,...,8},
+ 					minor ytick={-4,-3,...,4},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-6.37]{f};
+ 				\addplot[pccplot] expression[domain=-3.97:10]{f};
+ 				\addplot[soldot] coordinates{(2,0)};
+ 				\addplot[asymptote,domain=-6:6]({-5},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:which1}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=1/(x-3);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-5,ymax=6,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-4,4},
+ 					minor ytick={-3,...,5},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:2.8]{f};
+ 				\addplot[pccplot] expression[domain=3.17:10]{f};
+ 				\addplot[asymptote,domain=-6:6]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:which2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=1/((x-3)*(x+2));}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-5,ymax=5,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-4,4},
+ 					minor ytick={-3,...,3},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-2.03969]{f};
+ 				\addplot[pccplot] expression[domain=-1.95967:2.95967]{f};
+ 				\addplot[pccplot] expression[domain=3.03969:10]{f};
+ 				\addplot[asymptote,domain=-5:5]({-2},{x});
+ 				\addplot[asymptote,domain=-5:5]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:which3}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\label{rat:fig:whichiswhich}
+ \end{figure}
+  	  
+ \begin{pccsolution}
+ Let's start with the function $r$. Note that domain of $r$ is $(-\infty,3)\cup(0,3)$, so 
+ we search for a function that has a vertical asymptote at $3$. There 
+ are two possible choices: the functions graphed in \cref{rat:fig:which2,rat:fig:which3}, 
+ but note that the function in \cref{rat:fig:which3} also has a vertical asymptote at $-2$ 
+ which is not consistent with the formula for $r(x)$. Therefore, $y=r(x)$
+ is graphed in \cref{rat:fig:which2}. 
+  	  
+ The function $q$ has domain $(-\infty,-5)\cup(-5,\infty)$, so we search 
+ for a function that has a vertical asymptote at $-5$. The only candidate 
+ is the curve shown in \cref{rat:fig:which1}; note that the curve also goes through $(2,0)$, 
+ which is consistent with the formula for $q(x)$, since $q(2)=0$, i.e $q$
+ has a zero at $2$.
+  	  
+ The function $k$ has domain $(-\infty,-2)\cup(-2,3)\cup(3,\infty)$, and 
+ has vertical asymptotes at $-2$ and $3$. This is consistent with 
+ the graph in \cref{rat:fig:which3} (and is the only curve that 
+ has $3$ branches).
+  	  
+ We note that each function behaves like $\frac{1}{x}$ around its vertical asymptotes, 
+ because each linear factor in each denominator is raised to the power $1$; if (for example) 
+ the definition of $r$ was instead
+ \[
+ 	r(x)=\frac{1}{(x-3)^2}
+ \]
+ then we would see that $r$ behaves like $\frac{1}{x^2}$ around its vertical asymptote, and 
+ the graph of $r$ would be very different. We will deal with these cases in the examples that follow.
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Repeated factors in the denominator]
+ Consider the functions $f$, $g$, and $h$ that have formulas
+ \[
+ 	f(x)=\frac{x-2}{(x-3)(x+2)}, \qquad g(x)=\frac{x-2}{(x-3)^2(x+2)}, \qquad h(x)=\frac{x-2}{(x-3)(x+2)^2}
+ \]
+ which are graphed in \cref{rat:fig:repfactd}. Note that each function has $2$ 
+ vertical asymptotes, and the domain of each function is 
+ \[
+ 	(-\infty,-2)\cup(-2,3)\cup(3,\infty)
+ \]
+ so we are not surprised to see that each curve has $3$ branches. We also note that 
+ the numerator of each function is the same, which tells us that each function has 
+ only $1$ zero at $2$.
+  	  
+ The functions $g$ and $h$ are different from those that we have considered previously, 
+ because they have a repeated factor in the denominator. Notice in particular 
+ the way that the functions behave around their asymptotes:
+ \begin{itemize}
+ 	\item $f$ behaves like $\frac{1}{x}$ around both of its asymptotes;
+ 	\item $g$ behaves like $\frac{1}{x}$ around $-2$, and like $\frac{1}{x^2}$ around $3$;
+ 	\item $h$ behaves like $\frac{1}{x^2}$ around $-2$, and like $\frac{1}{x}$ around $3$.
+ \end{itemize}
+ \end{pccexample}
+ \begin{figure}[!htb]
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)*(x-3));}]
+ 			\begin{axis}[
+ 					%                    framed,
+ 					xmin=-5,xmax=5,
+ 					ymin=-4,ymax=4,
+ 					xtick={-4,-2,...,4},
+ 					ytick={-2,2},
+ 					%                    grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-5:-2.201]{f};
+ 				\addplot[pccplot] expression[domain=-1.802:2.951]{f};
+ 				\addplot[pccplot] expression[domain=3.052:5]{f};
+ 				\addplot[soldot] coordinates{(2,0)};
+ 				%                 \addplot[asymptote,domain=-6:6]({-2},{x});
+ 				%                 \addplot[asymptote,domain=-6:6]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=\dfrac{x-2}{(x+2)(x-3)}$}
+ 		\label{rat:fig:repfactd1}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)*(x-3)^2);}]
+ 			\begin{axis}[
+ 					%                    framed,
+ 					xmin=-5,xmax=5,
+ 					ymin=-4,ymax=4,
+ 					xtick={-4,-2,...,4},
+ 					ytick={-2,2},
+ 					%                    grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-5:-2.039]{f};
+ 				\addplot[pccplot] expression[domain=-1.959:2.796]{f};
+ 				\addplot[pccplot] expression[domain=3.243:5]{f};
+ 				\addplot[soldot] coordinates{(2,0)};
+ 				%                 \addplot[asymptote,domain=-4:4]({-2},{x});
+ 				%                 \addplot[asymptote,domain=-4:4]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=\dfrac{x-2}{(x+2)(x-3)^2}$}
+ 		\label{rat:fig:repfactd2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)^2*(x-3));}]
+ 			\begin{axis}[
+ 					%                    framed,
+ 					xmin=-5,xmax=5,
+ 					ymin=-4,ymax=4,
+ 					xtick={-4,-2,...,2},
+ 					ytick={-2,2},
+ 					%                    grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-5:-2.451]{f};
+ 				\addplot[pccplot] expression[domain=-1.558:2.990]{f};
+ 				\addplot[pccplot] expression[domain=3.010:6]{f};
+ 				\addplot[soldot] coordinates{(2,0)};
+ 				%                 \addplot[asymptote,domain=-4:4]({-2},{x});
+ 				%                 \addplot[asymptote,domain=-4:4]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=\dfrac{x-2}{(x+2)^2(x-3)}$}
+ 		\label{rat:fig:repfactd3}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\label{rat:fig:repfactd}
+ \end{figure}
+  	  
+ \Cref{rat:def:function} says that the zeros of 
+ the rational function $r$ that has formula $r(x)=\frac{p(x)}{q(x)}$ are 
+ the zeros of $p$. Let's explore this a little more.
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}[Zeros] Find the zeros of each of the following functions
+ \[
+ 	\alpha(x)=\frac{x+5}{3x-7}, \qquad \beta(x)=\frac{9-x}{x+1}, \qquad \gamma(x)=\frac{17x^2-10}{2x+1}
+ \]
+ \begin{pccsolution}
+ We find the zeros of each function in turn by setting the numerator equal to $0$. The zeros of 
+ $\alpha$ are found by solving 
+ \[
+ 	x+5=0
+ \]
+ The zero of $\alpha$ is $-5$.
+  	  
+ Similarly, we may solve $9-x=0$ to find the zero of $\beta$, which is clearly $9$.
+  	  
+ The zeros of $\gamma$ satisfy the equation
+ \[
+ 	17x^2-10=0
+ \]
+ which we can solve using the square root property to obtain
+ \[
+ 	x=\pm\frac{10}{17}
+ \]
+ The zeros of $\gamma$ are $\pm\frac{10}{17}$.
+ \end{pccsolution}
+ \end{pccexample}
+  	  
+ \subsection*{Long-run behavior}
+ Our focus so far has been on the behavior of rational functions around 
+ their \emph{vertical} asymptotes. In fact, rational functions also 
+ have interesting long-run behavior around their \emph{horizontal} or 
+ \emph{oblique} asymptotes. A rational function will always have either 
+ a horizontal or an oblique asymptote| the case is determined by the degree
+ of the numerator and the degree of the denominator.
+ \begin{pccdefinition}[Long-run behavior]\label{rat:def:longrun}
+ Let $r$ be the rational function that has formula
+ \[
+ 	r(x) = \frac{a_n x^n + a_{n-1}x^{n-1}+\ldots + a_0}{b_m x^m + b_{m-1}x^{m-1}+\ldots+b_0}
+ \]
+ We can classify the long-run behavior of the rational function $r$ 
+ according to the following criteria:
+ \begin{itemize}
+ 	\item if $n<m$ then  $r$ has a horizontal asymptote with equation $y=0$;
+ 	\item if $n=m$ then $r$ has a horizontal asymptote with equation $y=\dfrac{a_n}{b_m}$;
+ 	\item if $n>m$ then $r$ will have an oblique asymptote as $x\rightarrow\pm\infty$ (more on this in \cref{rat:sec:oblique})
+ \end{itemize} 
+ \end{pccdefinition}
+ We will concentrate on functions that have horizontal asymptotes until 
+ we reach \cref{rat:sec:oblique}.
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}[Long-run behavior graphically]\label{rat:ex:horizasymp}
+ \pccname{Kebede} has graphed the following functions in his graphing calculator
+ \[
+ 	r(x)=\frac{x+1}{x-3}, \qquad s(x)=\frac{2(x+1)}{x-3}, \qquad t(x)=\frac{3(x+1)}{x-3}
+ \]
+ and obtained the curves shown in \cref{rat:fig:horizasymp}. Kebede decides 
+ to test his knowledgeable friend \pccname{Oscar}, and asks him 
+ to match the formulas to the graphs.
+  	  
+ \begin{figure}[!htb]
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=2*(x+1)/(x-3);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-15,xmax=15,
+ 					ymin=-6,ymax=6,
+ 					xtick={-12,-8,...,12},
+ 					minor ytick={-4,-3,...,4},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-15:2]{f};
+ 				\addplot[pccplot] expression[domain=5:15]{f};
+ 				\addplot[soldot] coordinates{(-1,0)};
+ 				\addplot[asymptote,domain=-6:6]({3},{x});
+ 				\addplot[asymptote,domain=-15:15]({x},{2});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:horizasymp1}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x+1)/(x-3);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-15,xmax=15,
+ 					ymin=-6,ymax=6,
+ 					xtick={-12,-8,...,12},
+ 					minor ytick={-4,-3,...,4},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-15:2.42857,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=3.8:15,samples=50]{f};
+ 				\addplot[soldot] coordinates{(-1,0)};
+ 				\addplot[asymptote,domain=-6:6]({3},{x});
+ 				\addplot[asymptote,domain=-15:15]({x},{1});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:horizasymp2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=3*(x+1)/(x-3);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-15,xmax=15,
+ 					ymin=-6,ymax=6,
+ 					xtick={-12,-8,...,12},
+ 					minor ytick={-4,-3,...,4},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-15:1.6666,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=7:15]{f};
+ 				\addplot[soldot] coordinates{(-1,0)};
+ 				\addplot[asymptote,domain=-6:6]({3},{x});
+ 				\addplot[asymptote,domain=-15:15]({x},{3});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:horizasymp3}
+ 	\end{subfigure}
+ 	\caption{Horizontal asymptotes}
+ 	\label{rat:fig:horizasymp}
+ \end{figure}
+  	  
+ Oscar notices that each function has a vertical asymptote at $3$ and a zero at $-1$. 
+ The main thing that catches Oscar's eye is that each function has a different 
+ coefficient in the numerator, and that each curve has a different horizontal asymptote. 
+ In particular, Oscar notes that
+ \begin{itemize}
+ 	\item the curve shown in \cref{rat:fig:horizasymp1} has a horizontal asymptote with equation $y=2$;
+ 	\item the curve shown in \cref{rat:fig:horizasymp2} has a horizontal asymptote with equation $y=1$;
+ 	\item the curve shown in \cref{rat:fig:horizasymp3} has a horizontal asymptote with equation $y=3$.
+ \end{itemize}
+ Oscar is able to tie it all together for Kebede by referencing \cref{rat:def:longrun}. He says 
+ that since the degree of the numerator and the degree of the denominator is the same 
+ for each of the functions $r$, $s$, and $t$, the horizontal asymptote will be determined 
+ by evaluating the ratio of their leading coefficients. 
+  	  
+ Oscar therefore says that $r$ should have a horizontal asymptote $y=\frac{1}{1}=1$, $s$ should 
+ have a horizontal asymptote $y=\frac{2}{1}=2$, and $t$ should have a horizontal asymptote 
+ $y=\frac{3}{1}=3$. Kebede is able to finish the problem from here, and says that $r$ is 
+ shown in \cref{rat:fig:horizasymp2}, $s$ is shown in \cref{rat:fig:horizasymp1}, and 
+ $t$ is shown in \cref{rat:fig:horizasymp3}.
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}[Long-run behavior numerically]
+ \pccname{Xiao} and \pccname{Dwayne} saw \cref{rat:ex:horizasymp} but are a little confused 
+ about horizontal asymptotes. What does it mean to say that a function $r$ has a horizontal
+ asymptote? 
+  	   
+ They decide to explore the concept by 
+ constructing a table of values for the rational functions $R$ and  $S$ that have formulas
+ \[
+ 	R(x)=\frac{-5(x+1)}{x-3}, \qquad S(x)=\frac{7(x-5)}{2(x+1)}
+ \]
+ In \cref{rat:tab:plusinfty} they model the behavior of $R$ and $S$ as $x\rightarrow\infty$, 
+ and in \cref{rat:tab:minusinfty} they model the behavior of $R$ and $S$ as $x\rightarrow-\infty$
+ by substituting very large values of $|x|$ into each function.
+ \begin{table}[!htb]
+ 	\begin{minipage}{.5\textwidth}
+ 		\centering
+ 		\caption{$R$ and $S$ as $x\rightarrow\infty$}
+ 		\label{rat:tab:plusinfty}
+ 		\begin{tabular}{crr}
+ 			\beforeheading
+ 			$x$            & $R(x)$     & $S(x)$    \\ \afterheading
+ 			$1\times 10^2$ & $-5.20619$ & $3.29208$ \\\normalline
+ 			$1\times 10^3$ & $-5.02006$ & $3.47902$ \\\normalline
+ 			$1\times 10^4$ & $-5.00200$ & $3.49790$ \\\normalline
+ 			$1\times 10^5$ & $-5.00020$ & $3.49979$ \\\normalline
+ 			$1\times 10^6$ & $-5.00002$ & $3.49998$ \\\lastline
+ 		\end{tabular}
+ 	\end{minipage}%
+ 	\begin{minipage}{.5\textwidth}
+ 		\centering
+ 		\caption{$R$ and $S$ as $x\rightarrow-\infty$}
+ 		\label{rat:tab:minusinfty}
+ 		\begin{tabular}{crr}
+ 			\beforeheading
+ 			$x$             & $R(x)$     & $S(x)$    \\ \afterheading
+ 			$-1\times 10^2$ & $-4.80583$ & $3.71212$ \\\normalline
+ 			$-1\times 10^3$ & $-4.98006$ & $3.52102$ \\\normalline
+ 			$-1\times 10^4$ & $-4.99800$ & $3.50210$ \\\normalline
+ 			$-1\times 10^5$ & $-4.99980$ & $3.50021$ \\\normalline
+ 			$-1\times 10^6$ & $-4.99998$ & $3.50002$ \\\lastline
+ 		\end{tabular}
+ 	\end{minipage}
+ \end{table}
+  	  
+ Xiao and Dwayne study \cref{rat:tab:plusinfty,rat:tab:minusinfty} and decide that 
+ the functions $R$ and $S$ never actually touch their horizontal asymptotes, but they 
+ do get infinitely close. They also feel as if they have a better understanding of 
+ what it means to study the behavior of a function as $x\rightarrow\pm\infty$.
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}[Repeated factors in the numerator]
+ Consider the functions $f$, $g$, and $h$ that have formulas
+ \[
+ 	f(x)=\frac{(x-2)^2}{(x-3)(x+1)}, \qquad g(x)=\frac{x-2}{(x-3)(x+1)}, \qquad h(x)=\frac{(x-2)^3}{(x-3)(x+1)}
+ \]
+ which are graphed in \cref{rat:fig:repfactn}. We note that each function has vertical
+ asymptotes at $-1$ and $3$, and so the domain of each function is
+ \[
+ 	(-\infty,-1)\cup(-1,3)\cup(3,\infty)
+ \]
+ We also notice that the numerators of each function are quite similar| indeed, each 
+ function has a zero at $2$, but how does each function behave around their zero?
+  	  
+ Using \cref{rat:fig:repfactn} to guide us, we note that
+ \begin{itemize}
+ 	\item $f$ has a horizontal intercept $(2,0)$, but the curve of 
+ 	$f$ does not cut the horizontal axis| it bounces off it;
+ 	\item $g$ also has a horizontal intercept $(2,0)$, and the curve 
+ 	of $g$ \emph{does} cut the horizontal axis;
+ 	\item $h$ has a horizontal intercept $(2,0)$, and the curve of $h$ 
+ 	also cuts the axis, but appears flattened as it does so.
+ \end{itemize}
+  	  
+ We can further enrich our study by discussing the long-run behavior of each function. 
+ Using the tools of \cref{rat:def:longrun}, we can deduce that 
+ \begin{itemize}
+ 	\item $f$ has a horizontal asymptote with equation $y=1$;
+ 	\item $g$ has a horizontal asymptote with equation $y=0$;
+ 	\item $h$ does \emph{not} have a horizontal asymptote| it has an oblique asymptote (we'll 
+ 	study this more in \cref{rat:sec:oblique}).
+ \end{itemize}
+ \end{pccexample}
+  	  
+ \begin{figure}[!htb]
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-2)^2/((x+1)*(x-3));}]
+ 			\begin{axis}[
+ 					%                    framed,
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					xtick={-4,-2,...,4},
+ 					ytick={-8,-4,...,8},
+ 					%                    grid=both,
+ 					width=\figurewidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-5:-1.248,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=-0.794:2.976,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=3.026:5,samples=50]{f};
+ 				\addplot[soldot] coordinates{(2,0)};
+ 				%                 \addplot[asymptote,domain=-6:6]({-1},{x});
+ 				%                 \addplot[asymptote,domain=-6:6]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=\dfrac{(x-2)^2}{(x+1)(x-3)}$}
+ 		\label{rat:fig:repfactn1}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+1)*(x-3));}]
+ 			\begin{axis}[
+ 					%                    framed,
+ 					xmin=-5,xmax=5,
+ 					ymin=-10,ymax=10,
+ 					xtick={-4,-2,...,4},
+ 					ytick={-8,-4,...,8},
+ 					%                    grid=both,
+ 					width=\figurewidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-5:-1.075]{f};
+ 				\addplot[pccplot] expression[domain=-0.925:2.975]{f};
+ 				\addplot[pccplot] expression[domain=3.025:5]{f};
+ 				\addplot[soldot] coordinates{(2,0)};
+ 				%                 \addplot[asymptote,domain=-6:6]({-1},{x});
+ 				%                 \addplot[asymptote,domain=-6:6]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=\dfrac{x-2}{(x+1)(x-3)}$}
+ 		\label{rat:fig:repfactn2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-2)^3/((x+1)*(x-3));}]
+ 			\begin{axis}[
+ 					%                    framed,
+ 					xmin=-5,xmax=5,
+ 					xtick={-8,-6,...,8},
+ 					%                    grid=both,
+ 					ymin=-30,ymax=30,
+ 					width=\figurewidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-5:-1.27]{f};
+ 				\addplot[pccplot] expression[domain=-0.806:2.99185]{f};
+ 				\addplot[pccplot] expression[domain=3.0085:5]{f};
+ 				\addplot[soldot] coordinates{(2,0)};
+ 				%                 \addplot[asymptote,domain=-30:30]({-1},{x});
+ 				%                 \addplot[asymptote,domain=-30:30]({3},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=\dfrac{(x-2)^3}{(x+1)(x-3)}$}
+ 		\label{rat:fig:repfactn3}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\label{rat:fig:repfactn}
+ \end{figure}
+  	  
+ \subsection*{Holes}
+ Rational functions have a vertical asymptote at $a$ if the denominator is $0$ at $a$. 
+ What happens if the numerator is $0$ at the same place? In this case, we say that the rational 
+ function has a \emph{hole} at $a$.
+ \begin{pccdefinition}[Holes]
+ The rational function
+ \[
+ 	r(x)=\frac{p(x)}{q(x)}
+ \]
+ has a hole at $a$ if $p(a)=q(a)=0$. Note that holes are different from 
+ a vertical asymptotes. We represent that $r$ has a hole at the point 
+ $(a,r(a))$ on the curve $y=r(x)$ by 
+ using a hollow circle, $\circ$.
+ \end{pccdefinition}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{pccexample}
+ \pccname{Mohammed} and \pccname{Sue} have graphed the function $r$ that has formula
+ \[
+ 	r(x)=\frac{x^2+x-6}{(x-2)}
+ \]
+ in their calculators, and can not decide if the correct graph 
+ is \cref{rat:fig:hole} or \cref{rat:fig:hole1}.
+  	  
+ Luckily for them, Oscar is nearby, and can help them settle the debate. 
+ Oscar demonstrates that
+ \begin{align*}
+ 	r(x) & =\frac{(x+3)(x-2)}{(x-2)} \\
+ 	     & = x+3                     
+ \end{align*}
+ but only when $x\ne 2$, because the function is undefined at $2$. Oscar 
+ says that this necessarily means that the domain or $r$ is 
+ \[
+ 	(-\infty,2)\cup(2,\infty)
+ \]
+ and that $r$ must have a hole at $2$. 
+  	  
+ Mohammed and Sue are very grateful for the clarification, and conclude that 
+ the graph of $r$ is shown in \cref{rat:fig:hole1}.
+ \begin{figure}[!htb]
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-4,...,8},
+ 					ytick={-8,-4,...,8},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-10:7]{x+3};
+ 				\addplot[soldot] coordinates{(-3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:hole}
+ 	\end{minipage}%
+ 	\hfill
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-4,...,8},
+ 					ytick={-8,-4,...,8},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot expression[domain=-10:7]{x+3};
+ 				\addplot[holdot] coordinates{(2,5)};
+ 				\addplot[soldot] coordinates{(-3,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:hole1}
+ 	\end{minipage}%
+ \end{figure}
+ \end{pccexample}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{pccexample}
+ Consider the function $f$ that has formula
+ \[
+ 	f(x)=\frac{x(x+3)}{x^2-4x}
+ \]
+ The domain of $f$ is $(-\infty,0)\cup(0,4)\cup(4,\infty)$ because both $0$ and $4$ 
+ make the denominator equal to $0$. Notice that
+ \begin{align*}
+ 	f(x) & = \frac{x(x+3)}{x(x-4)} \\
+ 	     & = \frac{x+3}{x-4}       
+ \end{align*}
+ provided that $x\ne 0$. Since $0$ makes the numerator 
+ and the denominator 0 at the same time, we say that $f$ has a hole at $(0,-\nicefrac{3}{4})$. 
+ Note that this necessarily means that $f$ does not have a vertical intercept.
+  	  
+ We also note $f$ has a vertical asymptote at $4$; the function is graphed in \cref{rat:fig:holeex}.
+ \begin{figure}[!htb]
+ 	\centering
+ 	\begin{tikzpicture}[/pgf/declare function={f=(x+3)/(x-4);}]
+ 		\begin{axis}[
+ 				framed,
+ 				xmin=-10,xmax=10,
+ 				ymin=-10,ymax=10,
+ 				xtick={-8,-6,...,8},
+ 				ytick={-8,-6,...,8},
+ 				grid=both,
+ 			]
+ 			\addplot[pccplot] expression[domain=-10:3.36364,samples=50]{f};
+ 			\addplot[pccplot] expression[domain=4.77:10]{f};
+ 			\addplot[asymptote,domain=-10:10]({4},{x});
+ 			\addplot[holdot]coordinates{(0,-0.75)};
+ 			\addplot[soldot] coordinates{(-3,0)};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ 	\caption{$y=\dfrac{x(x+3)}{x^2-4x}$}
+ 	\label{rat:fig:holeex}
+ \end{figure}
+ \end{pccexample}
+  	  
+  	  
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{pccexample}[Minimums and maximums]
+ \pccname{Seamus} and \pccname{Trang} are discussing rational functions. Seamus says that 
+ if a rational function has a vertical asymptote, then it can 
+ not possibly have local minimums and maximums, nor can it have 
+ global minimums and maximums.
+  	  
+ Trang says this statement is not always true. She plots the functions 
+ $f$ and $g$ that have formulas
+ \[
+ 	f(x)=-\frac{32(x-1)(x+1)}{(x-2)^2(x+2)^2}, \qquad g(x)=\frac{32(x-1)(x+1)}{(x-2)^2(x+2)^2}
+ \]
+ in \cref{rat:fig:minmax1,rat:fig:minmax2} and shows them to Seamus. On seeing the graphs, 
+ Seamus quickly corrects himself, and says  that $f$ has a local (and global) 
+ maximum of $2$ at $0$, and that $g$ has a local (and global) minimum of $-2$ at $0$.
+  	  
+ \begin{figure}[!htb]
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=-32*(x-1)*(x+1)/(( x-2)^2*(x+2)^2);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-8,-6,...,8},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-3.01]{f};
+ 				\addplot[pccplot] expression[domain=-1.45:1.45]{f};
+ 				\addplot[pccplot] expression[domain=3.01:10]{f};
+ 				\addplot[soldot] coordinates{(-1,0)(1,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=f(x)$}
+ 		\label{rat:fig:minmax1}
+ 	\end{minipage}%
+ 	\hfill
+ 	\begin{minipage}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=32*(x-1)*(x+1)/(( x-2)^2*(x+2)^2);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-8,-6,...,8},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-3.01]{f};
+ 				\addplot[pccplot] expression[domain=-1.45:1.45]{f};
+ 				\addplot[pccplot] expression[domain=3.01:10]{f};
+ 				\addplot[soldot] coordinates{(-1,0)(1,0)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{$y=g(x)$}
+ 		\label{rat:fig:minmax2}
+ 	\end{minipage}%
+ \end{figure}
+  	  
+ Seamus also notes that (in its domain) the function $f$ is always concave down, and 
+ that (in its domain) the function $g$ is always concave up. Furthermore, Trang
+ observes that each function behaves like $\frac{1}{x^2}$ around each of its vertical 
+ asymptotes, because each linear factor in the denominator is raised to the power $2$.
+  	  
+ \pccname{Oscar} stops by and reminds both students about the long-run behavior; according 
+ to \cref{rat:def:longrun} since the degree of the denominator is greater than the
+ degree of the numerator (in both functions), each function has a horizontal asymptote
+ at $y=0$.
+ \end{pccexample}
+  	  
+  	  
+ \investigation*{}
+ %===================================
+ %   Author: Pettit/Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[The spaghetti incident]
+ The same Queen from \vref{exp:prob:queenschessboard} has recovered from 
+ the rice experiments, and has called her loyal jester for another challenge.
+  	  
+ The jester has an $11-$inch piece of uncooked spaghetti that he puts on a table; 
+ he uses a book to cover $\unit[1]{inch}$ of it so that 
+ $\unit[10]{inches}$ hang over the edge. The jester then produces a box of $\unit{mg}$
+ weights that can be hung from the spaghetti.
+  	  
+ The jester says it will take $\unit[y]{mg}$ to break the spaghetti when hung
+ $\unit[x]{inches}$ from the edge, according to the rule $y=\frac{100}{x}$.
+ \begin{margintable}
+ 	\centering
+ 	\captionof{table}{}
+ 	\label{rat:tab:spaghetti}
+ 	\begin{tabular}{cc}
+ 		\beforeheading
+ 		\heading{$x$} & \heading{$y$} \\
+ 		\afterheading
+ 		$1$           &               \\\normalline
+ 		$2$           &               \\\normalline
+ 		$3$           &               \\\normalline
+ 		$4$           &               \\\normalline
+ 		$5$           &               \\\normalline
+ 		$6$           &               \\\normalline
+ 		$7$           &               \\\normalline
+ 		$8$           &               \\\normalline
+ 		$9$           &               \\\normalline
+ 		$10$          &               \\\lastline
+ 	\end{tabular}
+ \end{margintable}
+ \begin{subproblem}\label{rat:prob:spaggt1}
+ 	Help the Queen complete \cref{rat:tab:spaghetti}, and use $2$ digits after the decimal
+ 	where appropriate.
+ 	\begin{shortsolution}
+ 		\begin{tabular}[t]{ld{2}}
+ 			\beforeheading
+ 			\heading{$x$} & \heading{$y$} \\
+ 			\afterheading
+ 			$1$           & 100           \\\normalline
+ 			$2$           & 50            \\\normalline
+ 			$3$           & 33.33         \\\normalline
+ 			$4$           & 25            \\\normalline
+ 			$5$           & 20            \\\normalline
+ 			$6$           & 16.67         \\\normalline
+ 			$7$           & 14.29         \\\normalline
+ 			$8$           & 12.50         \\\normalline
+ 			$9$           & 11.11         \\\normalline
+ 			$10$          & 10            \\\lastline
+ 		\end{tabular}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	What do you notice about the number of $\unit{mg}$ that it takes to break 
+ 	the spaghetti as $x$ increases?
+ 	\begin{shortsolution}
+ 		It seems that the number of $\unit{mg}$ that it takes to break the spaghetti decreases
+ 		as $x$ increases.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}\label{rat:prob:spaglt1}
+ 	The Queen wonders what happens when $x$ gets very small| help the Queen construct 
+ 	a table of values for $x$ and $y$ when $x=0.0001, 0.001, 0.01, 0.1, 0.5, 1$.
+ 	\begin{shortsolution}
+ 		\begin{tabular}[t]{d{2}l}
+ 			\beforeheading
+ 			\heading{$x$} & \heading{$y$} \\
+ 			\afterheading
+ 			0.0001        & $1000000$     \\\normalline
+ 			0.001         & $100000$      \\\normalline
+ 			0.01          & $10000$       \\\normalline
+ 			0.1           & $1000$        \\\normalline
+ 			0.5           & $200$         \\\normalline
+ 			1             & $100$         \\\lastline
+ 		\end{tabular}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	What do you notice about the number of $\unit{mg}$ that it takes to break the spaghetti
+ 	as $x\rightarrow 0$? Would it ever make sense to let $x=0$?
+ 	\begin{shortsolution}
+ 		The number of $\unit{mg}$ required to break the spaghetti increases as $x\rightarrow 0$.
+ 		We can not allow $x$ to be $0$, as we can not divide by $0$, and we can not 
+ 		be $0$ inches from the edge of the table.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Plot your results from \cref{rat:prob:spaggt1,rat:prob:spaglt1} on the same graph, 
+ 	and join the points using a smooth curve| set the maximum value of $y$ as $200$, and 
+ 	note that this necessarily means that you will not be able to plot all of the points.
+ 	\begin{shortsolution}
+ 		The graph of $y=\frac{100}{x}$ is shown below.
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-2,xmax=11,
+ 					ymin=-20,ymax=200,
+ 					xtick={2,4,...,10},
+ 					ytick={20,40,...,180},
+ 					grid=major,
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot+[-] expression[domain=0.5:10]{100/x};
+ 				\addplot[soldot] coordinates{(0.5,200)(1,100)(2,50)(3,33.33)
+ 				(4,25)(5,20)(16.67)(7,14.29)(8,12.50)(9,11.11)(10,10)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Using your graph, observe what happens to $y$ as $x$ increases. If we could somehow
+ 	construct a piece of uncooked spaghetti that was $\unit[101]{inches}$ long, how many 
+ 	$\unit{mg}$ would it take to break the spaghetti?
+ 	\begin{shortsolution}
+ 		As $x$ increases, $y\rightarrow 0$. If we could construct a piece of spaghetti 
+ 		$\unit[101]{inches}$ long, it would only take $\unit[1]{mg}$ to break it $\left(\frac{100}{100}=1\right)$. Of course, 
+ 		the weight of spaghetti would probably cause it to break without the weight.
+ 	\end{shortsolution}
+ \end{subproblem}
+ The Queen looks forward to more food-related investigations from her jester.
+ \end{problem}
+  	  
+  	  
+  	  
+ %===================================
+ %   Author: Adams (Hughes)
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Debt Amortization]
+ To amortize a debt means to pay it off in a given length of time using 
+ equal periodic payments. The payments include interest on the unpaid 
+ balance. The following formula gives the monthly payment, $M$, in dollars
+ that is necessary to amortize a debt of $P$ dollars in $n$ months 
+ at a monthly interest rate of $i$
+ \[
+ 	M=\frac{P\cdot i}{1-(1+i)^{-n}}
+ \]
+ Use this formula in each of the following problems.
+ \begin{subproblem}
+ 	What monthly payments are necessary on a credit card debt of \$2000 at 
+ 	$\unit[1.5]{\%}$ monthly if you want to pay off the debt in $2$ years?
+ 	In one year? How much money will you save by paying off the debt in the
+ 	shorter amount of time?
+ 	\begin{shortsolution}
+ 		Paying off the debt in $2$ years, we use
+ 		\begin{align*}
+ 			M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-24}} \\
+ 			  & \approx 99.85                               
+ 		\end{align*}
+ 		The monthly payments are \$99.85.
+ 		 			 		 		
+ 		Paying off the debt in $1$ year, we use
+ 		\begin{align*}
+ 			M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-12}} \\
+ 			  & \approx 183.36                              
+ 		\end{align*}
+ 		The monthly payments are \$183.36
+ 		 			 		 		
+ 		In the $2$-year model we would pay a total of $\$99.85\cdot 12=\$2396.40$. In the
+ 		$1$-year model we would pay a total of $\$183.36\cdot 12=\$2200.32$. We would therefore
+ 		save $\$196.08$ if we went with the $1$-year model instead of the $2$-year model.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	To purchase a home, a family needs a loan of \$300,000 at $\unit[5.2]{\%}$ 
+ 	annual interest.  Compare a $20$ year loan to a $30$ year loan and make 
+ 	a recommendation for the family.
+ 	(Note: when given an annual interest rate, it is a common business practice to divide by
+ 	$12$ to get a monthly rate.)
+ 	\begin{shortsolution}
+ 		For the $20$-year loan we use
+ 		\begin{align*}
+ 			M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 20}} \\
+ 			  & \approx 2013.16                                                                          
+ 		\end{align*}
+ 		The monthly payments are \$2013.16.
+ 		 			 		 		
+ 		For the $30$-year loan we use
+ 		\begin{align*}
+ 			M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 30}} \\
+ 			  & \approx 1647.33                                                                          
+ 		\end{align*}
+ 		The monthly payments are \$1647.33.
+ 		 			 		 		
+ 		The total amount paid during the $20$-year loan is $\$2013.16\cdot 12\cdot 20=\$483,158.40$. 
+ 		The total amount paid during the $30$-year loan is $\$1647.33\cdot 12\cdot 30=\$593,038.80$.
+ 		 			 		 		
+ 		Recommendation: if you can afford the payments, choose the $20$-year loan.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	\pccname{Ellen} wants to make monthly payments of \$100 to pay off a debt of \$3000 
+ 	at \unit[12]{\%} annual interest. How long will it take her to pay off the 
+ 	debt?
+ 	\begin{shortsolution}
+ 		We are given $M=100$, $P=3000$, $i=0.01$, and we need to find $n$
+ 		in the equation
+ 		\[
+ 			100 = \frac{3000\cdot 0.01}{1-(1+0.01)^{-n}}
+ 		\]
+ 		Using logarithms, we find that $n\approx 36$. It will take 
+ 		Ellen about $3$ years to pay off the debt.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	\pccname{Jake} is going to buy a new car. He puts \$2000 down and wants to finance the
+ 	remaining \$14,000. The dealer will offer him \unit[4]{\%} annual interest for 
+ 	$5$ years, or a \$2000
+ 	rebate which he can use to reduce the amount of the loan and \unit[8]{\%} 
+ 	annual interest for 5 years. Which should he choose?
+ 	\begin{shortsolution}
+ 		\begin{description}
+ 			\item[Option 1:] $\unit[4]{\%}$ annual interest for $5$ years on \$14,000.
+ 			This means that the monthly payments will be calculated using
+ 			\begin{align*}
+ 				M & = \frac{14000\cdot \frac{0.04}{12}}{1-\left( 1+\frac{0.04}{12} \right)^{-12\cdot 5}} \\
+ 				  & \approx 257.83                                                                       
+ 			\end{align*}
+ 			The monthly payments will be $\$257.83$. The total amount paid will be
+ 			$\$257.83\cdot 5\cdot 12=\$15,469.80$, of which $\$1469.80$ is interest.
+ 			\item[Option 2:] $\unit[8]{\%}$ annual interest for $5$ years on \$12,000.
+ 			This means that the monthly payments will be calculated using
+ 			\begin{align*}
+ 				M & = \frac{12000\cdot \frac{0.08}{12}}{1-\left( 1+\frac{0.08}{12} \right)^{-12\cdot 5}} \\
+ 				  & \approx 243.32                                                                       
+ 			\end{align*}
+ 			The monthly payments will be $\$243.32$. The total amount paid 
+ 			will be $\$243.32\cdot 5\cdot 12 =\$14,599.20$, of which $\$2599.2$ is 
+ 			interest.
+ 		\end{description}
+ 		Jake should choose option 1 to minimize the amount of interest 
+ 		he has to pay.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+  	  
+ \begin{exercises}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Rational or not]
+ Decide if each of the following functions are rational or not. If 
+ they are rational, state their domain.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$r(x)=\dfrac{3}{x}$    
+ 		\begin{shortsolution}
+ 			$r$ is rational; the domain of $r$ is $(-\infty,0)\cup (0,\infty)$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$s(y)=\dfrac{y}{6}$    
+ 		\begin{shortsolution}
+ 			$s$ is not rational ($s$ is linear). 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$t(z)=\dfrac{4-x}{7-8z}$    
+ 		\begin{shortsolution}
+ 			$t$ is rational; the domain of $t$ is $\left( -\infty,\dfrac{7}{8} \right)\cup \left( \dfrac{7}{8},\infty \right)$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$u(w)=\dfrac{w^2}{(w-3)(w+4)}$ 
+ 		\begin{shortsolution}
+ 			$u$ is rational; the domain of $w$ is $(-\infty,-4)\cup(-4,3)\cup(3,\infty)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$v(x)=\dfrac{4}{(x-2)^2}$ 
+ 		\begin{shortsolution}
+ 			$v$ is rational; the domain of $v$ is $(-\infty,2)\cup(2,\infty)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$w(x)=\dfrac{9-x}{x+17}$ 
+ 		\begin{shortsolution}
+ 			$w$ is rational; the domain of $w$ is $(-\infty,-17)\cup(-17,\infty)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$a(x)=x^2+4$ 
+ 		\begin{shortsolution}
+ 			$a$ is not rational ($a$ is quadratic, or a polynomial of degree $2$).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$b(y)=3^y$
+ 		\begin{shortsolution}
+ 			$b$ is not rational ($b$ is exponential).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$c(z)=\dfrac{z^2}{z^3}$ 
+ 		\begin{shortsolution}
+ 			$c$ is rational; the domain of $c$ is $(-\infty,0)\cup (0,\infty)$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$d(x)=x^2(x+3)(5x-7)$ 
+ 		\begin{shortsolution}
+ 			$d$ is not rational ($d$ is a polynomial).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$e(\alpha)=\dfrac{\alpha^2}{\alpha^2-1}$ 
+ 		\begin{shortsolution}
+ 			$e$ is rational; the domain of $e$ is $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$f(\beta)=\dfrac{3}{4}$ 
+ 		\begin{shortsolution}
+ 			$f$ is not rational ($f$ is constant).
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Function evaluation]
+ Let $r$ be the function that has formula 
+ \[
+ 	r(x)=\frac{(x-2)(x+3)}{(x+5)(x-7)}
+ \]
+ Evaluate each of the following (if possible); if the value is undefined, 
+ then state so.
+ \begin{multicols}{4}
+ 	\begin{subproblem}
+ 		$r(0)$ 
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r(0) & =\frac{(0-2)(0+3)}{(0+5)(0-7)} \\
+ 				     & =\frac{-6}{-35}                \\
+ 				     & =\frac{6}{35}                  
+ 			\end{aligned}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(1)$ 
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r(1) & =\frac{(1-2)(1+3)}{(1+5)(1-7)} \\
+ 				     & =\frac{-4}{-36}                \\
+ 				     & =\frac{1}{9}                   
+ 			\end{aligned}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(2)$ 
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r(2) & =\frac{(2-2)(2+3)}{(2+5)(2-7)} \\
+ 				     & = \frac{0}{-50}                \\
+ 				     & =0                             
+ 			\end{aligned}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(4)$ 
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r(4) & =\frac{(4-2)(4+3)}{(4+5)(4-7)} \\
+ 				     & =\frac{14}{-27}                \\
+ 				     & =-\frac{14}{27}                
+ 			\end{aligned}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(7)$ 
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r(7) & =\frac{(7-2)(7+3)}{(7+5)(7-7)} \\
+ 				     & =\frac{50}{0}                  
+ 			\end{aligned}$
+ 			 				 			 			         
+ 			$r(7)$ is undefined.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(-3)$ 
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r(-3) & =\frac{(-3-2)(-3+3)}{(-3+5)(-3-7)} \\
+ 				      & =\frac{0}{-20}                     \\
+ 				      & =0                                 
+ 			\end{aligned}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r(-5)$ 
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r(-5) & =\frac{(-5-2)(-5+3)}{(-5+5)(-5-7)} \\
+ 				      & =\frac{14}{0}                      
+ 			\end{aligned}$
+ 			 				 			 			
+ 			$r(-5)$ is undefined.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$r\left( \frac{1}{2} \right)$
+ 		\begin{shortsolution}
+ 			$\begin{aligned}[t]
+ 				r\left( \frac{1}{2} \right) & = \frac{\left( \frac{1}{2}-2 \right)\left( \frac{1}{2}+3 \right)}{\left( \frac{1}{2}+5 \right)\left( \frac{1}{2}-7 \right)} \\
+ 				                            & =\frac{-\frac{3}{2}\cdot\frac{7}{2}}{\frac{11}{2}\left( -\frac{13}{2} \right)}                                              \\
+ 				                            & =\frac{-\frac{21}{4}}{-\frac{143}{4}}                                                                                       \\
+ 				                            & =\frac{37}{143}                                                                                                             
+ 			\end{aligned}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Holes or asymptotes?]
+ State the domain of each of the following rational functions. Identify
+ any holes or asymptotes.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$f(x)=\dfrac{12}{x-2}$ 
+ 		\begin{shortsolution}
+ 			$f$ has a vertical asymptote at $2$; the domain of $f$ is $(-\infty,2)\cup (2,\infty)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$g(x)=\dfrac{x^2+x}{(x+1)(x-2)}$ 
+ 		\begin{shortsolution}
+ 			$g$ has a vertical asymptote at $2$, and a hole at $-1$; the domain of $g$ is $(-\infty,-1)\cup(-1,2)\cup(2,\infty)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$h(x)=\dfrac{x^2+5x+4}{x^2+x-12}$ 
+ 		\begin{shortsolution}
+ 			$h$ has a vertical asymptote at $3$, and a whole at $-4$; the domain of $h$ is $(-\infty,-4)\cup(-4,3)\cup(3,\infty)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$k(z)=\dfrac{z+2}{2z-3}$ 
+ 		\begin{shortsolution}
+ 			$k$ has a vertical asymptote at $\dfrac{3}{2}$; the domain of $k$ is $\left( -\infty,\dfrac{3}{2} \right)\cup\left( \dfrac{3}{2},\infty \right)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$l(w)=\dfrac{w}{w^2+1}$ 
+ 		\begin{shortsolution}
+ 			$l$ does not have any vertical asymptotes nor holes; the domain of $w$ is $(-\infty,\infty)$. 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$m(t)=\dfrac{14}{13-t^2}$ 
+ 		\begin{shortsolution}
+ 			$m$ has vertical asymptotes at $\pm\sqrt{13}$; the domain of $m$ is $(-\infty,\sqrt{13})\cup(-\sqrt{13},\sqrt{13})\cup(\sqrt{13},\infty)$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Find a formula from a graph]
+ Consider the rational functions graphed in \cref{rat:fig:findformula}. Find 
+ the vertical asymptotes for each function, together with any zeros, and 
+ give a possible formula for each. 
+ \begin{shortsolution}
+ 	\begin{itemize}
+ 		\item \Vref{rat:fig:formula1}: possible formula is $r(x)=\dfrac{1}{x+5}$ 
+ 		\item \Vref{rat:fig:formula2}: possible formula is $r(x)=\dfrac{(x+3)}{(x-5)}$
+ 		\item \Vref{rat:fig:formula3}: possible formula is $r(x)=\dfrac{1}{(x-4)(x+3)}$.
+ 	\end{itemize}
+ \end{shortsolution}
+ \end{problem}
+  	  
+ \begin{figure}[!htb]
+ 	\begin{widepage}
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=1/(x+4);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-6,ymax=6,
+ 					xtick={-8,-6,...,8},
+ 					minor ytick={-4,-3,...,4},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-4.16667,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=-3.83333:10,samples=50]{f};
+ 				\addplot[asymptote,domain=-6:6]({-4},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:formula1}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x+3)/(x-5);}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-6,ymax=6,
+ 					xtick={-8,-6,...,8},
+ 					minor ytick={-4,-3,...,4},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:3.85714]{f};
+ 				\addplot[pccplot] expression[domain=6.6:10]{f};
+ 				\addplot[soldot] coordinates{(-3,0)};
+ 				\addplot[asymptote,domain=-6:6]({5},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{1});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:formula2}
+ 	\end{subfigure}
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=1/((x-4)*(x+3));}]
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-3,ymax=3,
+ 					xtick={-8,-6,...,8},
+ 					minor ytick={-4,-3,...,4},
+ 					grid=both,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-3.0473]{f};
+ 				\addplot[pccplot] expression[domain=-2.95205:3.95205]{f};
+ 				\addplot[pccplot] expression[domain=4.0473:10]{f};
+ 				\addplot[asymptote,domain=-3:3]({-3},{x});
+ 				\addplot[asymptote,domain=-3:3]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{0});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:formula3}
+ 	\end{subfigure}
+ 	\caption{}
+ 	\label{rat:fig:findformula}
+ 	\end{widepage}
+ \end{figure}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Find a formula from a description]
+ In each of the following problems, give a formula of a rational 
+ function that has the listed properties.
+ \begin{subproblem}
+ 	Vertical asymptote at $2$.
+ 	\begin{shortsolution}
+ 		Possible option: $r(x)=\dfrac{1}{x-2}$. Note that we could multiply the 
+ 		numerator or denominator by any real number and still have the desired properties.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Vertical asymptote at $5$.
+ 	\begin{shortsolution}
+ 		Possible option: $r(x)=\dfrac{1}{x-5}$. Note that we could multiply the 
+ 		numerator or denominator by any real number and still have the desired properties.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Vertical asymptote at $-2$, and zero at $6$.
+ 	\begin{shortsolution}
+ 		Possible option: $r(x)=\dfrac{x-6}{x+2}$. Note that we could multiply the 
+ 		numerator or denominator by any real number and still have the desired properties.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Zeros at $2$ and $-5$ and vertical asymptotes at $1$ and $-7$.
+ 	\begin{shortsolution}
+ 		Possible option: $r(x)=\dfrac{(x-2)(x+5)}{(x-1)(x+7)}$. Note that we could multiply the 
+ 		numerator or denominator by any real number and still have the desired properties.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Given formula, find horizontal asymptotes]
+ Each of the following functions has a horizontal asymptote. Write the equation
+ of the horizontal asymptote for each function.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$f(x) = \dfrac{1}{x}$
+ 		\begin{shortsolution}
+ 			$y=0$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$g(x) = \dfrac{2x+3}{x}$
+ 		\begin{shortsolution}
+ 			$y=2$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$h(x) = \dfrac{x^2+2x}{x^2+3}$
+ 		\begin{shortsolution}
+ 			$y=1$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$k(x) = \dfrac{x^2+7}{x}$
+ 		\begin{shortsolution}
+ 			$y=1$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$l(x)=\dfrac{3x-2}{5x+8}$ 
+ 		\begin{shortsolution}
+ 			$y=\dfrac{3}{5}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$m(x)=\dfrac{3x-2}{5x^2+8}$ 
+ 		\begin{shortsolution}
+ 			$y=0$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$n(x)=\dfrac{(6x+1)(x-7)}{(11x-8)(x-5)}$ 
+ 		\begin{shortsolution}
+ 			$y=\dfrac{6}{11}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$p(x)=\dfrac{19x^3}{5-x^4}$ 
+ 		\begin{shortsolution}
+ 			$y=0$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$q(x)=\dfrac{14x^2+x}{1-7x^2}$ 
+ 		\begin{shortsolution}
+ 			$y=-2$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{problem}[Given horizontal asymptotes, find formula]
+ In each of the following problems, give a formula for a function that 
+ has the given horizontal asymptote. Note that there may be more than one option.
+ \begin{multicols}{4}
+ 	\begin{subproblem}
+ 		$y=7$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{7(x-2)}{x+1}$. Note that there
+ 			are other options, provided that the degree of the numerator is the same as the degree
+ 			of the denominator, and that the ratio of the leading 
+ 			coefficients is $7$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=-1$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{5-x^2}{x^2+10}$. Note that there
+ 			are other options, provided that the degree of the numerator is the same as the degree
+ 			of the denominator, and that the ratio of the leading 
+ 			coefficients is $10$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=53$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{53x^3}{x^3+4x^2-7}$. Note that there
+ 			are other options, provided that the degree of the numerator is the same as the degree
+ 			of the denominator, and that the ratio of the leading 
+ 			coefficients is $53$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=-17$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{34(x+2)}{7-2x}$. Note that there
+ 			are other options, provided that the degree of the numerator is the same as the degree
+ 			of the denominator, and that the ratio of the leading 
+ 			coefficients is $-17$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=\dfrac{3}{2}$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{3x+4}{2(x+1)}$. Note that there
+ 			are other options, provided that the degree of the numerator is the same as the degree
+ 			of the denominator, and that the ratio of the leading 
+ 			coefficients is $\dfrac{3}{2}$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=0$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{4}{x}$. Note that there
+ 			are other options, provided that the degree of the numerator is less than the degree
+ 			of the denominator.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=-1$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{10x}{5-10x}$. Note that there
+ 			are other options, provided that the degree of the numerator is the same as the degree
+ 			of the denominator, and that the ratio of the leading 
+ 			coefficients is $-1$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=2$ 
+ 		\begin{shortsolution}
+ 			Possible option: $f(x)=\dfrac{8x-3}{4x+1}$. Note that there
+ 			are other options, provided that the degree of the numerator is the same as the degree
+ 			of the denominator, and that the ratio of the leading 
+ 			coefficients is $2$.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Find a formula from a description]
+ In each of the following problems, give a formula for a function that 
+ has the prescribed properties. Note that there may be more than one option.
+ \begin{subproblem}
+ 	$f(x)\rightarrow 3$ as $x\rightarrow\pm\infty$.
+ 	\begin{shortsolution}
+ 		Possible option: $f(x)=\dfrac{3(x-2)}{x+7}$. Note that 
+ 		the zero and asymptote of $f$ could be changed, and $f$ would still have the desired properties.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	$r(x)\rightarrow -4$ as $x\rightarrow\pm\infty$.
+ 	\begin{shortsolution}
+ 		Possible option: $r(x)=\dfrac{-4(x-2)}{x+7}$. Note that 
+ 		the zero and asymptote of $r$ could be changed, and $r$ would still have the desired properties.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	$k(x)\rightarrow 2$ as $x\rightarrow\pm\infty$, and $k$ has vertical asymptotes at $-3$ and $5$.
+ 	\begin{shortsolution}
+ 		Possible option: $k(x)=\dfrac{2x^2}{(x+3)(x-5)}$. Note that the denominator 
+ 		must have the given factors; the numerator could be any degree $2$ polynomial, provided the 
+ 		leading coefficient is $2$.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   Feb 2011
+ %===================================
+ \begin{problem}
+ Let $r$ be the rational function that has
+ \[
+ 	r(x) = \frac{(x+2)(x-1)}{(x+3)(x-4)}
+ \]  
+ Each of the following questions are in relation to this function.
+ \begin{subproblem}
+ 	What is the vertical intercept of this function? State your answer as an 
+ 	ordered pair. \index{rational functions!vertical intercept}
+ 	\begin{shortsolution}
+ 		$\left(0,\frac{1}{6}\right)$
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}\label{rat:prob:rational}
+ 	What values of $x$ make the denominator equal to $0$?
+ 	\begin{shortsolution}
+ 		$-3,4$
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Use your answer to \cref{rat:prob:rational} to write the domain of the function in 
+ 	both interval, and set builder notation. %\index{rational functions!domain}\index{domain!rational functions}
+ 	\begin{shortsolution}
+ 		Interval notation: $(-\infty,-3)\cup (-3,4)\cup (4,\infty)$. 
+ 		Set builder: $\{x|x\ne -3, \mathrm{and}\, x\ne 4\}$
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	What are the vertical asymptotes of the function? State your answers in 
+ 	the form $x=$
+ 	\begin{shortsolution}
+ 		$x=-3$ and $x=4$
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}\label{rat:prob:zeroes}
+ 	What values of $x$ make the numerator equal to $0$?
+ 	\begin{shortsolution}
+ 		$-2,1$ 
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Use your answer to \cref{rat:prob:zeroes} to write the horizontal intercepts of 
+ 	$r$ as ordered pairs.
+ 	\begin{shortsolution}
+ 		$(-2,0)$ and $(1,0)$
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+  	  
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Holes]
+ \pccname{Josh} and \pccname{Pedro} are discussing the function
+ \[
+ 	r(x)=\frac{x^2-1}{(x+3)(x-1)}
+ \]
+ \begin{subproblem}
+ 	What is the domain of $r$?
+ 	\begin{shortsolution}
+ 		The domain of $r$ is $(-\infty,-3)\cup(-3,1)\cup(1,\infty)$.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Josh notices that the numerator can be factored- can you see how?
+ 	\begin{shortsolution}
+ 		$(x^2-1)=(x-1)(x+1)$
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Pedro asks, `Doesn't that just mean that 
+ 	\[
+ 		r(x)=\frac{x+1}{x+3}
+ 	\]
+ 	for all values of $x$?' Josh says, `Nearly\ldots but not for all values of $x$'. 
+ 	What does Josh mean?
+ 	\begin{shortsolution}
+ 		$r(x)=\dfrac{x+1}{x+3}$ provided that $x\ne -1$. 
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Where does $r$ have vertical asymptotes, and where does it have holes?
+ 	\begin{shortsolution}
+ 		The function $r$ has a vertical asymptote at $-3$, and a hole at $1$.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Sketch a graph of $r$.
+ 	\begin{shortsolution}
+ 		A graph of $r$ is shown below.
+ 		 			 		 		
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					framed,
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-8,-6,...,8},
+ 					grid=both,
+ 					width=\solutionfigurewidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-3.25]{(x+1)/(x+3)};
+ 				\addplot[pccplot] expression[domain=-2.75:10]{(x+1)/(x+3)};
+ 				\addplot[asymptote,domain=-10:10]({-3},{x});
+ 				\addplot[holdot]coordinates{(1,0.5)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   July 2012
+ %===================================
+ \begin{problem}[Function algebra]
+ Let $r$ and $s$ be the rational functions that have formulas
+ \[
+ 	r(x)=\frac{2-x}{x+3}, \qquad s(x)=\frac{x^2}{x-4}
+ \]
+ Evaluate each of the following (if possible).
+ \begin{multicols}{4}
+ 	\begin{subproblem}
+ 		$(r+s)(5)$ 
+ 		\begin{shortsolution}
+ 			$\frac{197}{8}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(r-s)(3)$
+ 		\begin{shortsolution}
+ 			$\frac{53}{6}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(r\cdot s)(4)$ 
+ 		\begin{shortsolution}
+ 			Undefined.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$\left( \frac{r}{s} \right)(1)$ 
+ 		\begin{shortsolution}
+ 			$-\frac{3}{4}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+  	  
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   July 2012
+ %===================================
+ \begin{problem}[Transformations: given the transformation, find the formula]
+ Let $r$ be the rational function that has formula.
+ \[
+ 	r(x)=\frac{x+5}{2x-3}
+ \]
+ In each of the following problems apply the given transformation to the function $r$ and 
+ write a formula for the transformed version of $r$.
+ \begin{multicols}{2}
+ 	\begin{subproblem}
+ 		Shift $r$ to the right by $3$ units. 
+ 		\begin{shortsolution}
+ 			$r(x-3)=\frac{x+2}{2x-9}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		Shift $r$ to the left by $4$ units. 
+ 		\begin{shortsolution}
+ 			$r(x+4)=\frac{x+9}{2x+5}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		Shift $r$ up by $\pi$ units. 
+ 		\begin{shortsolution}
+ 			$r(x)+\pi=\frac{x+5}{2x-3}+\pi$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		Shift $r$ down by $17$ units. 
+ 		\begin{shortsolution}
+ 			$r(x)-17=\frac{x+5}{2x-3}-17$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		Reflect $r$ over the horizontal axis.
+ 		\begin{shortsolution}
+ 			$-r(x)=-\frac{x+5}{2x-3}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		Reflect $r$ over the vertical axis.
+ 		\begin{shortsolution}
+ 			$r(-x)=\frac{x-5}{2x+3}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+  	  
+  	  
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2011
+ %===================================
+ \begin{problem}[Find a formula from a table]\label{rat:prob:findformula}
+ \Crefrange{rat:tab:findformular}{rat:tab:findformulau} show values of rational functions $r$, $q$, $s$, 
+ and $t$. Assume that any values marked with an X are undefined.
+  	  
+ \begin{table}[!htb]
+ 	\begin{widepage}
+ 	\centering
+ 	\caption{Tables for \cref{rat:prob:findformula}}
+ 	\label{rat:tab:findformula}
+ 	\begin{subtable}{.2\textwidth}
+ 		\centering
+ 		\caption{$y=r(x)$}
+ 		\label{rat:tab:findformular}
+ 		\begin{tabular}{rr}
+ 			\beforeheading
+ 			$x$  & $y$                \\ \afterheading
+ 			$-4$ & $\nicefrac{7}{2}$  \\\normalline
+ 			$-3$ & $-18$              \\\normalline
+ 			$-2$ & X                  \\\normalline
+ 			$-1$ & $-4$               \\\normalline
+ 			$0$  & $\nicefrac{-3}{2}$ \\\normalline
+ 			$1$  & $\nicefrac{-2}{3}$ \\\normalline
+ 			$2$  & $\nicefrac{-1}{4}$ \\\normalline
+ 			$3$  & $0$                \\\normalline
+ 			$4$  & $\nicefrac{1}{6}$  \\\lastline
+ 		\end{tabular}
+ 	\end{subtable}
+ 	\hfill
+ 	\begin{subtable}{.2\textwidth}
+ 		\centering
+ 		\caption{$y=s(x)$}
+ 		\label{rat:tab:findformulas}
+ 		\begin{tabular}{rr}
+ 			\beforeheading
+ 			$x$  & $y$                 \\ \afterheading
+ 			$-4$ & $\nicefrac{-2}{21}$ \\\normalline
+ 			$-3$ & $\nicefrac{-1}{12}$ \\\normalline
+ 			$-2$ & $0$                 \\\normalline
+ 			$-1$ & X                   \\\normalline
+ 			$0$  & $\nicefrac{-2}{3}$  \\\normalline
+ 			$1$  & $\nicefrac{-3}{4}$  \\\normalline
+ 			$2$  & $\nicefrac{-4}{3}$  \\\normalline
+ 			$3$  & X                   \\\normalline
+ 			$4$  & $\nicefrac{6}{5}$   \\\lastline
+ 		\end{tabular}
+ 	\end{subtable}
+ 	\hfill
+ 	\begin{subtable}{.2\textwidth}
+ 		\centering
+ 		\caption{$y=t(x)$}
+ 		\label{rat:tab:findformulat}
+ 		\begin{tabular}{rr}
+ 			\beforeheading
+ 			$x$  & $y$               \\ \afterheading
+ 			$-4$ & $\nicefrac{3}{5}$ \\\normalline
+ 			$-3$ & $0$               \\\normalline
+ 			$-2$ & X                 \\\normalline
+ 			$-1$ & $3$               \\\normalline
+ 			$0$  & $3$               \\\normalline
+ 			$1$  & X                 \\\normalline
+ 			$2$  & $0$               \\\normalline
+ 			$3$  & $\nicefrac{3}{5}$ \\\normalline
+ 			$4$  & $\nicefrac{7}{9}$ \\\lastline
+ 		\end{tabular}
+ 	\end{subtable}
+ 	\hfill
+ 	\begin{subtable}{.2\textwidth}
+ 		\centering
+ 		\caption{$y=u(x)$}
+ 		\label{rat:tab:findformulau}
+ 		\begin{tabular}{rr}
+ 			\beforeheading
+ 			$x$  & $y$                \\ \afterheading
+ 			$-4$ & $\nicefrac{16}{7}$ \\\normalline
+ 			$-3$ & X                  \\\normalline
+ 			$-2$ & $-\nicefrac{4}{5}$ \\\normalline
+ 			$-1$ & $-\nicefrac{1}{8}$ \\\normalline
+ 			$0$  & $0$                \\\normalline
+ 			$1$  & $-\nicefrac{1}{8}$ \\\normalline
+ 			$2$  & $-\nicefrac{4}{5}$ \\\normalline
+ 			$3$  & X                  \\\normalline
+ 			$4$  & $\nicefrac{16}{7}$ \\\lastline
+ 		\end{tabular}
+ 	\end{subtable}
+ 	\end{widepage}
+ \end{table}
+ \begin{subproblem}
+ 	Given that the formula for $r(x)$ has the form $r(x)=\dfrac{x-A}{x-B}$, use \cref{rat:tab:findformular}
+ 	to find values of $A$ and $B$.
+ 	\begin{shortsolution}
+ 		$A=3$ and $B=-2$, so $r(x)=\dfrac{x-3}{x+2}$.
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Check your formula by computing $r(x)$ at the values specified in the table.
+ 	\begin{shortsolution}
+ 		$\begin{aligned}[t]
+ 			r(-4) & = \frac{-4-3}{-4+2} \\
+ 			      & = \frac{7}{2}       \\
+ 		\end{aligned}$
+ 		 			 		 		
+ 		$r(-3)=\ldots$ etc
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	The function $s$ in \cref{rat:tab:findformulas} has two vertical asymptotes and one zero. 
+ 	Can you find a formula for $s(x)$?
+ 	\begin{shortsolution}
+ 		$s(x)=\dfrac{x+2}{(x-3)(x+1)}$
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Check your formula by computing $s(x)$ at the values specified in the table.
+ 	\begin{shortsolution}
+ 		$\begin{aligned}[t]
+ 			s(-4) & =\frac{-4+2}{(-4-3)(-4+1)} \\
+ 			      & =-\frac{2}{21}             
+ 		\end{aligned}$
+ 		 			 		 		
+ 		$s(-3)=\ldots$ etc
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Given that the formula for $t(x)$ has  the form $t(x)=\dfrac{(x-A)(x-B)}{(x-C)(x-D)}$, use \cref{rat:tab:findformulat} to find the 
+ 	values of $A$, $B$, $C$, and $D$; hence write a formula for $t(x)$.
+ 	\begin{shortsolution}
+ 		$t(x)=\dfrac{(x+3)(x-2)}{(x+2)(x+1)}$ 
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Given that the formula for $u(x)$ has  the form $u(x)=\dfrac{(x-A)^2}{(x-B)(x-C)}$, use \cref{rat:tab:findformulau} to find the 
+ 	values of $A$, $B$, and $C$; hence write a formula for $u(x)$.
+ 	\begin{shortsolution}
+ 		$u(x)=\dfrac{x^2}{(x+3)(x-3)}$ 
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+ \end{exercises}
+  	  
+\section{Graphing rational functions (horizontal asymptotes)}
+ \reformatstepslist{R} % the steps list should be R1, R2, \ldots
+ We studied rational functions in the previous section, but were 
+ not asked to graph them; in this section we will demonstrate the 
+ steps to be followed in order to sketch graphs of the functions. 
+    
+ Remember from \vref{rat:def:function} that rational functions have 
+ the form
+ \[
+ 	r(x)=\frac{p(x)}{q(x)}
+ \]
+ In this section we will restrict attention to the case when 
+ \[
+ 	\text{degree of }p\leq \text{degree of }q
+ \]
+ Note that this necessarily means that each function that we consider
+ in this section \emph{will have a horizontal asymptote} (see \vref{rat:def:longrun}).
+ The cases in which the degree of $p$ is greater than the degree of $q$ 
+ is covered in the next section.
+    
+ Before we begin, it is important to remember the following:
+ \begin{itemize}
+ 	\item Our sketches will give a good representation of the overall 
+ 	shape of the graph, but until we have the tools of calculus (from MTH 251)
+ 	we can not find local minimums, local maximums, and inflection points algebraically. This
+ 	means that we will make our best guess as to where these points are.
+ 	\item We will not concern ourselves too much with the vertical scale (because of 
+ 	our previous point)| we will, however, mark the vertical intercept (assuming there is one), 
+ 	and any horizontal asymptotes.
+ \end{itemize}
+ \begin{pccspecialcomment}[Steps to follow when sketching rational functions]\label{rat:def:stepsforsketch}
+ 	\begin{steps}
+ 		\item \label{rat:step:first} Find all vertical asymptotes and holes, and mark them on the 
+ 		graph using dashed vertical lines and open circles $\circ$ respectively.
+ 		\item Find any intercepts, and mark them using solid circles $\bullet$;
+ 		determine if the curve cuts the axis, or bounces off it at each zero.
+ 		\item Determine the behavior of the function around each asymptote| does
+ 		it behave like $\frac{1}{x}$ or $\frac{1}{x^2}$?
+ 		\item \label{rat:step:penultimate} Determine the long-run behavior of the function, and mark the horizontal 
+ 		asymptote using a dashed horizontal line.
+ 		\item \label{rat:step:last}  Deduce the overall shape of the curve, and sketch it. If there isn't
+ 		enough information from the previous steps, then construct a table of values
+ 		including sample points from each branch.
+ 	\end{steps}
+ 	Remember that until we have the tools of calculus, we won't be able to 
+ 	find the exact coordinates of local minimums, local maximums, and points
+ 	of inflection.
+ \end{pccspecialcomment}
+    
+ The examples that follow show how \crefrange{rat:step:first}{rat:step:last} can be
+ applied to a variety of different rational functions.
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{rat:ex:1overxminus2p2}
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $r$ 
+ that has formula
+ \[
+ 	r(x)=\frac{1}{x-2}
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $r$ has a vertical asymptote at $2$; $r$ does not have any holes. The curve of 
+ 	$r$ will have $2$ branches.
+ 	\item $r$ does not have any zeros since the numerator is never equal to $0$. The
+ 	vertical intercept of $r$ is $\left( 0,-\frac{1}{2} \right)$.
+ 	\item $r$ behaves like $\frac{1}{x}$ around its vertical asymptote since $(x-2)$
+ 	is raised to the power $1$.
+ 	\item Since the degree of the numerator is less than the degree of the denominator, 
+ 	according to \vref{rat:def:longrun} the horizontal asymptote of $r$ has equation $y=0$.
+ 	\item We put the details we have obtained so far on \cref{rat:fig:1overxminus2p1}. Notice 
+ 	that there is only one way to complete the graph, which we have done in \cref{rat:fig:1overxminus2p2}.
+ \end{steps}
+ \end{pccsolution}
+ \end{pccexample}
+    
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-5,ymax=5,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-5:5]({2},{x});
+ 				\addplot[asymptote,domain=-5:5]({x},{0});
+ 				\addplot[soldot] coordinates{(0,-0.5)}node[axisnode,anchor=north east]{$\left( 0,-\frac{1}{2} \right)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:1overxminus2p1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=1/(x-2);}]
+ 			\begin{axis}[
+ 					xmin=-5,xmax=5,
+ 					ymin=-5,ymax=5,
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-5:1.8,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=2.2:5]{f};
+ 				\addplot[asymptote,domain=-5:5]({2},{x});
+ 				\addplot[asymptote,domain=-5:5]({x},{0});
+ 				\addplot[soldot] coordinates{(0,-0.5)}node[axisnode,anchor=north east]{$\left( 0,-\frac{1}{2} \right)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:1overxminus2p2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{1}{x-2}$}
+ \end{figure}
+    
+ The function $r$ in \cref{rat:ex:1overxminus2p2} has a horizontal asymptote which has equation $y=0$.
+ This asymptote lies on the horizontal axis, and you might (understandably) find it hard 
+ to distinguish between the two lines (\cref{rat:fig:1overxminus2p2}). When faced
+ with such a situation, it is perfectly acceptable to draw the horizontal axis 
+ as a dashed line| just make sure to label it correctly. We will demonstrate this 
+ in the next example.
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{rat:ex:1overxp1}
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $v$ 
+ that has formula
+ \[
+ 	v(x)=\frac{10}{x}
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $v$ has a vertical asymptote at $0$. $v$ does not have
+ 	any holes. The curve of $v$ will have $2$ branches.
+ 	\item $v$ does not have any zeros (since $10\ne 0$). Furthermore, $v$ 
+ 	does not have a vertical intercept since $v(0)$ is undefined.
+ 	\item $v$ behaves like $\frac{1}{x}$ around its vertical asymptote.
+ 	\item $v$ has a horizontal asymptote with equation $y=0$.
+ 	\item We put the details we have obtained so far in \cref{rat:fig:1overxp1}. 
+ 	We do not have enough information to sketch $v$ yet (because $v$ does 
+ 	not have any intercepts), so let's pick a sample 
+ 	point in either of the $2$ branches| it doesn't matter where our sample point 
+ 	is, because we know what the overall shape will be. Let's compute $v(2)$
+ 	\begin{align*}
+ 		v(2) & =\dfrac{10}{2} \\
+ 		     & = 5            
+ 	\end{align*}
+ 	We therefore mark the point $(2,5)$ on \cref{rat:fig:1overxp2}, and then complete the sketch using 
+ 	the details we found in the previous steps.
+ \end{steps}
+    
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-5,5},
+ 					ytick={-5,5},
+ 					axis line style={color=white},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,<->,domain=-10:10]({0},{x});
+ 				\addplot[asymptote,<->,domain=-10:10]({x},{0});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:1overxp1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=10/x;}]
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-5,5},
+ 					ytick={-5,5},
+ 					axis line style={color=white},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:-1]{f};
+ 				\addplot[pccplot] expression[domain=1:10]{f};
+ 				\addplot[soldot] coordinates{(2,5)}node[axisnode,anchor=south west]{$(2,5)$};
+ 				\addplot[asymptote,<->,domain=-10:10]({0},{x});
+ 				\addplot[asymptote,<->,domain=-10:10]({x},{0});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:1overxp2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{10}{x}$}
+ \end{figure}
+ \end{pccsolution}
+ \end{pccexample}
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{rat:ex:asympandholep1}
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $u$ 
+ that has formula
+ \[
+ 	u(x)=\frac{-4(x^2-9)}{x^2-8x+15}
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item We begin by factoring both the numerator and denominator of $u$ to help
+ 	us find any vertical asymptotes or holes
+ 	\begin{align*}
+ 		u(x) & =\frac{-4(x^2-9)}{x^2-8x+15}     \\
+ 		     & =\frac{-4(x+3)(x-3)}{(x-5)(x-3)} \\
+ 		     & =\frac{-4(x+3)}{x-5}             
+ 	\end{align*}
+ 	provided that $x\ne 3$. Therefore $u$ has a vertical asymptote at $5$ and 
+ 	a hole at $3$. The curve of $u$ has $2$ branches.
+ 	\item $u$ has a simple zero at $-3$. The vertical intercept of $u$ is $\left( 0,\frac{12}{5} \right)$.
+ 	\item $u$ behaves like $\frac{1}{x}$ around its vertical asymptote at $4$.
+ 	\item Using \vref{rat:def:longrun} the equation of the horizontal asymptote of $u$ is $y=-4$.
+ 	\item We put the details we have obtained so far on \cref{rat:fig:1overxminus2p1}. Notice 
+ 	that there is only one way to complete the graph, which we have done in \cref{rat:fig:1overxminus2p2}.
+ \end{steps}
+    
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-20,ymax=20,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-10,10},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-20:20]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{-4});
+ 				\addplot[soldot] coordinates{(-3,0)(0,2.4)}node[axisnode,anchor=south east]{$\left( 0,\frac{12}{5} \right)$};
+ 				\addplot[holdot] coordinates{(3,12)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:asympandholep1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=-4*(x+3)/(x-5);}]
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-20,ymax=20,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-10,10},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[pccplot] expression[domain=-10:3.6666,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=7:10]{f};
+ 				\addplot[asymptote,domain=-20:20]({5},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{-4});
+ 				\addplot[soldot] coordinates{(-3,0)(0,2.4)}node[axisnode,anchor=south east]{$\left( 0,\frac{12}{5} \right)$};
+ 				\addplot[holdot] coordinates{(3,12)};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:asympandholep2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{-4(x+3)}{x-5}$}
+ \end{figure}
+ \end{pccsolution}
+ \end{pccexample}
+    
+ \Cref{rat:ex:1overxminus2p2,rat:ex:1overxp1,rat:ex:asympandholep1} have focused on functions
+ that only have one vertical asymptote; the remaining examples in this section
+ concern functions that have more than one vertical asymptote. We will demonstrate
+ that \crefrange{rat:step:first}{rat:step:last} still apply.
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{rat:ex:sketchtwoasymp}
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $w$ 
+ that has formula
+ \[
+ 	w(x)=\frac{2(x+3)(x-5)}{(x+5)(x-4)}
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $w$ has vertical asymptotes at $-5$ and $4$. $w$ does not have 
+ 	any holes. The curve of $w$ will have $3$ branches.
+ 	\item $w$ has simple zeros at $-3$ and $5$. The vertical intercept of $w$ 
+ 	is $\left( 0,\frac{3}{2} \right)$.
+ 	\item $w$ behaves like $\frac{1}{x}$ around both of its vertical 
+ 	asymptotes.
+ 	\item The degree of the numerator of $w$ is $2$ and the degree of the 
+ 	denominator of $w$ is also $2$. Using the ratio of the leading coefficients
+ 	of the numerator and denominator, we say that $w$ has a horizontal 
+ 	asymptote with equation $y=\frac{2}{1}=2$.
+ 	\item We put the details we have obtained so far on \cref{rat:fig:sketchtwoasymptp1}. 
+ 	 	 	 	         
+ 	The function $w$ is a little more complicated than the functions that 
+ 	we have considered in the previous examples because the curve has $3$
+ 	branches. When graphing such functions, it is generally a good idea to start with the branch
+ 	for which you have the most information| in this case, that is the \emph{middle} branch
+ 	on the interval $(-5,4)$.
+ 	 	 	 	        
+ 	Once we have drawn the middle branch, there is only one way to complete the graph 
+ 	(because of our observations about the behavior of $w$ around its vertical asymptotes), 
+ 	which we have done in \cref{rat:fig:sketchtwoasymptp2}.
+ \end{steps}
+ \end{pccsolution}
+ \end{pccexample}
+    
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-5},{x});
+ 				\addplot[asymptote,domain=-10:10]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{2});
+ 				\addplot[soldot] coordinates{(-3,0)(5,0)};
+ 				\addplot[soldot] coordinates{(0,1.5)}node[axisnode,anchor=north west]{$\left( 0,\frac{3}{2} \right)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:sketchtwoasymptp1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=2*(x+3)*(x-5)/( (x+5)*(x-4));}]
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-5},{x});
+ 				\addplot[asymptote,domain=-10:10]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{2});
+ 				\addplot[soldot] coordinates{(-3,0)(5,0)};
+ 				\addplot[soldot] coordinates{(0,1.5)}node[axisnode,anchor=north west]{$\left( 0,\frac{3}{2} \right)$};
+ 				\addplot[pccplot] expression[domain=-10:-5.56708]{f};
+ 				\addplot[pccplot] expression[domain=-4.63511:3.81708]{f};
+ 				\addplot[pccplot] expression[domain=4.13511:10]{f};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:sketchtwoasymptp2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{2(x+3)(x-5)}{(x+5)(x-4)}$}
+ \end{figure}
+    
+ The rational functions that we have considered so far have had simple
+ factors in the denominator; each function has behaved like $\frac{1}{x}$ 
+ around each of its vertical asymptotes. \Cref{rat:ex:2asympnozeros,rat:ex:2squaredasymp}
+ consider functions that have a repeated factor in the denominator.
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{rat:ex:2asympnozeros}
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $f$ 
+ that has formula
+ \[
+ 	f(x)=\frac{100}{(x+5)(x-4)^2}
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $f$ has vertical asymptotes at $-5$ and $4$. $f$ does not have 
+ 	any holes. The curve of $f$ will have $3$ branches.
+ 	\item $f$ does not have any zeros (since $100\ne 0$). The vertical intercept of $f$ 
+ 	is $\left( 0,\frac{5}{4} \right)$.
+ 	\item $f$ behaves like $\frac{1}{x}$ around $-5$ and behaves like $\frac{1}{x^2}$
+ 	around $4$.
+ 	\item The degree of the numerator of $f$ is $0$ and the degree of the 
+ 	denominator of $f$ is $2$. $f$ has a horizontal asymptote with 
+ 	equation $y=0$.
+ 	\item We put the details we have obtained so far on \cref{rat:fig:2asympnozerosp1}. 
+ 	 	 	 	
+ 	The function $f$ is similar to the function $w$ that we considered in \cref{rat:ex:sketchtwoasymp}|
+ 	it has two vertical asymptotes and $3$ branches, but in contrast to $w$ it does not have any zeros.
+ 	 	 	 	
+ 	We sketch $f$ in \cref{rat:fig:2asympnozerosp2}, using the middle branch as our guide
+ 	because we have the most information about the function on the interval $(-5,4)$.
+ 	 	 	 	        
+ 	Once we have drawn the middle branch, there is only one way to complete the graph 
+ 	because of our observations about the behavior of $f$ around its vertical asymptotes (it behaves like $\frac{1}{x}$), 
+ 	which we have done in \cref{rat:fig:2asympnozerosp2}.
+ 	 	 	 	
+ 	Note that we are not yet able to find the local minimum of $f$ algebraically on the interval $(-5,4)$, 
+ 	so we make a reasonable guess as to where it is| we can be confident that it is above the horizontal axis
+ 	since $f$ has no zeros. You may think that this is unsatisfactory, but once we have the tools of calculus, we will 
+ 	be able to find local minimums more precisely.
+ \end{steps}
+ \end{pccsolution}
+ \end{pccexample}
+    
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-5},{x});
+ 				\addplot[asymptote,domain=-10:10]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{0});
+ 				\addplot[soldot] coordinates{(0,1.25)}node[axisnode,anchor=south east]{$\left( 0,\frac{5}{4} \right)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:2asympnozerosp1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=100/( (x+5)*(x-4)^2);}]
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-5},{x});
+ 				\addplot[asymptote,domain=-10:10]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{0});
+ 				\addplot[soldot] coordinates{(0,1.25)}node[axisnode,anchor=south east]{$\left( 0,\frac{5}{4} \right)$};
+ 				\addplot[pccplot] expression[domain=-10:-5.12022]{f};
+ 				\addplot[pccplot] expression[domain=-4.87298:2.87298,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=5:10]{f};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:2asympnozerosp2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{100}{(x+5)(x-4)^2}$}
+ \end{figure}
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   May 2012
+ %===================================
+ \begin{pccexample}\label{rat:ex:2squaredasymp}
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $g$ 
+ that has formula
+ \[
+ 	g(x)=\frac{50(2-x)}{(x+3)^2(x-5)^2}
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $g$ has vertical asymptotes at $-3$ and $5$. $g$ does 
+ 	not have any holes. The curve of $g$ will have $3$ branches.
+ 	\item $g$ has a simple zero at $2$. The vertical intercept of $g$ is
+ 	$\left( 0,\frac{4}{9} \right)$.
+ 	\item $g$ behaves like $\frac{1}{x^2}$ around both of its 
+ 	vertical asymptotes.
+ 	\item The degree of the numerator of $g$ is $1$ and the degree of the denominator
+ 	of $g$ is $4$. Using \vref{rat:def:longrun}, we calculate that 
+ 	the horizontal asymptote of $g$ has equation $y=0$.
+ 	\item The details that we have found so far have been drawn in 
+ 	\cref{rat:fig:2squaredasymp1}. The function $g$ is similar to the functions 
+ 	we considered in \cref{rat:ex:sketchtwoasymp,rat:ex:2asympnozeros} because 
+ 	it has $2$ vertical asymptotes and $3$ branches. 
+ 	 	 	 	
+ 	We sketch $g$ using the middle branch as our guide because we have the most information
+ 	about $g$ on the interval $(-3,5)$. Note that there is no other way to draw this branch
+ 	without introducing other zeros which $g$ does not have.
+ 	 	 	 	
+ 	Once we have drawn the middle branch, there is only one way to complete the graph 
+ 	because of our observations about the behavior of $g$ around its vertical asymptotes| it 
+ 	behaves like $\frac{1}{x^2}$.
+ 	 	 	 	
+ \end{steps}
+ \end{pccsolution}
+ \end{pccexample}
+    
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-3},{x});
+ 				\addplot[asymptote,domain=-10:10]({5},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{0});
+ 				\addplot[soldot] coordinates{(2,0)(0,4/9)}node[axisnode,anchor=south west]{$\left( 0,\frac{4}{9} \right)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:2squaredasymp1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=50*(2-x)/( (x+3)^2*(x-5)^2);}]
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-5,5},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-3},{x});
+ 				\addplot[asymptote,domain=-10:10]({5},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{0});
+ 				\addplot[soldot] coordinates{(2,0)(0,4/9)}node[axisnode,anchor=south west]{$\left( 0,\frac{4}{9} \right)$};
+ 				\addplot[pccplot] expression[domain=-10:-3.61504]{f};
+ 				\addplot[pccplot] expression[domain=-2.3657:4.52773]{f};
+ 				\addplot[pccplot] expression[domain=5.49205:10]{f};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:2squaredasymp2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{50(2-x)}{(x+3)^2(x-5)^2}$}
+ \end{figure}
+    
+ Each of the rational functions that we have considered so far has had either 
+ a \emph{simple} zero, or no zeros at all. Remember from our work on polynomial 
+ functions, and particularly \vref{poly:def:multzero}, that a \emph{repeated} zero
+ corresponds to the curve of the function behaving differently at the zero
+ when compared to how the curve behaves at a simple zero. \Cref{rat:ex:doublezero} details a 
+ function that has a non-simple zero.
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{pccexample}\label{rat:ex:doublezero}
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $g$ 
+ that has formula
+ \[
+ 	h(x)=\frac{(x-3)^2}{(x+4)(x-6)}
+ \]
+ \begin{pccsolution}
+ \begin{steps}
+ 	\item $h$ has vertical asymptotes at $-4$ and $6$. $h$ does 
+ 	not have any holes. The curve of $h$ will have $3$ branches.
+ 	\item $h$ has a zero at $3$ that has \emph{multiplicity $2$}. 
+ 	The vertical intercept of $h$ is
+ 	$\left( 0,-\frac{3}{8} \right)$.
+ 	\item $h$ behaves like $\frac{1}{x}$ around both of its 
+ 	vertical asymptotes.
+ 	\item The degree of the numerator of $h$ is $2$ and the degree of the denominator
+ 	of $h$ is $2$. Using \vref{rat:def:longrun}, we calculate that 
+ 	the horizontal asymptote of $h$ has equation $y=1$.
+ 	\item The details that we have found so far have been drawn in 
+ 	\cref{rat:fig:doublezerop1}. The function $h$ is different 
+ 	from the functions that we have considered in previous examples because 
+ 	of the multiplicity of the zero at $3$.
+ 	 	 	 	
+ 	We sketch $h$ using the middle branch as our guide because we have the most information
+ 	about $h$ on the interval $(-4,6)$. Note that there is no other way to draw this branch
+ 	without introducing other zeros which $h$ does not have| also note how 
+ 	the curve bounces off the horizontal axis at $3$.
+ 	 	 	 	
+ 	Once we have drawn the middle branch, there is only one way to complete the graph 
+ 	because of our observations about the behavior of $h$ around its vertical asymptotes| it 
+ 	behaves like $\frac{1}{x}$.
+ 	 	 	 	
+ \end{steps}
+ \end{pccsolution}
+ \end{pccexample}
+    
+ \begin{figure}[!htbp]
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-5,ymax=5,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-3,3},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-4},{x});
+ 				\addplot[asymptote,domain=-10:10]({6},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{1});
+ 				\addplot[soldot] coordinates{(3,0)(0,-3/8)}node[axisnode,anchor=north west]{$\left( 0,-\frac{3}{8} \right)$};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:doublezerop1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{.45\textwidth}
+ 		\begin{tikzpicture}[/pgf/declare function={f=(x-3)^2/((x+4)*(x-6));}]
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-5,ymax=5,
+ 					xtick={-8,-6,...,8},
+ 					ytick={-3,3},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[asymptote,domain=-10:10]({-4},{x});
+ 				\addplot[asymptote,domain=-10:10]({6},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{1});
+ 				\addplot[soldot] coordinates{(3,0)(0,-3/8)}node[axisnode,anchor=north west]{$\left( 0,-\frac{3}{8} \right)$};
+ 				\addplot[pccplot] expression[domain=-10:-5.20088]{f};
+ 				\addplot[pccplot] expression[domain=-3.16975:5.83642,samples=50]{f};
+ 				\addplot[pccplot] expression[domain=6.20088:10]{f};
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:doublezerop2}
+ 	\end{subfigure}%
+ 	\caption{$y=\dfrac{(x-3)^2}{(x+4)(x-6)}$}
+ \end{figure}
+ \begin{exercises}
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{problem}[\Cref{rat:step:last}]\label{rat:prob:deduce}
+ \pccname{Katie} is working on graphing rational functions. She 
+ has been concentrating on functions that have the form
+ \begin{equation}\label{rat:eq:deducecurve}
+ 	f(x)=\frac{a(x-b)}{x-c}
+ \end{equation}
+ Katie notes that functions with this type of formula have a zero
+ at $b$, and a vertical asymptote at $c$. Furthermore, these functions
+ behave like $\frac{1}{x}$ around their vertical asymptote, and the 
+ curve of each function will have $2$ branches.
+    
+ Katie has been working with $3$ functions that have the form given 
+ in \cref{rat:eq:deducecurve}, and has followed \crefrange{rat:step:first}{rat:step:penultimate};
+ her results are shown in \cref{rat:fig:deducecurve}. There is just one
+ more thing to do to complete the graphs| follow \cref{rat:step:last}.
+ Help Katie finish each graph by deducing the curve of each function.
+ \begin{shortsolution}
+ 	\Vref{rat:fig:deducecurve1}
+ 	 	 	 	    
+ 	\begin{tikzpicture}[/pgf/declare function={f=3*(x+4)/(x+5);}]
+ 		\begin{axis}[
+ 				xmin=-10,xmax=10,
+ 				ymin=-10,ymax=10,
+ 				xtick={-8,-6,...,8},
+ 				width=\solutionfigurewidth,
+ 			]
+ 			\addplot[soldot] coordinates{(-4,0)(0,12/5)};
+ 			\addplot[asymptote,domain=-10:10]({-5},{x});
+ 			\addplot[asymptote,domain=-10:10]({x},{3});
+ 			\addplot[pccplot] expression[domain=-10:-5.42857]{f};
+ 			\addplot[pccplot] expression[domain=-4.76923:10,samples=50]{f};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ 	 	 	 	    
+ 	\Vref{rat:fig:deducecurve2}
+ 	 	 	 	    
+ 	\begin{tikzpicture}[/pgf/declare function={f=-3*(x-2)/(x-4);}]
+ 		\begin{axis}[
+ 				xmin=-10,xmax=10,
+ 				ymin=-10,ymax=10,
+ 				xtick={-8,-6,...,8},
+ 				width=\solutionfigurewidth,
+ 			]
+ 			\addplot[soldot] coordinates{(2,0)(0,-3/2)};
+ 			\addplot[asymptote,domain=-10:10]({4},{x});
+ 			\addplot[asymptote,domain=-10:10]({x},{-3});
+ 			\addplot[pccplot] expression[domain=-10:3.53846,samples=50]{f};
+ 			\addplot[pccplot] expression[domain=4.85714:10]{f};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ 	 	 	 	    
+ 	\Vref{rat:fig:deducecurve4}
+ 	 	 	 	    
+ 	\begin{tikzpicture}[/pgf/declare function={f=2*(x-6)/(x-4);}]
+ 		\begin{axis}[
+ 				xmin=-10,xmax=10,
+ 				ymin=-10,ymax=10,
+ 				xtick={-8,-6,...,8},
+ 				width=\solutionfigurewidth,
+ 			]
+ 			\addplot[soldot] coordinates{(6,0)(0,3)};
+ 			\addplot[asymptote,domain=-10:10]({x},{2});
+ 			\addplot[asymptote,domain=-10:10]({4},{x});
+ 			\addplot[pccplot] expression[domain=-10:3.5,samples=50]{f};
+ 			\addplot[pccplot] expression[domain=4.3333:10]{f};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ \end{shortsolution}
+ \end{problem}
+    
+ \begin{figure}[!htb]
+ 	\begin{widepage}
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-4,0)(0,12/5)};
+ 				\addplot[asymptote,domain=-10:10]({-5},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{3});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:deducecurve1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(2,0)(0,-3/2)};
+ 				\addplot[asymptote,domain=-10:10]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{-3});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:deducecurve2}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(6,0)(0,3)};
+ 				\addplot[asymptote,domain=-10:10]({x},{2});
+ 				\addplot[asymptote,domain=-10:10]({4},{x});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:deducecurve4}
+ 	\end{subfigure}
+ 	\caption{Graphs for \cref{rat:prob:deduce}}
+ 	\label{rat:fig:deducecurve}
+ 	\end{widepage}
+ \end{figure}
+    
+ %===================================
+ %   Author: Hughes
+ %   Date:   June 2012
+ %===================================
+ \begin{problem}[\Cref{rat:step:last} for  more complicated rational functions]\label{rat:prob:deducehard}
+ \pccname{David} is also working on graphing rational functions, and 
+ has been concentrating on functions that have the form
+ \[
+ 	r(x)=\frac{a(x-b)(x-c)}{(x-d)(x-e)}
+ \]
+ David notices that functions with this type of formula have simple zeros 
+ at $b$ and $c$, and vertical asymptotes at $d$ and $e$. Furthermore, 
+ these functions behave like $\frac{1}{x}$ around both vertical asymptotes,
+ and the curve of the function will have $3$ branches. 
+    
+ David has followed \crefrange{rat:step:first}{rat:step:penultimate} for 
+ $3$ separate functions, and drawn the results in \cref{rat:fig:deducehard}.
+ Help David finish each graph by deducing the curve of each function.
+ \begin{shortsolution}
+ 	\Vref{rat:fig:deducehard1}
+ 	 	 	 	
+ 	\begin{tikzpicture}[/pgf/declare function={f=(x-6)*(x+3)/( (x-4)*(x+1));}]
+ 		\begin{axis}[
+ 				xmin=-10,xmax=10,
+ 				ymin=-10,ymax=10,
+ 				xtick={-8,-6,...,8},
+ 				width=\solutionfigurewidth,
+ 			]
+ 			\addplot[soldot] coordinates{(-3,0)(6,0)(0,9/2)};
+ 			\addplot[asymptote,domain=-10:10]({-1},{x});
+ 			\addplot[asymptote,domain=-10:10]({4},{x});
+ 			\addplot[asymptote,domain=-10:10]({x},{2});
+ 			\addplot[pccplot] expression[domain=-10:-1.24276]{f};
+ 			\addplot[pccplot] expression[domain=-0.6666:3.66667]{f};
+ 			\addplot[pccplot] expression[domain=4.24276:10]{f};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ 	 	 	 	
+ 	\Vref{rat:fig:deducehard2}
+ 	 	 	 	
+ 	\begin{tikzpicture}[/pgf/declare function={f=3*(x-2)*(x+3)/( (x-6)*(x+5));}]
+ 		\begin{axis}[
+ 				xmin=-10,xmax=10,
+ 				ymin=-10,ymax=10,
+ 				xtick={-8,-6,...,8},
+ 				width=\solutionfigurewidth,
+ 			]
+ 			\addplot[soldot] coordinates{(-3,0)(2,0)(0,3/5)};
+ 			\addplot[asymptote,domain=-10:10]({-5},{x});
+ 			\addplot[asymptote,domain=-10:10]({6},{x});
+ 			\addplot[asymptote,domain=-10:10]({x},{3});
+ 			\addplot[pccplot] expression[domain=-10:-5.4861]{f};
+ 			\addplot[pccplot] expression[domain=-4.68395:5.22241]{f};
+ 			\addplot[pccplot] expression[domain=7.34324:10]{f};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ 	 	 	 	
+ 	\Vref{rat:fig:deducehard3}
+ 	 	 	 	
+ 	\begin{tikzpicture}[/pgf/declare function={f=2*(x-7)*(x+3)/( (x+6)*(x-5));}]
+ 		\begin{axis}[
+ 				xmin=-10,xmax=10,
+ 				ymin=-10,ymax=10,
+ 				xtick={-8,-6,...,8},
+ 				width=\solutionfigurewidth,
+ 			]
+ 			\addplot[soldot] coordinates{(-3,0)(7,0)(0,1.4)};
+ 			\addplot[asymptote,domain=-10:10]({-6},{x});
+ 			\addplot[asymptote,domain=-10:10]({5},{x});
+ 			\addplot[asymptote,domain=-10:10]({x},{2});
+ 			\addplot[pccplot] expression[domain=-10:-6.91427]{f};
+ 			\addplot[pccplot] expression[domain=-5.42252:4.66427]{f};
+ 			\addplot[pccplot] expression[domain=5.25586:10]{f};
+ 		\end{axis}
+ 	\end{tikzpicture}
+ 	 	 	 	
+ \end{shortsolution}
+ \end{problem}
+    
+ \begin{figure}[!htb]
+ 	\begin{widepage}
+ 	\setlength{\figurewidth}{0.3\textwidth}
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-3,0)(6,0)(0,9/2)};
+ 				\addplot[asymptote,domain=-10:10]({-1},{x});
+ 				\addplot[asymptote,domain=-10:10]({4},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{2});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:deducehard1}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-3,0)(2,0)(0,3/5)};
+ 				\addplot[asymptote,domain=-10:10]({-5},{x});
+ 				\addplot[asymptote,domain=-10:10]({6},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{3});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:deducehard2}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\begin{subfigure}{\figurewidth}
+ 		\begin{tikzpicture}
+ 			\begin{axis}[
+ 					xmin=-10,xmax=10,
+ 					ymin=-10,ymax=10,
+ 					xtick={-8,-6,...,8},
+ 					width=\textwidth,
+ 				]
+ 				\addplot[soldot] coordinates{(-3,0)(7,0)(0,1.4)};
+ 				\addplot[asymptote,domain=-10:10]({-6},{x});
+ 				\addplot[asymptote,domain=-10:10]({5},{x});
+ 				\addplot[asymptote,domain=-10:10]({x},{2});
+ 			\end{axis}
+ 		\end{tikzpicture}
+ 		\caption{}
+ 		\label{rat:fig:deducehard3}
+ 	\end{subfigure}%
+ 	\hfill
+ 	\caption{Graphs for \cref{rat:prob:deducehard}}
+ 	\label{rat:fig:deducehard}
+ 	\end{widepage}
+ \end{figure}
+ %===================================
+ %   Author: Adams (Hughes)
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[\Crefrange{rat:step:first}{rat:step:last}]
+ Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of 
+ each of the following functions
+ \fixthis{need 2 more subproblems here}
+ \begin{multicols}{4}
+ 	\begin{subproblem}
+ 		$y=\dfrac{4}{x+2}$ 
+ 		\begin{shortsolution}
+ 			Vertical intercept: $(0,2)$; vertical asymptote: $x=-2$, horizontal asymptote: $y=0$.
+ 			 			 			 			
+ 			\begin{tikzpicture}
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-5,xmax=5,
+ 						ymin=-5,ymax=5,
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot] expression[domain=-5:-2.8]{4/(x+2)};
+ 					\addplot[pccplot] expression[domain=-1.2:5]{4/(x+2)};
+ 					\addplot[soldot]coordinates{(0,2)};
+ 					\addplot[asymptote,domain=-5:5]({-2},{x});
+ 					\addplot[asymptote,domain=-5:5]({x},{0});
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=\dfrac{2x-1}{x^2-9}$ 
+ 		\begin{shortsolution}
+ 			Vertical intercept:$\left( 0,\frac{1}{9} \right)$; 
+ 			horizontal intercept: $\left( \frac{1}{2},0 \right)$;
+ 			vertical asymptotes: $x=-3$, $x=3$, horizontal asymptote: $y=0$.
+ 			 			 			 			
+ 			\begin{tikzpicture}
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-5,xmax=5,
+ 						ymin=-5,ymax=5,
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot] expression[domain=-5:-3.23974]{(2*x-1)/(x^2-9)};
+ 					\addplot[pccplot,samples=50] expression[domain=-2.77321:2.83974]{(2*x-1)/(x^2-9)};
+ 					\addplot[pccplot] expression[domain=3.17321:5]{(2*x-1)/(x^2-9)};
+ 					\addplot[soldot]coordinates{(0,1/9)(1/2,0)};
+ 					\addplot[asymptote,domain=-5:5]({-3},{x});
+ 					\addplot[asymptote,domain=-5:5]({3},{x});
+ 					\addplot[asymptote,domain=-5:5]({x},{0});
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=\dfrac{x+3}{x-5}$ 
+ 		\begin{shortsolution}
+ 			Vertical intercept $\left( 0,-\frac{3}{5} \right)$; horizontal
+ 			intercept: $(-3,0)$; vertical asymptote: $x=5$; horizontal asymptote: $y=1$.
+ 			 			 			 			
+ 			\begin{tikzpicture}
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-10,xmax=10,
+ 						ymin=-5,ymax=5,
+ 						xtick={-8,-6,...,8},
+ 						minor ytick={-3,-1,...,3},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot] expression[domain=-10:3.666]{(x+3)/(x-5)};
+ 					\addplot[pccplot] expression[domain=7:10]{(x+3)/(x-5)};
+ 					\addplot[asymptote,domain=-5:5]({5},{x});
+ 					\addplot[asymptote,domain=-10:10]({x},{1});
+ 					\addplot[soldot]coordinates{(0,-3/5)(-3,0)};
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=\dfrac{2x+3}{3x-1}$ 
+ 		\begin{shortsolution}
+ 			Vertical intercept: $(0,-3)$; horizontal intercept: $\left( -\frac{3}{2},0 \right)$;
+ 			vertical asymptote: $x=\frac{1}{3}$, horizontal asymptote: $y=\frac{2}{3}$.
+ 			 			 			 			
+ 			\begin{tikzpicture}[/pgf/declare function={f=(2*x+3)/(3*x-1);}]
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-5,xmax=5,
+ 						ymin=-5,ymax=5,
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot] expression[domain=-5:0.1176]{f};
+ 					\addplot[pccplot] expression[domain=0.6153:5]{f};
+ 					\addplot[asymptote,domain=-5:5]({1/3},{x});
+ 					\addplot[asymptote,domain=-5:5]({x},{2/3});
+ 					\addplot[soldot]coordinates{(0,-3)(-3/2,0)};
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=\dfrac{4-x^2}{x^2-9}$ 
+ 		\begin{shortsolution}
+ 			Vertical intercept: $\left( 0,-\frac{4}{9} \right)$;
+ 			horizontal intercepts: $(2,0)$, $(-2,0)$; 
+ 			vertical asymptotes: $x=-3$, $x=3$; horizontal asymptote: $y=-1$.
+ 			 			 			 			
+ 			\begin{tikzpicture}[/pgf/declare function={f=(4-x^2)/(x^2-9);}]
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-5,xmax=5,
+ 						ymin=-5,ymax=5,
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot] expression[domain=-5:-3.20156]{f};
+ 					\addplot[pccplot,samples=50] expression[domain=-2.85774:2.85774]{f};
+ 					\addplot[pccplot] expression[domain=3.20156:5]{f};
+ 					\addplot[asymptote,domain=-5:5]({-3},{x});
+ 					\addplot[asymptote,domain=-5:5]({3},{x});
+ 					\addplot[asymptote,domain=-5:5]({x},{-1});
+ 					\addplot[soldot] coordinates{(-2,0)(2,0)(0,-4/9)};
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$y=\dfrac{(4x+5)(3x-4)}{(2x+5)(x-5)}$ 
+ 		\begin{shortsolution}
+ 			Vertical intercept: $\left( 0,\frac{4}{5} \right)$; 
+ 			horizontal intercepts: $\left( -\frac{5}{4},0 \right)$, $\left( \frac{4}{3},0 \right)$; 
+ 			vertical asymptotes: $x=-\frac{5}{2}$, $x=5$; horizontal asymptote: $y=6$.
+ 			 			 			 			
+ 			\begin{tikzpicture}[/pgf/declare function={f=(4*x+5)*(3*x-4)/((2*x+5)*(x-5));}]
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-10,xmax=10,
+ 						ymin=-20,ymax=20,
+ 						xtick={-8,-6,...,8},
+ 						ytick={-10,0,...,10},
+ 						minor ytick={-15,-5,...,15},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot] expression[domain=-10:-2.73416]{f};
+ 					\addplot[pccplot] expression[domain=-2.33689:4.2792]{f};
+ 					\addplot[pccplot] expression[domain=6.26988:10]{f};
+ 					\addplot[asymptote,domain=-20:20]({-5/2},{x});
+ 					\addplot[asymptote,domain=-20:20]({5},{x});
+ 					\addplot[asymptote,domain=-10:10]({x},{6});
+ 					\addplot[soldot]coordinates{(0,4/5)(-5/4,0)(4/3,0)};
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Inverse functions]
+ Each of the following rational functions are invertible 
+ \[
+ 	F(x)=\frac{2x+1}{x-3}, \qquad G(x)= \frac{1-4x}{x+3}
+ \]
+ \begin{subproblem}
+ 	State the domain of each function. 
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item The domain of $F$ is $(-\infty,3)\cup(3,\infty)$.
+ 			\item The domain of $G$ is $(-\infty,-3)\cup(-3,\infty)$.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Find the inverse of each function, and state its domain.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item $F^{-1}(x)=\frac{3x+1}{x-2}$; the domain of $F^{-1}$ is $(-\infty,2)\cup(2,\infty)$.
+ 			\item $G^{-1}(x)=\frac{3x+1}{x+4}$; the domain of $G^{-1}$ is $(-\infty,-4)\cup(-4,\infty)$.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	Hence state the range of the original functions.
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item The range of $F$ is the domain of $F^{-1}$, which is $(-\infty,2)\cup(2,\infty)$.
+ 			\item The range of $G$ is the domain of $G^{-1}$, which is $(-\infty,-4)\cup(-4,\infty)$.
+ 		\end{itemize}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	State the range of each inverse function. 
+ 	\begin{shortsolution}
+ 		\begin{itemize}
+ 			\item The range of $F^{-1}$ is the domain of $F$, which is $(-\infty,3)\cup(3,\infty)$.
+ 			\item The range of $G^{-1}$ is the domain of $G$, which is $(-\infty,-3)\cup(-3,\infty)$.
+ 		\end{itemize}<++>
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Composition]
+ Let $r$ and $s$ be the rational functions that have formulas
+ \[
+ 	r(x)=\frac{3}{x^2},\qquad s(x)=\frac{4-x}{x+5}
+ \]
+ Evaluate each of the following.
+ \begin{multicols}{3}
+ 	\begin{subproblem}
+ 		$(r\circ s)(0)$
+ 		\begin{shortsolution}
+ 			$\frac{75}{16}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(s\circ r)(0)$ 
+ 		\begin{shortsolution}
+ 			$(s\circ r)(0)$ is undefined.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(r\circ s)(2)$ 
+ 		\begin{shortsolution}
+ 			$\frac{147}{4}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(s\circ r)(3)$ 
+ 		\begin{shortsolution}
+ 			$192$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(s\circ r)(4)$ 
+ 		\begin{shortsolution}
+ 			$(s\circ r)(4)$ is undefined.
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$(s\circ r)(x)$ 
+ 		\begin{shortsolution}
+ 			$\dfrac{4x^2-3}{1+5x^2}$
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \end{problem}
+ %===================================
+ %   Author: Hughes
+ %   Date:   March 2012
+ %===================================
+ \begin{problem}[Piecewise rational functions]
+ The function $R$ has formula
+ \[
+ 	R(x)=
+ 	\begin{dcases}
+ 		\frac{2}{x+3},    & x<-5     \\
+ 		\frac{x-4}{x-10}, & x\geq -5 
+ 	\end{dcases}
+ \]
+ Evaluate each of the following.
+ \begin{multicols}{4}
+ 	\begin{subproblem}
+ 		$R(-6)$    
+ 		\begin{shortsolution}
+ 			$-\frac{2}{3}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$R(-5)$ 
+ 		\begin{shortsolution}
+ 			$\frac{3}{5}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$R(-3)$ 
+ 		\begin{shortsolution}
+ 			$\frac{7}{13}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ 	\begin{subproblem}
+ 		$R(5)$ 
+ 		\begin{shortsolution}
+ 			$-\frac{1}{5}$ 
+ 		\end{shortsolution}
+ 	\end{subproblem}
+ \end{multicols}
+ \begin{subproblem}
+ 	What is the domain of $R$? 
+ 	\begin{shortsolution}
+ 		$(-\infty,10)\cup(10,\infty)$ 
+ 	\end{shortsolution}
+ \end{subproblem}
+ \end{problem}
+ \end{exercises}
+    
+\section{Graphing rational functions (oblique asymptotes)}\label{rat:sec:oblique}
+ \begin{subproblem}
+ 	$y=\dfrac{x^2+1}{x-4}$ 
+ 	\begin{shortsolution}
+ 		\begin{enumerate}
+ 			\item $\left( 0,-\frac{1}{4} \right)$
+ 			\item Vertical asymptote: $x=4$.
+ 			\item A graph of the function is shown below
+ 			 			 			 			
+ 			\begin{tikzpicture}[/pgf/declare function={f=(x^2+1)/(x-4);}]
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-20,xmax=20,
+ 						ymin=-30,ymax=30,
+ 						xtick={-10,10},
+ 						minor xtick={-15,-5,...,15},
+ 						minor ytick={-10,10},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot,samples=50] expression[domain=-20:3.54724]{f};
+ 					\addplot[pccplot,samples=50] expression[domain=4.80196:20]{f};
+ 					\addplot[asymptote,domain=-30:30]({4},{x});
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{enumerate}
+ 	\end{shortsolution}
+ \end{subproblem}
+ \begin{subproblem}
+ 	$y=\dfrac{x^3(x+3)}{x-5}$ 
+ 	\begin{shortsolution}
+ 		\begin{enumerate}
+ 			\item $(0,0)$, $(-3,0)$
+ 			\item Vertical asymptote: $x=5$, horizontal asymptote: none.
+ 			\item A graph of the function is shown below
+ 			 			 			 			
+ 			\begin{tikzpicture}[/pgf/declare function={f=x^3*(x+3)/(x-5);}]
+ 				\begin{axis}[
+ 						framed,
+ 						xmin=-10,xmax=10,
+ 						ymin=-500,ymax=2500,
+ 						xtick={-8,-6,...,8},
+ 						ytick={500,1000,1500,2000},
+ 						grid=both,
+ 						width=\solutionfigurewidth,
+ 					]
+ 					\addplot[pccplot,samples=50] expression[domain=-10:4]{f};
+ 					\addplot[pccplot] expression[domain=5.6068:9.777]{f};
+ 					\addplot[asymptote,domain=-500:2500]({5},{x});
+ 				\end{axis}
+ 			\end{tikzpicture}
+ 		\end{enumerate}
+ 	\end{shortsolution}
+ \end{subproblem}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/stylefile.tex b/Master/texmf-dist/doc/support/latexindent/success/stylefile.tex
new file mode 100644
index 00000000000..fb1154d3420
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/stylefile.tex
@@ -0,0 +1,132 @@
+% http://tex.stackexchange.com/questions/106244/using-a-lot-of-marginpars
+\ProvidesPackage{tabto}[2013/03/25 \space v 1.3 \space 
+Another tabbing mechanism]\relax
+
+\newdimen\CurrentLineWidth
+\let\TabPrevPos\z@
+
+\newcommand\tabto[1]{%
+	\leavevmode
+	\begingroup
+	\def\@tempa{*}\def\@tempb{#1}%
+	\ifx\@tempa\@tempb % \tab* 
+	\endgroup
+	\TTo@overlaptrue % ... set a flag and re-issue \tabto to get argument
+	\expandafter\tabto
+	\else
+	\ifinner % in a \hbox, so ignore
+	\else % unrestricted horizontal mode
+	\null% \predisplaysize will tell the position of this box (must be box)
+	\parfillskip\fill
+	\everydisplay{}\everymath{}%
+	\predisplaypenalty\@M \postdisplaypenalty\@M
+	$$% math display so we can test \predisplaysize
+	\lineskiplimit=-999pt % so we get pure \baselineskip
+	\abovedisplayskip=-\baselineskip \abovedisplayshortskip=-\baselineskip
+	\belowdisplayskip\z@skip \belowdisplayshortskip\z@skip
+	\halign{##\cr\noalign{%
+		% get the width of the line above
+		%\message{>>> Line \the\inputlineno\space -- \predisplaydirection\the\predisplaydirection, \predisplaysize\the\predisplaysize, \displayindent\the\displayindent, \leftskip\the\leftskip, \linewidth\the\linewidth. }%
+		\ifdim\predisplaysize=\maxdimen % mixed R and L; call the line full
+		\message{Mixed R and L, so line is full. }%
+		\CurrentLineWidth\linewidth
+		\else
+		\ifdim\predisplaysize=-\maxdimen % impossible, in vmode; call the line empty
+		\message{Not in paragraph, so line is empty. }%
+		\CurrentLineWidth\z@
+		\else
+		\ifnum\TTo@Direction<\z@
+		\CurrentLineWidth\linewidth \advance\CurrentLineWidth\predisplaysize
+		\else
+		\CurrentLineWidth\predisplaysize 
+		\fi
+		% Correct the 2em offset
+		\advance\CurrentLineWidth -2em
+		\advance\CurrentLineWidth -\displayindent
+		\advance\CurrentLineWidth -\leftskip
+		\fi\fi
+		\ifdim\CurrentLineWidth<\z@ \CurrentLineWidth\z@\fi
+		% Enshrine the tab-to position; #1 might reference \CurrentLineWidth
+		\@tempdimb=#1\relax
+		\message{*** Tab to \the\@tempdimb, previous width is \the\CurrentLineWidth. ***}%
+		% Save width for possible return use
+		\xdef\TabPrevPos{\the\CurrentLineWidth}%
+		% Build the action to perform
+		\protected@xdef\TTo@action{%
+			\vrule\@width\z@\@depth\the\prevdepth
+			\ifdim\CurrentLineWidth>\@tempdimb
+			\ifTTo@overlap\else
+			\protect\newline \protect\null
+			\fi\fi
+			\protect\nobreak
+			\protect\hskip\the\@tempdimb\relax
+		}%
+		%\message{\string\TTo@action: \meaning \TTo@action. }%
+		% get back to the baseline, regardless of its depth.
+		\vskip-\prevdepth
+		\prevdepth-99\p@
+		\vskip\prevdepth
+	}}%
+	$$
+	% Don't count the display as lines in the paragraph
+	\count@\prevgraf \advance\count@-4 \prevgraf\count@
+	\TTo@action
+	%%   \penalty\@m % to allow a penalized line break
+	\fi
+	\endgroup
+	\TTo@overlapfalse
+	\ignorespaces
+	\fi
+}
+
+% \tab -- to the next position
+% \hskip so \tab\tab moves two positions
+% Allow a (penalized but flexible) line-break right after the tab.
+%
+\newcommand\tab{\leavevmode\hskip2sp\tabto{\NextTabStop}%
+	\nobreak\hskip\z@\@plus 30\p@\penalty4000\hskip\z@\@plus-30\p@\relax}
+
+
+% Expandable macro to select the next tab position from the list
+
+\newcommand\NextTabStop{%
+	\expandafter \TTo@nexttabstop \TabStopList,\maxdimen,>%
+}
+
+\def\TTo@nexttabstop #1,{%
+	\ifdim#1<\CurrentLineWidth
+	\expandafter\TTo@nexttabstop
+	\else
+	\ifdim#1<0.9999\linewidth#1\else\z@\fi
+	\expandafter\strip@prefix
+	\fi
+}
+\def\TTo@foundtabstop#1>{}
+
+\newcommand\TabPositions[1]{\def\TabStopList{\z@,#1}}
+
+\newcommand\NumTabs[1]{%
+	\def\TabStopList{}%
+	\@tempdimb\linewidth 
+	\divide\@tempdimb by#1\relax
+	\advance\@tempdimb 1sp % counteract rounding-down by \divide
+	\CurrentLineWidth\z@
+	\@whiledim\CurrentLineWidth<\linewidth\do {%
+		\edef\TabStopList{\TabStopList\the\CurrentLineWidth,}%
+		\advance\CurrentLineWidth\@tempdimb
+	}%
+	\edef\TabStopList{\TabStopList\linewidth}%
+}
+
+% default setting of tab positions:
+\TabPositions{\parindent,.5\linewidth}
+
+\newif\ifTTo@overlap \TTo@overlapfalse
+
+\@ifundefined{predisplaydirection}{
+	\let\TTo@Direction\predisplaysize
+	\let\predisplaydirection\@undefined
+}
+{
+	\let\TTo@Direction\predisplaydirection
+}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/table1.tex b/Master/texmf-dist/doc/support/latexindent/success/table1.tex
new file mode 100644
index 00000000000..5002f8b81a6
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/table1.tex
@@ -0,0 +1,22 @@
+% arara: indent: {overwrite: true, silent: on}
+\documentclass{article}
+\usepackage{multirow}
+\usepackage{booktabs}
+\begin{document}
+\begin{table}[h!]
+	\centering
+	\caption{mycaption}
+	\label{tab:test}
+	\begin{tabular}{llll}
+		\toprule
+		\textbf{headerone} & \textbf{headertwo} & \textbf{headerthree} & \textbf{headerfour}                           \\\midrule
+		r1c1               & r1c2               & r1c3                 & \multirow{4}{*}{norowlinesinthefirstfourrows} \\\cmidrule{1-3}
+		r2c1               & r2c2               & r2c3                 &                                               \\\cmidrule{1-3}
+		r3c1               & r3c2               & r3c3                 &                                               \\\cmidrule{1-3}
+		r4c1               & r4c2               & r4c3                 &                                               \\\midrule
+		r5c1               & r5c2               & r5c3                 &                                               \\\midrule
+		r6c1               & r6c2               & r6c3                 &                                               \\\midrule
+		r7c1               & r7c2               & r7c3                 &                                               \\\bottomrule
+	\end{tabular}
+\end{table}
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/table2.tex b/Master/texmf-dist/doc/support/latexindent/success/table2.tex
new file mode 100644
index 00000000000..cc0c12cc763
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/table2.tex
@@ -0,0 +1,26 @@
+% arara: indent: {overwrite: true, silent: yes}
+\documentclass{article}
+\usepackage{array} % Thanks to Heiko for catching the redundant package loading
+\newcolumntype{M}{>{$}c<{$}}
+
+\begin{document}
+
+\begin{table}%
+	\centering
+	\begin{tabular}{M|MMMMMMMMM}
+		    & A_1 & A_2 & A_3 & A_4 & A_5 & A_6 & A_7 & A_8 & A_9 \\\hline
+		A_1 & 0   &     &     &     &     &     &     &     &     \\
+		A_2 &     & 0   &     &     &     &     &     &     &     \\
+		A_3 &     &     & 0   &     &     &     &     &     &     \\
+		A_4 &     &     &     & 0   &     &     &     &     &     \\
+		A_5 &     &     &     &     & 0   &     &     &     &     \\
+		A_6 &     &     &     &     &     & 0   &     &     &     \\
+		A_7 &     &     &     &     &     &     & 0   &     &     \\
+		A_8 &     &     &     &     &     &     &     & 0   &     \\
+		A_9 &     &     &     &     &     &     &     &     & 0   \\
+	\end{tabular}
+	\caption{Some caption}
+	\label{table:mytable}
+\end{table}
+\end{document}
+
diff --git a/Master/texmf-dist/doc/support/latexindent/success/table3.tex b/Master/texmf-dist/doc/support/latexindent/success/table3.tex
new file mode 100644
index 00000000000..e4e0940fb80
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/table3.tex
@@ -0,0 +1,26 @@
+% arara: pdflatex
+% !arara: indent: {overwrite: yes, trace: on}
+\documentclass{article}
+\usepackage{multirow}
+
+\begin{document}
+\begin{figure*}
+	\centering
+	\begin{tabular}{|c|c|c c c c|c|}
+		\hline
+		\multicolumn{2}{|c|}{\multirow{2}{*}{$V_{\rm rot}/{\sigma}$}}&\multicolumn{4}{c|}{W1}\\
+		\cline{3-6}
+		\multicolumn{2}{|c|}{}&3&6&9&12\\
+		\hline
+		\multirow{6}{*}{W2} & \multirow{3}{*}{3} & $0.090475\pm 0.011115$ & \multirow{3}{*}{21} & \multirow{3}{*}{6} & \multirow{3}{*}{3} \\
+		                    &                    & $0.14861\pm 0.03562$   &                     &                    &                    \\
+		                    &                    & $0.1861 \pm 0.01728$   &                     &                    &                    \\
+		                    & 6                  & 8                      & 14                  & 5                  & 2                  \\
+		                    & 9                  & 8                      & 14                  & 5                  & 2                  \\
+		                    & 12                 & 8                      & 14                  & 5                  & 2                  \\
+		\hline
+	\end{tabular}
+	\caption{Multirow in multirow}   
+	\label{ta.Multirow}                             
+\end{figure*}
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/testHeadings.tex b/Master/texmf-dist/doc/support/latexindent/success/testHeadings.tex
new file mode 100644
index 00000000000..3c766b3e77f
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/testHeadings.tex
@@ -0,0 +1,59 @@
+% arara: indent: {onlyDefault: no, overwrite: true, trace: on, silent: yes, localSettings: true}
+\part{part}
+	part text
+	part text
+	\chapter{chapter long title}
+	 chapter text
+	 chapter text
+	 \[
+	 	f(x)=x^2
+	 \]
+	 \section[for the toc]{section}
+	  section text
+	  section text
+	 \section[for the toc]{section}
+	  section text
+	  section text
+	  \subsection[for the toc]{subsection}
+	  	subsection text
+	  	subsection text
+	  \subsection[for the toc]{subsection}
+	  	subsection text
+	  	subsection text
+	 \section[for the toc]{section}
+	  section text
+	  section text
+	\chapter{chapter}
+	 chapter text
+	 chapter text
+\part{part}
+	part text
+	part text
+	\chapter[toc]{chapter title}
+	 chapter text
+	 chapter text
+	 \section[for the toc]{section}
+	  section text
+	  section text
+	  \subsubsection[for the toc]{subsubsection}
+	  	subsubsection text
+	  	subsubsection text
+	  	\paragraph{paragraph}
+	  		paragraph text
+	  		paragraph text
+	  		\subparagraph{subparagraph}
+	  			subparagraph text
+	  			subparagraph text
+	 \section[for the toc]{section}
+	  section text
+	  section text
+	  \subsubsection[for the toc]{subsubsection}
+	  	subsubsection text
+	  	subsubsection text
+	  	\paragraph{paragraph}
+	  		paragraph text
+	  		paragraph text
+	  		\subparagraph{subparagraph}
+	  			subparagraph text
+	  			subparagraph text
+	\chapter[somethingelse]{goes here}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/testHeadings1.tex b/Master/texmf-dist/doc/support/latexindent/success/testHeadings1.tex
new file mode 100644
index 00000000000..de90829fb5f
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/testHeadings1.tex
@@ -0,0 +1,25 @@
+% arara: indent: {overwrite: yes, localSettings: yes, trace: on}
+\documentclass[a4paper]{article}
+\usepackage{filecontents}
+\begin{filecontents}
+	\begin{document}
+	hello world
+	\end{document}
+\end{filecontents}
+\begin{document}
+\section{}
+ \subsection{}
+ 	\subsubsection{}
+ 		some text goes here
+ 		some text goes here
+ 		some text goes here
+ 		\begin{verbatim}
+                \documentclass[<+options+>]{<+class+>}
+
+           \begin{document}
+        <++>
+                            \end{document}
+                     \end{document}
+                            more text here
+ 		\end{verbatim}
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/theorem.tex b/Master/texmf-dist/doc/support/latexindent/success/theorem.tex
new file mode 100644
index 00000000000..524b9dfc6f1
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/theorem.tex
@@ -0,0 +1,47 @@
+% arara: indent: {overwrite: on}
+\documentclass[12pt,twoside]{report}
+\usepackage[margin=2cm]{geometry} 
+\usepackage{amsmath,amsthm,amssymb}
+\usepackage{thmtools}
+\usepackage{tikz}
+\usepackage[framemethod=TikZ]{mdframed}
+
+\declaretheoremstyle
+[
+	spaceabove=0pt, spacebelow=0pt, headfont=\normalfont\bfseries,
+	notefont=\mdseries, notebraces={(}{)}, headpunct={\newline}, headindent={},
+	postheadspace={ }, postheadspace=4pt, bodyfont=\normalfont, qed=$\blacktriangle$,
+	preheadhook={\begin{mdframed}[style=myframedstyle]},
+	postfoothook=\end{mdframed},
+]{mystyle}
+
+\declaretheorem[style=mystyle,numberwithin=chapter,title=Exemplo]{example}
+\mdfdefinestyle{myframedstyle}{%
+	outermargin = 1.3cm , %
+	leftmargin = 0pt , rightmargin = 0pt , %
+	innerleftmargin = 5pt , innerrightmargin = 5pt , %
+	innertopmargin = 5pt, innerbottommargin = 5pt , %
+	backgroundcolor = blue!10 , %
+	align = center , % align the environment itself (left, center, rigth)
+	nobreak = true, % prevent a frame from splitting
+	hidealllines = true , %
+	topline = true , bottomline = true , %
+	splittopskip = \topskip , splitbottomskip = 0pt , %
+	skipabove = 0.5\baselineskip ,  skipbelow = 0.3\baselineskip}
+
+\begin{document}
+\section{Introduction}
+Lorem ipsum sed nulla id risus adipiscing vulputate.
+
+\begin{example}
+	Um consumidor financiou a compra de um veículo pagando 48 parcelas de \$800,00 mensais e a taxa de juros cobrada pela concessionária foi de 1,2\% a.m.. Qual era o valor à vista do automóvel adquirido?
+	\newline
+	\textbf{Solução:}
+	\newline
+	$PV = 800 \times \left[ \dfrac{1,012^{48}-1}{1,012^{48}\times 0,012} \right] \newline
+	PV = 800 \times \left[ \dfrac{0,772820}{0,021274} \right] \newline
+	PV = \$29.061,79$
+\end{example}
+
+Lorem ipsum sed nulla id risus adipiscing vulputate.
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/tikz1.tex b/Master/texmf-dist/doc/support/latexindent/success/tikz1.tex
new file mode 100644
index 00000000000..36624b91bcb
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/tikz1.tex
@@ -0,0 +1,88 @@
+% arara: indent: {overwrite: true, silent: on}
+\documentclass[png,border=10pt,tikz]{standalone}
+\usepackage{xstring}
+\usepackage{tikz}
+\usetikzlibrary{calc}
+
+\pgfkeys{/tikz/.cd,
+	vertical factor/.initial=0.5,
+	vertical factor/.get=\vertfactor,
+	vertical factor/.store in=\vertfactor,
+	start coordinate/.initial={0,\vertfactor},
+	start coordinate/.get=\startcoord,
+	start coordinate/.store in=\startcoord,
+	sample color/.initial=black,
+	sample color/.get=\samplecol,
+	sample color/.store in=\samplecol,
+	sample size/.initial=1pt,
+	sample size/.get=\samplesize,
+	sample size/.store in=\samplesize,
+	sample line width/.initial=very thick,
+	sample line width/.get=\samplelinewidth,
+	sample line width/.store in=\samplelinewidth,
+}
+
+
+\newcommand{\samplepath}[1]{%
+	\coordinate (start) at (\startcoord) ;
+	\foreach \samples[count=\xi from 1] in {#1}{%
+		\StrCut{\samples}{|}{\vertdir}{\hordir}
+		\ifnum\xi=1
+		\draw[\samplelinewidth,\samplecol](start)
+		--++(\hordir,0) coordinate (start);
+		\else
+		\IfStrEq{\vertdir}{+}{%true
+			\draw[\samplelinewidth,\samplecol]($(start)+(0,\vertfactor)$)
+			--++(\hordir,0)coordinate(start);
+			}{%false
+			\relax
+		}
+		\IfStrEq{\vertdir}{-}{%true
+			\draw[\samplelinewidth,\samplecol]($(start)+(0,-\vertfactor)$)
+			--++(\hordir,0)coordinate(start);
+			}{%false
+			\relax
+		}
+		\fi
+	}
+}
+
+\tikzset{sample/.style={
+	circle,
+	inner sep=\samplesize,
+	fill=\samplecol,
+	}
+}
+
+\newcommand{\discretesamplepath}[1]{%
+	\coordinate (start) at (\startcoord) ;
+	\foreach \samples[count=\xi from 1] in {#1}{%
+		\StrCut{\samples}{|}{\vertdir}{\hordir}
+		\ifnum\xi=1
+		\path(start)node[sample]{}
+		--++(\hordir,0) coordinate (start);
+		\else
+		\IfStrEq{\vertdir}{+}{%true
+			\path($(start)+(0,\vertfactor)$)node[sample]{}
+			--++(\hordir,0)coordinate(start);
+			}{%false
+			\relax
+		}
+		\IfStrEq{\vertdir}{-}{%true
+			\path($(start)+(0,-\vertfactor)$)node[sample]{}
+			--++(\hordir,0)coordinate(start);
+			}{%false
+			\relax
+		}
+		\fi
+	}
+}
+
+\begin{document}
+\begin{tikzpicture}
+	% axis
+	\draw[-stealth] (0,-1)--(0,4) node[left]{$X(t)$};
+	\draw[-stealth] (-1,0)--(5,0) node[below]{$t$};
+	\samplepath{+|0.5,+|0.25,-|1.5,+|1,+|0.5,+|0.75}
+\end{tikzpicture}
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/tikz2.tex b/Master/texmf-dist/doc/support/latexindent/success/tikz2.tex
new file mode 100644
index 00000000000..5ba03622079
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/tikz2.tex
@@ -0,0 +1,61 @@
+% arara: indent: {overwrite: true, silent: on}
+% http://tex.stackexchange.com/questions/104528/tikz-shade-also-the-border-of-a-node
+\documentclass[tikz,border=10pt,png]{standalone}
+\usepackage{tikz}
+\usetikzlibrary{calc}
+\begin{document}
+\tikzset{
+	shrink inner sep/.code={
+		\pgfkeysgetvalue{/pgf/inner xsep}{\currentinnerxsep}
+		\pgfkeysgetvalue{/pgf/inner ysep}{\currentinnerysep}
+		\pgfkeyssetvalue{/pgf/inner xsep}{\currentinnerxsep - 0.5\pgflinewidth}
+		\pgfkeyssetvalue{/pgf/inner ysep}{\currentinnerysep - 0.5\pgflinewidth}
+	}
+}
+
+\tikzset{horizontal shaded border/.style args={#1 and #2}{
+	append after command={
+		\pgfextra{%
+			\begin{pgfinterruptpath}
+				\path[rounded corners,left color=#1,right color=#2]
+				($(\tikzlastnode.south west)+(-\pgflinewidth,-\pgflinewidth)$)
+				rectangle
+				($(\tikzlastnode.north east)+(\pgflinewidth,\pgflinewidth)$);
+			\end{pgfinterruptpath}
+		}
+	}
+	},
+	vertical shaded border/.style args={#1 and #2}{
+		append after command={
+			\pgfextra{%
+				\begin{pgfinterruptpath}
+					\path[rounded corners,top color=#1,bottom color=#2]
+					($(\tikzlastnode.south west)+(-\pgflinewidth,-\pgflinewidth)$)
+					rectangle
+					($(\tikzlastnode.north east)+(\pgflinewidth,\pgflinewidth)$);
+				\end{pgfinterruptpath}
+			}
+		}
+	}
+}
+\begin{tikzpicture}
+	\draw (0,0) node[rectangle,
+		rounded corners,
+		thick,
+		outer sep=0pt,
+		shrink inner sep,
+		left color=red!50!white,
+		right color=green!50!white,
+		horizontal shaded border=red and green
+	](A){abcabc abc};
+	\draw (2.5,0) node[rectangle,
+		rounded corners,
+		thick,
+		outer sep=0pt,
+		shrink inner sep,
+		top color=cyan!50,
+		bottom color=orange!50,
+		vertical shaded border=blue and orange
+	](A){abcabc abc};
+\end{tikzpicture}
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/tikz3.tex b/Master/texmf-dist/doc/support/latexindent/success/tikz3.tex
new file mode 100644
index 00000000000..f50646ac5c8
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/tikz3.tex
@@ -0,0 +1,53 @@
+% arara: indent: {overwrite: true, silent: on}
+\documentclass[11pt]{article}
+\usepackage{tikz}
+\usetikzlibrary{trees}
+\usetikzlibrary{decorations.pathmorphing}
+\usetikzlibrary{decorations.markings}
+
+\begin{document}
+
+\tikzset{
+	photon/.style={decorate, decoration={snake}, draw=red},
+	particle/.style={draw=blue, postaction={decorate},
+		decoration={markings,mark=at position .5 with {\arrow[draw=blue]{>}}}},
+	antiparticle/.style={draw=blue, postaction={decorate},
+		decoration={markings,mark=at position .5 with {\arrow[draw=blue]{<}}}},
+	gluon/.style={decorate, draw=black,
+		decoration={coil,amplitude=4pt, segment length=5pt}}
+}
+
+\begin{tikzpicture}[
+		thick,
+		% Set the overall layout of the tree
+		level/.style={level distance=1.5cm},
+		level 2/.style={sibling distance=3.5cm},
+	]
+	\coordinate
+	child[grow=down]{
+		edge from parent [antiparticle]
+		child {
+			node{$E$}
+			edge from parent [particle]
+		}
+		child {
+			node{$D$}
+			edge from parent [gluon]
+		}
+		node [above=3pt] {$C$}
+	}
+	% I have to insert a dummy child to get the tree to grow
+	% correctly to the right.
+	child[grow=right, level distance=0pt] {
+		child {
+			node{$A$}
+			edge from parent [gluon]
+		}
+		child {
+			node{$B$}
+			edge from parent [particle]
+		}
+	};
+\end{tikzpicture}
+
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/tikz4.tex b/Master/texmf-dist/doc/support/latexindent/success/tikz4.tex
new file mode 100644
index 00000000000..97830271b3f
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/tikz4.tex
@@ -0,0 +1,27 @@
+% arara: indent: {overwrite: true, silent: on}
+\documentclass{article}
+
+% in the preamble
+% nothing
+% should happend
+\foreach \x in {0,1,2,3,4}{
+	\foreach \y in {0,1,2,3,4}{
+		\foreach \z in {0,1,2,3,4}{
+			\fill[black] (\x, \y, \z) circle (0.1);
+		}
+	}
+};
+\usepackage{tikz}
+
+\begin{document}
+\begin{tikzpicture}
+	\foreach \x in {0,1,2,3,4}{
+		\foreach \y in {0,1,2,3,4}{
+			\foreach \z in {0,1,2,3,4}{
+				\fill[black] (\x, \y, \z) circle (0.1);
+			}
+		}
+	};
+\end{tikzpicture}
+\end{document}
+
diff --git a/Master/texmf-dist/doc/support/latexindent/success/tikz5.tex b/Master/texmf-dist/doc/support/latexindent/success/tikz5.tex
new file mode 100644
index 00000000000..b24b429e318
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/tikz5.tex
@@ -0,0 +1,34 @@
+% arara: indent: {overwrite: yes}
+\documentclass[professionalfont, fleqn]{beamer}
+\mode<presentation>
+\usetheme{Warsaw}
+\usetheme{CambridgeUS}
+
+\usepackage{pgfplots}
+\usetikzlibrary{arrows,shapes,positioning}
+\graphicspath{{graphics/}}
+
+\begin{document}
+\frame
+{
+	\frametitle{Frame Title}
+	\begin{tikzpicture}
+		\begin{axis}
+			[
+				axis x line  = bottom,
+				axis y line  = left,
+				width        = 1.01\textwidth,
+				height       = .63\textwidth,          % Adjusted
+				ymax         = 93,
+				ymin         = 27,
+				ytick        = {30,40,...,90},
+				xmax         = 1993,
+				xmin         = 1967,
+				xtick        = {1970, 1980, ..., 1990},
+			]
+			\node[anchor=west] at (axis cs:1968.5,89.5){%
+				\textbullet\ Comment here about data
+			};
+		\end{axis}
+	\end{tikzpicture}
+}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/torusPGF.tex b/Master/texmf-dist/doc/support/latexindent/success/torusPGF.tex
new file mode 100644
index 00000000000..2f8ac9b7d5f
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/torusPGF.tex
@@ -0,0 +1,20 @@
+% arara: indent: {overwrite: yes}
+\documentclass{article}
+\usepackage{pgfplots}
+
+\begin{document}
+
+\begin{tikzpicture}
+	\begin{axis}
+		\addplot3[surf,
+			colormap/cool,
+			samples=20,
+			domain=0:2*pi,y domain=0:2*pi,
+		z buffer=sort]
+		({(2+cos(deg(x)))*cos(deg(y+pi/2))}, 
+		{(2+cos(deg(x)))*sin(deg(y+pi/2))}, 
+		{sin(deg(x))});
+	\end{axis}
+\end{tikzpicture}
+
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/torusPSTricks.tex b/Master/texmf-dist/doc/support/latexindent/success/torusPSTricks.tex
new file mode 100644
index 00000000000..7e3fc4bd501
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/torusPSTricks.tex
@@ -0,0 +1,18 @@
+% arara: indent: {overwrite: on, silent: yes}
+\documentclass{article}
+\usepackage{pst-solides3d}
+\begin{document}
+
+\begin{pspicture}(-3,-4)(3,6)
+	\psset{viewpoint=20 40 40 rtp2xyz,Decran=30,lightsrc=20 10 10}
+	\defFunction[algebraic]{torus}(u,v)
+	{(2+cos(u))*cos(v+\Pi)}
+	{(2+cos(u))*sin(v+\Pi)}
+	{sin(u)}  
+	\psSolid[object=surfaceparametree,
+		base=-10 10 0 6.28,fillcolor=black!70,incolor=orange,
+		function=torus,ngrid=60 0.4,
+	opacity=0.25]
+\end{pspicture}
+
+\end{document}
diff --git a/Master/texmf-dist/doc/support/latexindent/success/trailingComments.tex b/Master/texmf-dist/doc/support/latexindent/success/trailingComments.tex
new file mode 100644
index 00000000000..bc0d613c717
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/success/trailingComments.tex
@@ -0,0 +1,7 @@
+% arara: indent: {overwrite: yes, trace: true}
+\parbox{% more comments here
+	some stuff% comments go here 
+	some \% stuff this is not a comment
+}
+some stuff
+some stuff
diff --git a/Master/texmf-dist/scripts/latexindent/defaultSettings.yaml b/Master/texmf-dist/scripts/latexindent/defaultSettings.yaml
new file mode 100755
index 00000000000..457629ce863
--- /dev/null
+++ b/Master/texmf-dist/scripts/latexindent/defaultSettings.yaml
@@ -0,0 +1,181 @@
+# 
+# defaultSettings.yaml
+#
+# You're welcome to change anything you like in here, but 
+# it would probably be better to have your own user settings 
+# files somewhere else- remember that this file may be overwritten
+# anytime that you update your distribution. Please see the manual
+# for details of how to setup your own settings files.
+#
+# Please read the manual first to understand what each switch does :)
+
+# Default value of indentation
+defaultIndent: "\t"
+
+# default file extension of backup file (if original is overwritten with -w switch)
+# for example, if your .tex file is called
+#       myfile.tex
+# and you specify the backupExtension as BACKUP.bak then your 
+# backup file will be
+#       myfileBACKUP.bak
+backupExtension: .bak
+
+# only one backup per file; if onlyOneBackUp is 0 then the
+# number on the extension increments by 1 each time
+# (this is in place as a safety measure) myfile.bak0, myfile.bak1, myfile.bak2
+#
+# if you set onlyOnebackUp to 1, then the backup file will
+# be overwritten each time (not recommended until you trust the script)
+onlyOneBackUp: 0
+
+# some users may only want a set number of backup files, 
+# say at most 3; in which case, they can change this switch.
+# If maxNumberOfBackUps is set to 0 (or less) then infinitely 
+# many backups are possible, unless onlyOneBackUp is switched on
+maxNumberOfBackUps: 0
+
+# indent preamble
+indentPreamble: 0
+
+# always look for split { }, which means that the user doesn't
+# have to complete checkunmatched, checkunmatchedELSE
+alwaysLookforSplitBraces: 1
+
+# always look for split [ ], which means that the user doesn't
+# have to complete checkunmatchedbracket
+alwaysLookforSplitBrackets: 1
+
+# remove trailing whitespace from all lines 
+removeTrailingWhitespace: 0
+
+# environments that have tab delimiters, add more 
+# as needed
+lookForAlignDelims:
+   tabular: 1
+   tabularx: 1
+   array: 1
+   matrix: 1
+   bmatrix: 1
+   pmatrix: 1
+   align: 1
+   align*: 1
+   alignat: 1
+   alignat*: 1
+   aligned: 1
+   cases: 1
+   dcases: 1
+   pmatrix: 1
+   listabla: 1
+
+# if you have indent rules for particular environments
+# or commands, put them in here; for example, you might just want 
+# to use a space " " or maybe a double tab "\t\t"
+indentRules:
+   myenvironment: "\t\t"
+   anotherenvironment: "\t\t\t\t"
+   chapter: " "
+   section: " "
+
+#  verbatim environments- environments specified 
+#  in this hash table will not be changed at all!
+verbatimEnvironments:
+    verbatim: 1
+    lstlisting: 1
+
+#  no indent blocks (not necessarily verbatim 
+#  environments) which are marked as %\begin{noindent}
+#  or anything else that the user puts in this hash
+#  table
+noIndentBlock:
+    noindent: 1
+    cmhtest: 1
+
+# if you don't want to have additional indentation 
+# in an environment put it in this hash table; note that
+# environments in this hash table will inherit 
+# the *current* level of indentation they just won't
+# get any *additional*.
+noAdditionalIndent:
+    myexample: 1
+    mydefinition: 1
+    problem: 1
+    exercises: 1
+    mysolution: 1
+    foreach: 0
+    widepage: 1
+    comment: 1
+    \[: 0
+    \]: 0
+    document: 1
+    frame: 0
+
+# if you want to add indentation after
+# a heading, such as \part, \chapter, etc
+# then populate it in here - you can add 
+# an indent rule to indentRules if you would 
+# like something other than defaultIndent
+#
+# you can also change the level if you like, 
+# or add your own title command
+indentAfterHeadings:
+    part:
+       indent: 0
+       level: 1
+    chapter: 
+       indent: 0
+       level: 2
+    section:
+       indent: 0
+       level: 3
+    subsection:
+       indent: 0
+       level: 4
+    subsection*:
+       indent: 0
+       level: 4
+    subsubsection:
+       indent: 0
+       level: 5
+    paragraph:
+       indent: 0
+       level: 6
+    subparagraph:
+       indent: 0
+       level: 7
+
+# *** NOTE ***
+# If you have specified alwaysLookforSplitBraces: 1
+# and alwaysLookforSplitBrackets: 1 then you don't need
+# to worry about completing
+#
+#       checkunmatched
+#       checkunmatchedELSE
+#       checkunmatchedbracket
+#
+# in other words, you don't really need to edit anything 
+# below this line- it used to be necessary for older 
+# versions of the script, but not anymore :)
+#***      ***
+
+# commands that might split {} across lines
+# such as \parbox, \marginpar, etc
+checkunmatched:
+    parbox: 1
+    vbox: 1
+
+# very similar to %checkunmatched except these 
+# commands might have an else construct
+checkunmatchedELSE:
+    pgfkeysifdefined: 1
+    DTLforeach: 1
+    ifthenelse: 1
+
+# commands that might split []  across lines
+# such as \pgfplotstablecreatecol, etc
+checkunmatchedbracket:
+    pgfplotstablecreatecol: 1
+    pgfplotstablesave: 1
+    pgfplotstabletypeset: 1
+    mycommand: 1
+    psSolid: 1
+
diff --git a/Master/texmf-dist/scripts/latexindent/latexindent.pl b/Master/texmf-dist/scripts/latexindent/latexindent.pl
new file mode 100755
index 00000000000..74ab7ebb3e5
--- /dev/null
+++ b/Master/texmf-dist/scripts/latexindent/latexindent.pl
@@ -0,0 +1,1784 @@
+#!/usr/bin/perl
+#	This program is free software: you can redistribute it and/or modify
+#	it under the terms of the GNU General Public License as published by
+#	the Free Software Foundation, either version 3 of the License, or
+#	(at your option) any later version.
+#
+#	This program is distributed in the hope that it will be useful,
+#	but WITHOUT ANY WARRANTY; without even the implied warranty of
+#	MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
+#	GNU General Public License for more details.
+#
+#	See http://www.gnu.org/licenses/.
+#
+#	For details of how to use this file, please see readme.txt
+
+# load packages/modules
+use strict;
+use warnings;           
+use FindBin;            # help find defaultSettings.yaml 
+use YAML::Tiny;         # interpret defaultSettings.yaml
+use File::Copy;         # to copy the original file to backup (if overwrite option set)
+use File::Basename;     # to get the filename and directory path
+#use Getopt::Std;        
+use Getopt::Long;       # to get the switches/options/flags
+use File::HomeDir;      # to get users home directory, regardless of OS
+
+# get the options
+my $overwrite;
+my $outputToFile;
+my $silentMode;
+my $tracingMode;
+my $readLocalSettings;
+my $onlyDefault;
+my $showhelp;
+my $cruftDirectory='./';
+
+GetOptions ("w"=>\$overwrite,
+"o"=>\$outputToFile,
+"s"=>\$silentMode,
+"t"=>\$tracingMode,
+"l"=>\$readLocalSettings,
+"d"=>\$onlyDefault,
+"h"=>\$showhelp,
+"c=s"=>\$cruftDirectory,
+);
+
+die "Could not find directory $cruftDirectory\nExiting, no indentation done." if(!(-d $cruftDirectory));
+
+# version number
+my $versionNumber = "1.1R";
+
+# Check the number of input arguments- if it is 0 then simply 
+# display the list of options (like a manual)
+if(scalar(@ARGV) < 1 or $showhelp)
+{
+    print <<ENDQUOTE
+latexindent.pl version $versionNumber
+usage: latexindent.pl [options] [file][.tex]
+      -h  help (see the documentation for detailed instructions and examples)
+      -o  output to another file; sample usage
+                latexindent.pl -o myfile.tex outputfile.tex
+      -w  overwrite the current file- a backup will be made, but still be careful
+      -s  silent mode- no output will be given to the terminal
+      -t  tracing mode- verbose information given to the log file
+      -l  use localSettings.yaml (assuming it exists in the directory of your file)
+      -d  ONLY use defaultSettings.yaml, ignore ALL user files
+      -c=cruft directory used to specify the location of backup files and indent.log
+ENDQUOTE
+    ;
+    exit(2);
+}
+
+
+# we'll be outputting to the logfile and to standard output
+my $logfile;
+my $out = *STDOUT;
+
+# open the log file
+open($logfile,">","$cruftDirectory/indent.log") or die "Can't open indent.log";
+
+# output time to log file
+my $time = localtime();
+print $logfile $time;
+
+# output version to log file
+print $logfile <<ENDQUOTE
+
+latexindent.pl version $versionNumber, a script to indent .tex files
+latexindent.pl lives here: $FindBin::RealBin/
+
+Directory for backup files and indent.log: $cruftDirectory
+
+file: $ARGV[0]
+ENDQUOTE
+;
+
+# a quick options check
+if($outputToFile and $overwrite)
+{
+    print $logfile <<ENDQUOTE 
+
+WARNING: 
+\t You have called latexindent.pl with both -o and -w
+\t -o (output to file) will take priority, and -w (over write) will be ignored
+ 
+ENDQUOTE
+;
+    $overwrite = 0;
+}
+
+# can't call the script with MORE THAN 2 files
+if(scalar(@ARGV)>2)
+{
+    for my $fh ($out,$logfile) {print $fh <<ENDQUOTE
+
+ERROR:
+\t You're calling latexindent.pl with more than two file names
+\t The script can take at MOST two file names, but you 
+\t need to call it with the -o switch; for example
+
+\t latexindent.pl -o originalfile.tex outputfile.tex
+
+No indentation done :(
+Exiting...
+ENDQUOTE
+    };
+    exit(2);
+}
+
+# don't call the script with 2 files unless the -o flag is active
+if(!$outputToFile and scalar(@ARGV)==2)
+{
+for my $fh ($out,$logfile) {
+print $fh <<ENDQUOTE
+
+ERROR:
+\t You're calling latexindent.pl with two file names, but not the -o flag.
+\t Did you mean to use the -o flag ?
+
+No indentation done :(
+Exiting...
+ENDQUOTE
+};
+    exit(2);
+}
+
+# if the script is called with the -o switch, then check that 
+# a second file is present in the call, e.g
+#           latexindent.pl -o myfile.tex output.tex
+if($outputToFile and scalar(@ARGV)==1)
+{
+    for my $fh ($out,$logfile) {print $fh <<ENDQUOTE
+ERROR: When using the -o flag you need to call latexindent.pl with 2 arguments
+
+latexindent.pl -o "$ARGV[0]" [needs another name here]
+
+No indentation done :(
+Exiting...
+ENDQUOTE
+};
+    exit(2);
+}
+
+
+# Read in YAML file
+my $defaultSettings = YAML::Tiny->new;
+
+print $logfile "Reading defaultSettings.yaml from $FindBin::Bin/defaultSettings.yaml\n\n";
+
+# Open defaultSettings.yaml
+$defaultSettings = YAML::Tiny->read( "$FindBin::Bin/defaultSettings.yaml" );
+
+if(!$defaultSettings)
+{
+  for my $fh ($out,$logfile) {
+ print $fh <<ENDQUOTE 
+ ERROR  There seems to be a yaml formatting error in defaultSettings.yaml
+        Please check it for mistakes- you can find a working version at https://github.com/cmhughes/latexindent.plx
+        if you would like to overwrite your current version
+
+        Exiting, no indendation done.
+ENDQUOTE
+};
+ exit(2);
+}
+
+# setup the DEFAULT variables and hashes from the YAML file
+
+# scalar variables
+my $defaultIndent = $defaultSettings->[0]->{defaultIndent};
+my $alwaysLookforSplitBraces = $defaultSettings->[0]->{alwaysLookforSplitBraces};
+my $alwaysLookforSplitBrackets = $defaultSettings->[0]->{alwaysLookforSplitBrackets};
+my $backupExtension = $defaultSettings->[0]->{backupExtension};
+my $indentPreamble = $defaultSettings->[0]->{indentPreamble};
+my $onlyOneBackUp = $defaultSettings->[0]->{onlyOneBackUp};
+my $maxNumberOfBackUps = $defaultSettings->[0]->{maxNumberOfBackUps};
+my $removeTrailingWhitespace = $defaultSettings->[0]->{removeTrailingWhitespace};
+
+# hash variables
+my %lookForAlignDelims= %{$defaultSettings->[0]->{lookForAlignDelims}};
+my %indentRules= %{$defaultSettings->[0]->{indentRules}};
+my %verbatimEnvironments= %{$defaultSettings->[0]->{verbatimEnvironments}};
+my %noIndentBlock= %{$defaultSettings->[0]->{noIndentBlock}};
+my %checkunmatched= %{$defaultSettings->[0]->{checkunmatched}};
+my %checkunmatchedELSE= %{$defaultSettings->[0]->{checkunmatchedELSE}};
+my %checkunmatchedbracket= %{$defaultSettings->[0]->{checkunmatchedbracket}};
+my %noAdditionalIndent= %{$defaultSettings->[0]->{noAdditionalIndent}};
+my %indentAfterHeadings= %{$defaultSettings->[0]->{indentAfterHeadings}};
+
+# need new hashes to store the user and local data before
+# overwriting the default
+my %lookForAlignDelimsUSER;
+my %indentRulesUSER;
+my %verbatimEnvironmentsUSER;
+my %noIndentBlockUSER;
+my %checkunmatchedUSER;
+my %checkunmatchedELSEUSER;
+my %checkunmatchedbracketUSER;
+my %noAdditionalIndentUSER;
+my %indentAfterHeadingsUSER;
+
+# for printing the user and local settings to the log file
+my %dataDump;
+
+# empty array to store the paths
+my @absPaths;
+
+# scalar to read user settings
+my $userSettings;
+
+# get information about user settings- first check if indentconfig.yaml exists
+my $indentconfig = File::HomeDir->my_home . "/indentconfig.yaml";
+if ( -e $indentconfig and !$onlyDefault ) 
+{
+      print $logfile "Reading path information from ",File::HomeDir->my_home,"/indentconfig.yaml\n";
+
+      # read the absolute paths from indentconfig.yaml
+      $userSettings = YAML::Tiny->read( "$indentconfig" );
+
+      # integrity check
+      if($userSettings)
+      {
+        %dataDump = %{$userSettings->[0]};
+        print $logfile Dump \%dataDump;
+        print $logfile "\n";
+        @absPaths = @{$userSettings->[0]->{paths}};
+      }
+      else
+      {
+        print $logfile <<ENDQUOTE
+WARNING:  $indentconfig 
+          contains some invalid .yaml formatting- unable to read from it.
+          No user settings loaded. 
+ENDQUOTE
+;
+      }
+} 
+else
+{
+      if($onlyDefault)
+      {
+        print $logfile "Only default settings requested, not reading USER settings from indentconfig.yaml \n";
+        print $logfile "Ignoring localSettings.yaml\n" if($readLocalSettings);
+        $readLocalSettings = 0;
+      }
+      else
+      {
+        # give the user instructions on where to put indentconfig.yaml
+        print $logfile "Home directory is ",File::HomeDir->my_home,"\n";
+        print $logfile "To specify user settings you would put indentconfig.yaml here: \n\t",File::HomeDir->my_home,"/indentconfig.yaml\n\n";
+      }
+}
+
+# get information about LOCAL settings, assuming that localSettings.yaml exists
+my $directoryName = dirname $ARGV[0];
+
+# add local settings to the paths, if appropriate
+if ( (-e "$directoryName/localSettings.yaml") and $readLocalSettings and !(-z "$directoryName/localSettings.yaml")) 
+{
+    print $logfile "\nAdding ./localSettings.yaml to paths\n\n";
+    push(@absPaths,"$directoryName/localSettings.yaml");
+}
+elsif ( !(-e "$directoryName/localSettings.yaml") and $readLocalSettings) 
+{
+      print $logfile "WARNING\n\t$directoryName/localSettings.yaml not found\n";
+      print $logfile "\tcarrying on without it.\n";
+} 
+
+# read in the settings from each file
+foreach my $settings (@absPaths)
+{
+  # check that the settings file exists and that it isn't empty
+  if (-e $settings and !(-z $settings))
+  {
+      print $logfile "Reading USER settings from $settings\n";
+      $userSettings = YAML::Tiny->read( "$settings" );
+
+      # if we can read userSettings
+      if($userSettings)
+      {
+            # output settings to $logfile
+            %dataDump = %{$userSettings->[0]};
+            print $logfile Dump \%dataDump;
+            print $logfile "\n";
+
+            # scalar variables
+            $defaultIndent = $userSettings->[0]->{defaultIndent} if defined($userSettings->[0]->{defaultIndent});
+            $alwaysLookforSplitBraces = $userSettings->[0]->{alwaysLookforSplitBraces} if defined($userSettings->[0]->{alwaysLookforSplitBraces});
+            $alwaysLookforSplitBrackets = $userSettings->[0]->{alwaysLookforSplitBrackets} if defined($userSettings->[0]->{alwaysLookforSplitBrackets});
+            $backupExtension = $userSettings->[0]->{backupExtension} if defined($userSettings->[0]->{backupExtension});
+            $indentPreamble = $userSettings->[0]->{indentPreamble} if defined($userSettings->[0]->{indentPreamble});
+            $onlyOneBackUp = $userSettings->[0]->{onlyOneBackUp} if defined($userSettings->[0]->{onlyOneBackUp});
+            $maxNumberOfBackUps = $userSettings->[0]->{maxNumberOfBackUps} if defined($userSettings->[0]->{maxNumberOfBackUps});
+            $removeTrailingWhitespace = $userSettings->[0]->{removeTrailingWhitespace} if defined($userSettings->[0]->{removeTrailingWhitespace});
+
+            # hash variables - note that each one requires two lines, 
+            # one to read in the data, one to put the keys&values in correctly
+
+            %lookForAlignDelimsUSER= %{$userSettings->[0]->{lookForAlignDelims}} if defined($userSettings->[0]->{lookForAlignDelims});
+            @lookForAlignDelims{ keys %lookForAlignDelimsUSER } = values %lookForAlignDelimsUSER if (%lookForAlignDelimsUSER);
+
+            %indentRulesUSER= %{$userSettings->[0]->{indentRules}} if defined($userSettings->[0]->{indentRules});
+            @indentRules{ keys %indentRulesUSER } = values %indentRulesUSER if (%indentRulesUSER);
+
+            %verbatimEnvironmentsUSER= %{$userSettings->[0]->{verbatimEnvironments}} if defined($userSettings->[0]->{verbatimEnvironments});
+            @verbatimEnvironments{ keys %verbatimEnvironmentsUSER } = values %verbatimEnvironmentsUSER if (%verbatimEnvironmentsUSER);
+
+            %noIndentBlockUSER= %{$userSettings->[0]->{noIndentBlock}} if defined($userSettings->[0]->{noIndentBlock});
+            @noIndentBlock{ keys %noIndentBlockUSER } = values %noIndentBlockUSER if (%noIndentBlockUSER);
+
+            %checkunmatchedUSER= %{$userSettings->[0]->{checkunmatched}} if defined($userSettings->[0]->{checkunmatched});
+            @checkunmatched{ keys %checkunmatchedUSER } = values %checkunmatchedUSER if (%checkunmatchedUSER);
+
+            %checkunmatchedbracketUSER= %{$userSettings->[0]->{checkunmatchedbracket}} if defined($userSettings->[0]->{checkunmatchedbracket});
+            @checkunmatchedbracket{ keys %checkunmatchedbracketUSER } = values %checkunmatchedbracketUSER if (%checkunmatchedbracketUSER);
+
+            %noAdditionalIndentUSER= %{$userSettings->[0]->{noAdditionalIndent}} if defined($userSettings->[0]->{noAdditionalIndent});
+            @noAdditionalIndent{ keys %noAdditionalIndentUSER } = values %noAdditionalIndentUSER if (%noAdditionalIndentUSER);
+
+            %indentAfterHeadingsUSER= %{$userSettings->[0]->{indentAfterHeadings}} if defined($userSettings->[0]->{indentAfterHeadings});
+            @indentAfterHeadings{ keys %indentAfterHeadingsUSER } = values %indentAfterHeadingsUSER if (%indentAfterHeadingsUSER);
+
+       }
+       else
+       {
+             # otherwise print a warning that we can not read userSettings.yaml
+             print $logfile "WARNING\n\t$settings \n\t contains invalid yaml format- not reading from it\n";
+       }
+
+  }
+  else
+  {
+      # otherwise keep going, but put a warning in the log file
+      print $logfile "\nWARNING\n\t",File::HomeDir->my_home,"/indentconfig.yaml\n";
+      if (-z $settings)
+      {
+          print $logfile "\tspecifies $settings \n\tbut this file is EMPTY- not reading from it\n\n"
+      }
+      else
+      {
+          print $logfile "\tspecifies $settings \n\tbut this file does not exist- unable to read settings from this file\n\n"
+      }
+  }
+}
+
+
+# if we want to over write the current file
+# create a backup first
+if ($overwrite)
+{
+    print $logfile "\nBackup procedure:\n";
+    # original name of file
+    my $filename = $ARGV[0];
+    # copy it
+    my $backupFile = $filename;
+
+    # add the user's backup directory to the backup path
+    $backupFile = "$cruftDirectory/$backupFile";
+
+    # if both ($onlyOneBackUp and $maxNumberOfBackUps) then we have 
+    # a conflict- er on the side of caution and turn off onlyOneBackUp
+    if($onlyOneBackUp and $maxNumberOfBackUps>1)
+    {
+        print $logfile "\t WARNING: onlyOneBackUp=$onlyOneBackUp and maxNumberOfBackUps: $maxNumberOfBackUps\n";
+        print $logfile "\t\t setting onlyOneBackUp=0 which will allow you to reach $maxNumberOfBackUps back ups\n";
+        $onlyOneBackUp = 0;
+    }
+
+    # if the user has specified that $maxNumberOfBackUps = 1 then 
+    # they only want one backup
+    if($maxNumberOfBackUps==1)
+    {
+        $onlyOneBackUp=1 ;
+        print $logfile "\t FYI: you set maxNumberOfBackUps=1, so I'm setting onlyOneBackUp: 1 \n";
+    }
+    elsif($maxNumberOfBackUps<=0 and !$onlyOneBackUp)
+    {
+#        print $logfile "\t FYI: maxNumberOfBackUps=$maxNumberOfBackUps which won't have any effect\n";
+#        print $logfile "\t      on the script- at least ONE backup is made when the -w flag is invoked.\n";
+#        print $logfile "\t      I'm setting onlyOneBackUp: 0, which means that you'll get a new back up file \n";
+#        print $logfile "\t      every time you run the script.\n";
+        $onlyOneBackUp=0 ;
+        $maxNumberOfBackUps=-1;
+    }
+
+    # if onlyOneBackUp is set, then the backup file will
+    # be overwritten each time
+    if($onlyOneBackUp)
+    {
+        $backupFile =~ s/\.tex/$backupExtension/;
+        print $logfile "\t copying $filename to $backupFile\n";
+        print $logfile "\t $backupFile was overwritten\n\n" if (-e $backupFile);
+    }
+    else
+    {
+        # start with a backup file .bak0 (or whatever $backupExtension is present)
+        my $backupCounter = 0;
+        $backupFile =~ s/\.tex$/$backupExtension$backupCounter/;
+
+        # if it exists, then keep going: .bak0, .bak1, ...
+        while (-e $backupFile or $maxNumberOfBackUps>1)
+        {
+            if($backupCounter==$maxNumberOfBackUps)
+            {
+                print $logfile "\t maxNumberOfBackUps reached ($maxNumberOfBackUps)\n";
+                $maxNumberOfBackUps=1 ;
+                last; # break out of the loop
+            }
+            elsif(!(-e $backupFile))
+            {
+                $maxNumberOfBackUps=1 ;
+                last; # break out of the loop
+            }
+            print $logfile "\t $backupFile already exists, incrementing by 1...\n";
+            $backupCounter++;
+            $backupFile =~ s/$backupExtension.*/$backupExtension$backupCounter/;
+        }
+        print $logfile "\n\t copying $filename to $backupFile\n\n";
+    }
+
+    # output these lines to the log file
+    print $logfile "\t Backup file: ",$backupFile,"\n";
+    print $logfile "\t Overwriting file: ",$filename,"\n\n";
+    copy($filename,$backupFile) or die "Could not write to backup file $backupFile. Please check permissions. Exiting.\n";
+}
+
+if(!($outputToFile or $overwrite)) 
+{
+    print $logfile "Just out put to the terminal :)\n\n" if !$silentMode  ;
+}
+
+
+# scalar variables
+my $line;                   # $line: takes the $line of the file
+my $inpreamble=!$indentPreamble;
+                            # $inpreamble: switch to determine if in
+                            #               preamble or not
+my $inverbatim=0;           # $inverbatim: switch to determine if in
+                            #               a verbatim environment or not
+my $delimiters=0;           # $delimiters: switch that governs if
+                            #              we need to check for & or not
+my $matchedbraces=0;        # $matchedbraces: counter to see if { }
+                            #               are matched; it will be 
+                            #               positive if too many { 
+                            #               negative if too many }
+                            #               0 if matched
+my $matchedBRACKETS=0;      # $matchedBRACKETS: counter to see if [ ]
+                            #               are matched; it will be 
+                            #               positive if too many { 
+                            #               negative if too many }
+                            #               0 if matched
+my $commandname;            # $commandname: either \parbox, \marginpar,
+                            #               or anything else from %checkunmatched
+my $commanddetails;         # $command details: a scalar that stores
+                            #               details about the command 
+                            #               that splits {} across lines
+my $countzeros;             # $countzeros:  a counter that helps 
+                            #               when determining if we're in
+                            #               an else construct
+my $lookforelse=0;          # $lookforelse: a boolean to help determine 
+                            #               if we need to look for an 
+                            #               else construct
+my $trailingcomments;       # $trailingcomments stores the comments at the end of 
+                            #           a line 
+my $lineCounter=0;          # $lineCounter keeps track of the line number
+my $inIndentBlock=0;        # $inindentblock: switch to determine if in
+                            #               a inindentblock or not
+my $headingLevel=0;         # $headingLevel: scalar that stores which heading
+                            #               we are under: \part, \chapter, etc
+
+# array variables
+my @indent;                 # @indent: stores current level of indentation
+my @lines;                  # @lines: stores the newly indented lines
+my @block;                  # @block: stores blocks that have & delimiters
+my @commandstore;           # @commandstore: stores commands that 
+                            #           have split {} across lines
+my @commandstorebrackets;   # @commandstorebrackets: stores commands that 
+                            #           have split [] across lines
+my @mainfile;               # @mainfile: stores input file; used to 
+                            #            grep for \documentclass
+my @headingStore;           # @headingStore: stores headings: chapter, section, etc
+my @indentNames;            # @indentNames: keeps names of commands and 
+                            #               environments that have caused
+                            #               indentation to increase
+my @environmentStack;       # @environmentStack: stores the (nested) names 
+                            #                    of environments
+
+# check to see if the current file has \documentclass, if so, then 
+# it's the main file, if not, then it doesn't have preamble
+open(MAINFILE, $ARGV[0]) or die "Could not open input file";
+    @mainfile=<MAINFILE>;
+close(MAINFILE);
+
+# if the MAINFILE doesn't have a \documentclass statement, then 
+# it shouldn't have preamble
+if(scalar(@{[grep(m/^\s*\\documentclass/, @mainfile)]})==0)
+{
+    $inpreamble=0;
+
+    print $logfile "Trace:\tNo documentclass detected, assuming no preamble\n" if($tracingMode);
+}
+else
+{
+    print $logfile "Trace:\t documentclass detected, assuming preamble\n" if($tracingMode);
+}
+
+# the previous OPEN command puts us at the END of the file
+open(MAINFILE, $ARGV[0]) or die "Could not open input file";
+
+# loop through the lines in the INPUT file
+while(<MAINFILE>)
+{
+    # increment the line counter
+    $lineCounter++;
+    
+    # tracing mode
+    print $logfile "\n" if($tracingMode and !($inpreamble or $inverbatim or $inIndentBlock));
+
+    # check to see if we're still in the preamble
+    # or in a verbatim environment or in IndentBlock
+    if(!($inpreamble or $inverbatim or $inIndentBlock))
+    {
+        # if not, remove all leading spaces and tabs
+        # from the current line, assuming it isn't empty
+        #s/^\ *//; 
+        #s/^\s+//; 
+        #s/^\t+//; 
+        s/^\t*// if($_ !~ /^((\s*)|(\t*))*$/); 
+        s/^\s*// if($_ !~ /^((\s*)|(\t*))*$/);
+
+        # tracing mode
+        print $logfile "Line $lineCounter\t removing leading spaces\n" if($tracingMode);
+    }
+    else
+    {
+        # otherwise check to see if we've reached the main
+        # part of the document
+        if(m/^\s*\\begin{document}/)
+        {
+            $inpreamble = 0;
+
+            # tracing mode
+            print $logfile "Line $lineCounter\t \\begin{document} found \n" if($tracingMode);
+        }
+        else
+        {
+            # tracing mode
+            if($inpreamble)
+            {
+                print $logfile "Line $lineCounter\t still in PREAMBLE, doing nothing\n" if($tracingMode);
+            }
+            elsif($inverbatim)
+            {
+                print $logfile "Line $lineCounter\t in VERBATIM-LIKE environment, doing nothing\n" if($tracingMode);
+            }
+            elsif($inIndentBlock)
+            {
+                print $logfile "Line $lineCounter\t in NO INDENT BLOCK, doing nothing\n" if($tracingMode);
+            }
+        }
+    }
+
+    # check to see if we have \end{something} or \]
+    &at_end_of_env_or_eq() unless ($inpreamble or $inIndentBlock);
+
+    # check to see if we're at the end of a noindent 
+    # block %\end{noindent}
+    &at_end_noindent();
+
+    # only check for unmatched braces if we're not in
+    # a verbatim-like environment or in the preamble or in a
+    # noIndentBlock
+    if(!($inverbatim or $inpreamble or $inIndentBlock))
+    {
+        # The check for closing } and ] relies on counting, so 
+        # we have to remove trailing comments so that any {, }, [, ]
+        # that are found after % are not counted
+        #
+        # note that these lines are NOT in @lines, so we
+        # have to store the $trailingcomments to put
+        # back on after the counting
+        #
+        # note the use of (?<!\\)% so that we don't match \%
+        if ( $_=~ m/(?<!\\)%.*/)
+        {
+            s/((?<!\\)%.*)//;
+            $trailingcomments=$1;
+
+            # tracing mode
+            print $logfile "Line $lineCounter\t Removed trailing comments to count braces and brackets: $1\n" if($tracingMode);
+        }
+
+        # check to see if we're at the end of a \parbox, \marginpar
+        # or other split-across-lines command and check that
+        # we're not starting another command that has split braces (nesting)
+        &end_command_or_key_unmatched_braces();
+
+        # check to see if we're at the end of a command that splits 
+        # [ ] across lines
+        &end_command_or_key_unmatched_brackets();
+
+        # check for a heading
+        &indent_heading();
+
+        # add the trailing comments back to the end of the line
+        if(scalar($trailingcomments))
+        {
+            # some line break magic, http://stackoverflow.com/questions/881779/neatest-way-to-remove-linebreaks-in-perl
+            s/\R//;
+            $_ = $_ . $trailingcomments."\n" ;
+
+            # tracing mode
+            print $logfile "Line $lineCounter\t counting braces/brackets complete: added trailing comments back on $trailingcomments\n" if($tracingMode);
+
+            # empty the trailingcomments
+            $trailingcomments='';
+
+        }
+    }
+
+    # remove trailing whitespace
+    if ($removeTrailingWhitespace)
+    {
+        print $logfile "Line $lineCounter\t removing trailing whitespace\n" if ($tracingMode);
+        s/\s+$/\n/;
+    }
+
+    # ADD CURRENT LEVEL OF INDENTATION
+    # (unless we're in a delimiter-aligned block)
+    if(!$delimiters)
+    {
+        # make sure we're not in a verbatim block or in the preamble
+        if($inverbatim or $inpreamble or $inIndentBlock)
+        {
+           # just push the current line as is
+           push(@lines,$_);
+        }
+        else
+        {
+            # add current value of indentation to the current line
+            # and output it
+            # unless this would only create trailing whitespace and the
+            # corresponding option is set
+            unless ($_ =~ m/^$/ and $removeTrailingWhitespace){
+                $_ = join("",@indent).$_;
+            }
+            push(@lines,$_);
+            # tracing mode
+            print $logfile "Line $lineCounter\t Adding current level of indentation: ",join(", ",@indentNames),"\n" if($tracingMode);
+        }
+    }
+    else
+    {
+        # output to @block if we're in a delimiter block
+        push(@block,$_);
+
+        # tracing mode
+        print $logfile "Line $lineCounter\t In delimeter block, waiting for block formatting\n" if($tracingMode);
+    }
+
+    # only check for new environments or commands if we're 
+    # not in a verbatim-like environment or in the preamble
+    # or in a noIndentBlock, or delimiter block
+    if(!($inverbatim or $inpreamble or $inIndentBlock or $delimiters))
+    {
+
+        # check if we are in a 
+        #   % \begin{noindent}
+        # block; this is similar to a verbatim block, the user
+        # may not want some blocks of code to be touched 
+        #
+        # IMPORTANT: this needs to go before the trailing comments
+        # are removed!
+        &at_beg_noindent();
+
+        # remove trailing comments so that any {, }, [, ]
+        # that are found after % are not counted
+        #
+        # note that these lines are already in @lines, so we
+        # can remove the trailing comments WITHOUT having
+        # to put them back in
+        #
+        # Note that this won't match \%
+        s/(?<!\\)%.*// if( $_=~ m/(?<!\\)%.*/);
+
+        # tracing mode
+        print $logfile "Line $lineCounter\t Removing trailing comments for brace count (line is already stored)\n" if($tracingMode);
+
+        # check to see if we have \begin{something} or \[ 
+        &at_beg_of_env_or_eq();
+
+        # check to see if we have \parbox, \marginpar, or
+        # something similar that might split braces {} across lines,
+        # specified in %checkunmatched hash table
+        &start_command_or_key_unmatched_braces();
+
+        # check for an else statement
+        &check_for_else();
+
+        # check for a command that splits [] across lines
+        &start_command_or_key_unmatched_brackets();
+
+        # check for a heading
+        &indent_after_heading();
+
+        # tracing mode
+        print $logfile "Line $lineCounter\t Environments: ",join(", ",@environmentStack),"\n" if($tracingMode and scalar(@environmentStack));
+    }
+}
+
+# close the main file
+close(MAINFILE);
+
+# put line count information in the log file
+print $logfile "Line Count of $ARGV[0]: ",scalar(@mainfile),"\n";
+print $logfile "Line Count of indented $ARGV[0]: ",scalar(@lines);
+if(scalar(@mainfile) != scalar(@lines))
+{
+  print $logfile <<ENDQUOTE
+WARNING: \t line count of original file and indented file does 
+\t not match- consider reverting to a back up, see $backupExtension;
+ENDQUOTE
+;
+}
+else
+{
+    print $logfile "\n\nLine counts of original file and indented file match";
+}
+
+# output the formatted lines to the terminal
+print @lines if(!$silentMode);
+
+# if -w is active then output to $ARGV[0]
+if($overwrite)
+{
+    open(OUTPUTFILE,">",$ARGV[0]);
+    print OUTPUTFILE @lines;
+    close(OUTPUTFILE);
+}
+
+# if -o is active then output to $ARGV[1]
+if($outputToFile)
+{
+    open(OUTPUTFILE,">",$ARGV[1]);
+    print OUTPUTFILE @lines;
+    close(OUTPUTFILE);
+}
+
+# close the log file
+close($logfile);
+
+exit;
+
+sub indent_heading{
+    # PURPOSE: This matches
+    #           \part
+    #           \chapter
+    #           \section
+    #           \subsection
+    #           \subsubsection
+    #           \paragraph
+    #           \subparagraph
+    #
+    #           and anything else listed in indentAfterHeadings
+    #
+    #           This subroutine specifies the indentation for the 
+    #           heading itself, i.e the line that has \chapter, \section etc
+    if( $_ =~ m/^\s*\\(.*?)(\[|{)/ and $indentAfterHeadings{$1})
+    {
+       # tracing mode
+       print $logfile "Line $lineCounter\t Heading found: $1 \n" if($tracingMode);
+
+       # get the heading settings- it's a hash within a hash
+       my %currentHeading = %{$indentAfterHeadings{$1}};
+
+       # local scalar
+       my $previousHeadingLevel = $headingLevel;
+
+       # if current heading level < old heading level, 
+       if($currentHeading{level}<$previousHeadingLevel)
+       {
+            # decrease indentation, but only if 
+            # specified in indentHeadings. Note that this check 
+            # needs to be done here- decrease_indent won't 
+            # check a nested hash
+          
+            if(scalar(@headingStore))  
+            {
+               while($currentHeading{level}<$previousHeadingLevel and scalar(@headingStore))
+               {
+                    my $higherHeadingName = pop(@headingStore);
+                    my %higherLevelHeading = %{$indentAfterHeadings{$higherHeadingName}};
+
+                    # tracing mode
+                    print $logfile "Line $lineCounter\t stepping UP heading level from $higherHeadingName \n" if($tracingMode);
+
+                    &decrease_indent($higherHeadingName) if($higherLevelHeading{indent});
+                    $previousHeadingLevel=$higherLevelHeading{level};
+               }
+               # put the heading name back in to storage
+               push(@headingStore,$1);
+            }
+       }
+       elsif($currentHeading{level}==$previousHeadingLevel)
+       {
+            if(scalar(@headingStore))  
+            {
+                 my $higherHeadingName = pop(@headingStore);
+                 my %higherLevelHeading = %{$indentAfterHeadings{$higherHeadingName}};
+                 &decrease_indent($higherHeadingName) if($higherLevelHeading{indent});
+            }
+            # put the heading name back in to storage
+            push(@headingStore,$1);
+       }
+       else
+       {
+            # put the heading name into storage
+            push(@headingStore,$1);
+       }
+    }
+}
+
+sub indent_after_heading{
+    # PURPOSE: This matches
+    #           \part
+    #           \chapter
+    #           \section
+    #           \subsection
+    #           \subsubsection
+    #           \paragraph
+    #           \subparagraph
+    #
+    #           and anything else listed in indentAfterHeadings
+    #
+    #           This subroutine is specifies the indentation for 
+    #           the text AFTER the heading, i.e the body of conent
+    #           in each \chapter, \section, etc
+    if( $_ =~ m/^\s*\\(.*?)(\[|{)/ and $indentAfterHeadings{$1})
+    {
+       # get the heading settings- it's a hash within a hash
+       my %currentHeading = %{$indentAfterHeadings{$1}};
+
+       &increase_indent($1) if($currentHeading{indent});
+
+       # update heading level
+       $headingLevel = $currentHeading{level};
+    }
+}
+
+
+
+sub at_end_noindent{
+    # PURPOSE: This matches
+    #           % \end{noindent}
+    #          with any number of spaces (possibly none) between
+    #          the comment and \end{noindent}.
+    #
+    #          the comment symbol IS indended!
+    #
+    #          This is for blocks of code that the user wants
+    #          to leave untouched- similar to verbatim blocks
+
+    if( $_ =~ m/^%\s*\\end{(.*?)}/ and $noIndentBlock{$1}) 
+    {
+            $inIndentBlock=0;
+            # tracing mode
+            print $logfile "Line $lineCounter\t % \\end{no indent block} found, switching inIndentBlock OFF \n" if($tracingMode);
+    }
+}
+
+sub at_beg_noindent{
+    # PURPOSE: This matches
+    #           % \begin{noindent}
+    #          with any number of spaces (possibly none) between
+    #          the comment and \begin{noindent}.
+    #
+    #          the comment symbol IS indended!
+    #
+    #          This is for blocks of code that the user wants
+    #          to leave untouched- similar to verbatim blocks
+
+    if( $_ =~ m/^%\s*\\begin{(.*?)}/ and $noIndentBlock{$1}) 
+    {
+           $inIndentBlock = 1;
+           # tracing mode
+           print $logfile "Line $lineCounter\t % \\begin{no indent block} found, switching inIndentBlock ON \n" if($tracingMode);
+    }
+}
+
+sub start_command_or_key_unmatched_brackets{
+    # PURPOSE: This matches 
+    #              \pgfplotstablecreatecol[...
+    #
+    #              or any other command/key that has brackets [ ] 
+    #              split across lines specified in the 
+    #              hash tables, %checkunmatchedbracket
+    #
+    # How to read: ^\s*(\\)?(.*?)(\[\s*)
+    #
+    #       ^       line begins with
+    #       \s*     any (or no)spaces
+    #       (\\)?   matches a \ backslash but not necessarily
+    #       (.*?)   non-greedy character match and store the result
+    #       ((?<!\\)\[\s*) match [ possibly leading with spaces
+    #                      but it WON'T match \[
+
+    if ($_ =~ m/^\s*(\\)?(.*?)(\s*(?<!\\)\[)/ 
+        and (scalar($checkunmatchedbracket{$2})
+             or $alwaysLookforSplitBrackets)
+        )
+        {
+            # store the command name, because $2
+            # will not exist after the next match
+            $commandname = $2;
+            $matchedBRACKETS=0;
+
+            # match [ but don't match \[
+            $matchedBRACKETS++ while ($_ =~ /(?<!\\)\[/g);
+            # match ] but don't match \]
+            $matchedBRACKETS-- while ($_ =~ /(?<!\\)\]/g);
+
+            # set the indentation
+            if($matchedBRACKETS != 0 )
+            {
+                  # tracing mode
+                  print $logfile "Line $lineCounter\t Found opening BRACKET [ $commandname\n" if($tracingMode);
+
+                  &increase_indent($commandname);
+
+                  # store the command name
+                  # and the value of $matchedBRACKETS
+                  push(@commandstorebrackets,{commandname=>$commandname,
+                                      matchedBRACKETS=>$matchedBRACKETS});
+
+            }
+        }
+}
+
+sub end_command_or_key_unmatched_brackets{
+    # PURPOSE:  Check for the closing BRACKET of a command that 
+    #           splits its BRACKETS across lines, such as
+    #
+    #               \pgfplotstablecreatecol[ ...
+    #
+    #           It works by checking if we have any entries
+    #           in the array @commandstorebrackets, and making 
+    #           sure that we're not starting another command/key
+    #           that has split BRACKETS (nesting).
+    #
+    #           It also checks that the line is not commented.
+    #
+    #           We count the number of [ and ADD to the counter
+    #                                  ] and SUBTRACT to the counter
+    if(scalar(@commandstorebrackets) 
+        and  !($_ =~ m/^\s*(\\)?(.*?)(\s*\[)/ 
+                    and (scalar($checkunmatchedbracket{$2})
+                         or $alwaysLookforSplitBrackets))
+        and $_ !~ m/^\s*%/
+       )
+    {
+       # get the details of the most recent command name
+       $commanddetails = pop(@commandstorebrackets);
+       $commandname = $commanddetails->{'commandname'};
+       $matchedBRACKETS = $commanddetails->{'matchedBRACKETS'};
+
+       # match [ but don't match \[
+       $matchedBRACKETS++ while ($_ =~ m/(?<!\\)\[/g);
+
+       # match ] but don't match \]
+       $matchedBRACKETS-- while ($_ =~ m/(?<!\\)\]/g);
+
+       # if we've matched up the BRACKETS then
+       # we can decrease the indent by 1 level
+       if($matchedBRACKETS == 0)
+       {
+            # tracing mode
+            print $logfile "Line $lineCounter\t Found closing BRACKET ] $commandname\n" if($tracingMode);
+
+            # decrease the indentation (if appropriate)
+            &decrease_indent($commandname);
+       }
+       else
+       {
+           # otherwise we need to enter the new value
+           # of $matchedBRACKETS and the value of $command
+           # back into storage
+           push(@commandstorebrackets,{commandname=>$commandname,
+                                       matchedBRACKETS=>$matchedBRACKETS});
+           # tracing mode
+           print $logfile "Line $lineCounter\t Searching for closing BRACKET ] $commandname\n" if($tracingMode);
+       }
+     }
+}
+
+sub start_command_or_key_unmatched_braces{
+    # PURPOSE: This matches 
+    #              \parbox{...
+    #              \parbox[..]..{
+    #              empty header/.style={
+    #              \foreach \something
+    #              etc
+    #
+    #              or any other command/key that has BRACES
+    #              split across lines specified in the 
+    #              hash tables, %checkunmatched, %checkunmatchedELSE
+    #
+    # How to read: ^\s*(\\)?(.*?)(\[|{|\s)
+    #
+    #       ^                  line begins with
+    #       \s*                any (or no) spaces
+    #       (\\)?              matches a \ backslash but not necessarily
+    #       (.*?)              non-greedy character match and store the result
+    #       (\[|}|=|(\s*\\))   either [ or { or = or space \
+
+    if ($_ =~ m/^\s*(\\)?(.*?)(\[|{|=|(\s*\\))/ 
+            and (scalar($checkunmatched{$2}) 
+                 or scalar($checkunmatchedELSE{$2})
+                 or $alwaysLookforSplitBraces)
+        )
+        {
+            # store the command name, because $2
+            # will not exist after the next match
+            $commandname = $2;
+            $matchedbraces=0;
+
+            # by default, don't look for an else construct
+            $lookforelse=0;
+            if(scalar($checkunmatchedELSE{$2}))
+            {
+                $lookforelse=1;
+            }
+
+            # match { but don't match \{
+            $matchedbraces++ while ($_ =~ /(?<!\\){/g);
+
+            # match } but don't match \}
+            $matchedbraces-- while ($_ =~ /(?<!\\)}/g);
+
+            # tracing mode
+            print $logfile "Line $lineCounter\t matchedbraces = $matchedbraces\n" if($tracingMode);
+
+            # set the indentation
+            if($matchedbraces > 0 )
+            {
+                  # tracing mode
+                  print $logfile "Line $lineCounter\t Found opening BRACE { $commandname\n" if($tracingMode);
+
+                  &increase_indent($commandname);
+
+                  # store the command name
+                  # and the value of $matchedbraces
+                  push(@commandstore,{commandname=>$commandname,
+                                      matchedbraces=>$matchedbraces,
+                                      lookforelse=>$lookforelse,
+                                      countzeros=>0});
+
+            }
+            elsif($matchedbraces<0)
+            {
+                # if $matchedbraces < 0 then we must be matching
+                # braces from a previous split-braces command
+                
+                # keep matching { OR }, and don't match \{ or \}
+                while ($_ =~ m/(((?<!\\){)|((?<!\\)}))/g)
+                {
+
+                     # store the match, either { or }
+                     my $braceType = $1;
+
+                     # get the details of the most recent command name
+                     $commanddetails = pop(@commandstore);
+                     $commandname = $commanddetails->{'commandname'};
+                     $matchedbraces = $commanddetails->{'matchedbraces'};
+                     $countzeros = $commanddetails->{'countzeros'};
+                     $lookforelse= $commanddetails->{'lookforelse'};
+
+                     $matchedbraces++ if($1 eq "{");
+                     $matchedbraces-- if($1 eq "}");
+
+                     # if we've matched up the braces then
+                     # we can decrease the indent by 1 level
+                     if($matchedbraces == 0)
+                     {
+                          $countzeros++ if $lookforelse;
+
+                          # tracing mode
+                          print $logfile "Line $lineCounter\t Found closing BRACE } $1\n" if($tracingMode);
+
+                          # decrease the indentation (if appropriate)
+                          &decrease_indent($commandname);
+
+                         if($countzeros==1)
+                         {
+                              push(@commandstore,{commandname=>$commandname,
+                                                  matchedbraces=>$matchedbraces,
+                                                  lookforelse=>$lookforelse,
+                                                  countzeros=>$countzeros});
+                         }
+                     }
+                     else
+                     {
+                            # otherwise we need to put the command back for the 
+                            # next brace count
+                            push(@commandstore,{commandname=>$commandname,
+                                                matchedbraces=>$matchedbraces,
+                                                lookforelse=>$lookforelse,
+                                                countzeros=>$countzeros});
+                     }
+                }
+            }
+        }
+}
+
+sub end_command_or_key_unmatched_braces{
+    # PURPOSE:  Check for the closing BRACE of a command that 
+    #           splits its BRACES across lines, such as
+    #
+    #               \parbox{ ...
+    #
+    #           or one of the tikz keys, such as
+    #           
+    #              empty header/.style={
+    #
+    #           It works by checking if we have any entries
+    #           in the array @commandstore, and making 
+    #           sure that we're not starting another command/key
+    #           that has split BRACES (nesting).
+    #
+    #           It also checks that the line is not commented.
+    #
+    #           We count the number of { and ADD to the counter
+    #                                  } and SUBTRACT to the counter
+    if(scalar(@commandstore) 
+      and  !($_ =~ m/^\s*(\\)?(.*?)(\[|{|=|(\s*\\))/ 
+                    and (scalar($checkunmatched{$2}) 
+                         or scalar($checkunmatchedELSE{$2})
+                         or $alwaysLookforSplitBraces))
+        and $_ !~ m/^\s*%/
+       )
+    {
+       # keep matching { OR }, and don't match \{ or \}
+       while ($_ =~ m/(((?<!\\){)|((?<!\\)}))/g)
+       {
+            # store the match, either { or }
+            my $braceType = $1;
+
+            # get the details of the most recent command name
+            $commanddetails = pop(@commandstore);
+            $commandname = $commanddetails->{'commandname'};
+            $matchedbraces = $commanddetails->{'matchedbraces'};
+            $countzeros = $commanddetails->{'countzeros'};
+            $lookforelse= $commanddetails->{'lookforelse'};
+
+            $matchedbraces++ if($1 eq "{");
+            $matchedbraces-- if($1 eq "}");
+
+            # if we've matched up the braces then
+            # we can decrease the indent by 1 level
+            if($matchedbraces == 0)
+            {
+                 $countzeros++ if $lookforelse;
+
+                 # tracing mode
+                 print $logfile "Line $lineCounter\t Found closing BRACE } $commandname\n" if($tracingMode);
+
+                 # decrease the indentation (if appropriate)
+                 &decrease_indent($commandname);
+
+                if($countzeros==1)
+                {
+                     push(@commandstore,{commandname=>$commandname,
+                                         matchedbraces=>$matchedbraces,
+                                         lookforelse=>$lookforelse,
+                                         countzeros=>$countzeros});
+                }
+            }
+            else
+            {
+                # otherwise we need to enter the new value
+                # of $matchedbraces and the value of $command
+                # back into storage
+                push(@commandstore,{commandname=>$commandname,
+                                    matchedbraces=>$matchedbraces,
+                                    lookforelse=>$lookforelse,
+                                    countzeros=>$countzeros});
+
+               # tracing mode
+               print $logfile "Line $lineCounter\t Searching for closing BRACE } $commandname\n" if($tracingMode);
+            }
+     }
+     }
+}
+
+sub check_for_else{
+    # PURPOSE: Check for an else clause
+    #
+    #          Some commands have the form
+    #
+    #               \mycommand{
+    #                   if this
+    #               }
+    #               {
+    #                   else this
+    #               }
+    #
+    #          so we need to look for the else bit, and set 
+    #          the indentation appropriately.
+    #
+    #          We only perform this check if there's something
+    #          in the array @commandstore, and if 
+    #          the line itself is not a command, or comment, 
+    #          and if it begins with {
+
+    if(scalar(@commandstore) 
+        and  !($_ =~ m/^\s*(\\)?(.*?)(\[|{|=)/ 
+                    and (scalar($checkunmatched{$2}) 
+                         or scalar($checkunmatchedELSE{$2})
+                         or $alwaysLookforSplitBraces))
+        and $_ =~ m/^\s*{/
+        and $_ !~ m/^\s*%/
+       )
+    {
+       # get the details of the most recent command name
+       $commanddetails = pop(@commandstore);
+       $commandname = $commanddetails->{'commandname'};
+       $matchedbraces = $commanddetails->{'matchedbraces'};
+       $countzeros = $commanddetails->{'countzeros'};
+       $lookforelse= $commanddetails->{'lookforelse'};
+
+       # increase indentation
+       if($lookforelse and $countzeros==1)
+       {
+         &increase_indent($commandname);
+       }
+
+       # put the array back together
+       push(@commandstore,{commandname=>$commandname,
+                           matchedbraces=>$matchedbraces,
+                           lookforelse=>$lookforelse,
+                           countzeros=>$countzeros});
+    }
+}
+
+sub at_beg_of_env_or_eq{
+    # PURPOSE: Check if we're at the BEGINning of an environment
+    #          or at the BEGINning of a displayed equation \[
+    #
+    #          This subroutine checks for matches of the form
+    #
+    #               \begin{environmentname}
+    #          or
+    #               \[
+    #
+    #          It also checks to see if the current environment
+    #          should have alignment delimiters; if so, we need to turn 
+    #          ON the $delimiter switch 
+
+    # How to read
+    #  m/^\s*(\$)?\\begin{(.*?)}/ 
+    #
+    #   ^               beginning of a line
+    #   \s*             any white spaces (possibly none)
+    #   (\$)?           possibly a $ symbol, but not required
+    #   \\begin{(.*)?}  \begin{environmentname}
+    #
+    # How to read
+    #  m/^\s*()(\\\[)/
+    #
+    #  ^        beginning of a line
+    #  \s*      any white spaces (possibly none)
+    #  ()       empty just so that $1 and $2 are defined
+    #  (\\\[)   \[  there are lots of \ because both \ and [ need escaping 
+
+    if( (   ( $_ =~ m/^\s*(\$)?\\begin{(.*?)}/ and $_ !~ m/\\end{$2}/) 
+         or ($_=~ m/^\s*()(\\\[)/ and $_ !~ m/\\\]/) )
+        and $_ !~ m/^\s*%/ )
+    {
+       # tracing mode
+       print $logfile "Line $lineCounter\t \\begin{environment} found: $2 \n" if($tracingMode);
+
+       # increase the indentation 
+       &increase_indent($2);
+
+       # check for verbatim-like environments
+       if($verbatimEnvironments{$2})
+       {
+           $inverbatim = 1;
+           # tracing mode
+           print $logfile "Line $lineCounter\t \\begin{verbatim-like} found, $2, switching ON verbatim \n" if($tracingMode);
+
+           # remove the key and value from %lookForAlignDelims hash
+           # to avoid any further confusion
+           if($lookForAlignDelims{$2})
+           {
+                print $logfile "WARNING\n\t Line $lineCounter\t $2 is in *both* lookForAlignDelims and verbatimEnvironments\n";
+                print $logfile "\t\t\t ignoring lookForAlignDelims and prioritizing verbatimEnvironments\n";
+                print $logfile "\t\t\t Note that you only get this message once per environment\n";
+                delete $lookForAlignDelims{$2};
+           }
+
+       }
+
+       # check to see if we need to look for alignment
+       # delimiters
+       if($lookForAlignDelims{$2})
+       {
+           $delimiters=1;
+            # tracing mode
+            print $logfile "Line $lineCounter\t Delimiter environment started: $2 (see lookForAlignDelims)\n" if($tracingMode);
+       }
+
+       # store the name of the environment
+       push(@environmentStack,$2);
+
+    }
+}
+
+sub at_end_of_env_or_eq{
+    # PURPOSE: Check if we're at the END of an environment
+    #          or at the END of a displayed equation \]
+    #
+    #          This subroutine checks for matches of the form
+    #
+    #               \end{environmentname}
+    #          or
+    #               \]
+    #
+    #          It also checks to see if the current environment
+    #          had alignment delimiters; if so, we need to turn 
+    #          OFF the $delimiter switch 
+    
+    if( ($_ =~ m/^\s*\\end{(.*?)}/ or $_=~ m/^(\\\])/)
+         and $_ !~ m/\s*^%/)
+    {
+
+       # check if we're at the end of a verbatim-like environment
+       if($verbatimEnvironments{$1})
+       {
+           $inverbatim = 0;
+            # tracing mode
+
+            print $logfile "Line $lineCounter\t \\end{verbatim-like} found: $1, switching off verbatim \n" if($tracingMode);
+            print $logfile "Line $lineCounter\t removing leading spaces \n" if($tracingMode);
+            #s/^\ *//; 
+            s/^\t+// if($_ ne ""); 
+            s/^\s+// if($_ ne ""); 
+       }
+
+       # check to see if \end{environment} fits with most recent \begin{...}
+       my $previousEnvironment = pop(@environmentStack);
+
+       # check to see if we need to turn off alignment
+       # delimiters and output the current block
+       if($lookForAlignDelims{$1} and ($previousEnvironment eq $1))
+       {
+           $delimiters=0;
+
+           # tracing mode
+           print $logfile "Line $lineCounter\t Delimiter body FINISHED: $1\n" if($tracingMode);
+
+           # print the current FORMATTED block
+           @block = &format_block(@block);
+           foreach $line (@block)
+           {
+                # add the indentation and add the 
+                # each line of the formatted block
+                # to the output
+                # unless this would only create trailing whitespace and the
+                # corresponding option is set
+                unless ($line =~ m/^$/ and $removeTrailingWhitespace)
+                {
+                    $line = join("",@indent).$line;
+                }
+                push(@lines,$line);
+           }
+           # empty the @block, very important!
+           @block=();
+       }
+
+       # tracing mode
+       print $logfile "Line $lineCounter\t \\end{envrionment} found: $1 \n" if($tracingMode and !$verbatimEnvironments{$1});
+
+       # check to see if \end{environment} fits with most recent \begin{...}
+       if($previousEnvironment eq $1)
+       {
+            # decrease the indentation (if appropriate)
+            &decrease_indent($1);
+       }
+       else
+       {
+           # otherwise put the environment name back on the stack
+           push(@environmentStack,$previousEnvironment);
+           print $logfile "Line $lineCounter\t WARNING: \\end{$1} found on its own line, not matched to \\begin{$previousEnvironment}\n" unless ($delimiters or $inverbatim or $inIndentBlock or $1 eq "\\\]");
+       }
+
+       # need a special check for \[ and \]
+       if($1 eq "\\\]")
+       {
+            &decrease_indent($1);
+            pop(@environmentStack);
+       }
+
+       # if we're at the end of the document, we remove all current 
+       # indentation- this is especially prominent in examples that 
+       # have headings, and the user has chosen to indentAfterHeadings
+       if($1 eq "document" and !(grep(/filecontents/, @indentNames)) and !$inpreamble and !$delimiters and !$inverbatim and !$inIndentBlock)
+       {
+            @indent=(); 
+            @indentNames=(); 
+            print $logfile "Line $lineCounter\t \\end{$1} found- emptying indent array \n" unless ($delimiters or $inverbatim or $inIndentBlock or $1 eq "\\\]");
+       }
+    }
+}
+
+sub format_block{
+    #   PURPOSE: Format a delimited environment such as the 
+    #            tabular or align environment that contains &
+    #
+    #   INPUT: @block               array containing unformatted block
+    #                               from, for example, align, or tabular
+    #   OUTPUT: @formattedblock     array containing FORMATTED block
+
+
+    # @block is the input
+    my @block=@_;
+
+    # tracing mode
+    print $logfile "\t\tFormatting alignment block\n" if($tracingMode);
+
+    # step the line counter back to the beginning of the block-
+    # it will be increased back to the end of the block in the 
+    # loop later on:  foreach $row (@tmpblock)
+    $lineCounter -= scalar(@block);
+
+
+    # local array variables
+    my @formattedblock;
+    my @tmprow=();
+    my @tmpblock=();
+    my @maxmstringsize=();
+    my @ampersandCount=();
+
+    # local scalar variables
+    my $alignrowcounter=-1;
+    my $aligncolcounter=-1;
+    my $tmpstring;
+    my $row;
+    my $column;
+    my $maxmcolstrlength;
+    my $i;
+    my $j;
+    my $fmtstring;
+    my $linebreak;
+    my $maxNumberAmpersands = 0;
+    my $currentNumberAmpersands;
+    my $trailingcomments;
+
+    # local hash table
+    my %stringsize=();
+
+    # loop through the block and count & per line- store the biggest
+    # NOTE: this needs to be done in its own block so that 
+    # we can know what the maximum number of & in the block is
+    foreach $row (@block)
+    {
+       # delete trailing comments
+       $trailingcomments='';
+       if($row =~ m/((?<!\\)%.*$)/)
+       {
+            $row =~ s/((?<!\\)%.*)/%TC/;
+            $trailingcomments=$1;
+       }
+
+       # reset temporary counter
+       $currentNumberAmpersands=0;
+
+       # count & in current row (exclude \&)
+       $currentNumberAmpersands++ while ($row =~ /(?<!\\)&/g);
+
+       # store the ampersand count for future
+       push(@ampersandCount,$currentNumberAmpersands);
+
+       # overwrite maximum count if the temp count is higher
+       $maxNumberAmpersands = $currentNumberAmpersands if($currentNumberAmpersands > $maxNumberAmpersands );
+
+       # put trailing comments back on
+       if($trailingcomments)
+       {
+            $row =~ s/%TC/$trailingcomments/;
+       }
+    }
+
+    # tracing mode
+    print $logfile "\t\tmaximum number of & in any row: $maxNumberAmpersands\n" if($tracingMode);
+    
+    # loop through the lines in the @block
+    foreach $row (@block)
+    {
+        # get the ampersand count
+        $currentNumberAmpersands = shift(@ampersandCount);
+
+        # increment row counter
+        $alignrowcounter++;
+
+        # clear the $linebreak variable
+        $linebreak='';
+
+        # check for line break \\
+        # and don't mess with a line that doesn't have the maximum
+        # number of &
+        if($row =~ m/\\\\/ and $currentNumberAmpersands==$maxNumberAmpersands )
+        {
+          # remove \\ and all characters that follow
+          # and put it back in later, once the measurement
+          # has been done
+          $row =~ s/(\\\\.*)//;
+          $linebreak = $1;
+        }
+
+        if($currentNumberAmpersands==$maxNumberAmpersands)
+        {
+
+            # remove trailing comments
+            $trailingcomments='';
+            if($row =~ m/((?<!\\)%.*$)/)
+            {
+                 $row =~ s/((?<!\\)%.*)/%TC/;
+                 $trailingcomments=$1;
+            }
+
+            # separate the row at each &, but not at \&
+            @tmprow = split(/(?<!\\)&/,$row);
+
+            # reset column counter
+            $aligncolcounter=-1;
+
+            # loop through each column element
+            # removing leading and trailing space
+            foreach $column (@tmprow)
+            {
+               # increment column counter
+               $aligncolcounter++;
+    
+               # remove leading and trailing space from element
+    	       $column =~ s/^\s+//;
+               $column =~ s/\s+$//;
+    
+               # assign string size to the array
+               $stringsize{$alignrowcounter.$aligncolcounter}=length($column);
+               if(length($column)==0)
+               {
+                 $column=" ";
+               }
+
+               # put the row back together
+               if ($aligncolcounter ==0)
+               {
+                 $tmpstring = $column;
+               }
+               else
+               {
+                 $tmpstring .= "&".$column;
+               }
+            }
+
+            
+            # put $linebreak back on the string, now that
+            # the measurement has been done
+            $tmpstring .= $linebreak;
+
+            # put trailing comments back on
+            if($trailingcomments)
+            {
+                 $tmpstring =~ s/%TC/$trailingcomments/;
+            }
+
+            push(@tmpblock,$tmpstring);
+        }
+        else
+        {
+               # if there are no & then use the 
+               # NOFORMATTING token
+               # remove leading space
+    	       s/^\s+//;
+               push(@tmpblock,$row."NOFORMATTING");
+        }
+    }
+
+    # calculate the maximum string size of each column
+    for($j=0;$j<=$aligncolcounter;$j++)
+    {
+        $maxmcolstrlength=0;
+        for($i=0; $i<=$alignrowcounter;$i++)
+        {
+            # make sure the stringsize is defined
+            if(defined $stringsize{$i.$j})
+            {
+                if ($stringsize{$i.$j}>$maxmcolstrlength)
+                {
+                    $maxmcolstrlength = $stringsize{$i.$j};
+                }
+            }
+        }
+        push(@maxmstringsize,$maxmcolstrlength);
+    }
+
+    # README: printf( formatting, expression)
+    #
+    #   formatting has the form %-50s & %-20s & %-19s
+    #   (the numbers have been made up for example)
+    #       the - symbols mean that each column should be left-aligned
+    #       the numbers represent how wide each column is
+    #       the s represents string
+    #       the & needs to be inserted
+    
+    # join up the maximum string lengths using "s %-"
+    $fmtstring = join("s & %-",@maxmstringsize);
+    
+    # add an s to the end, and a newline
+    $fmtstring .= "s ";
+    
+    # add %- to the beginning
+    $fmtstring = "%-".$fmtstring;
+    
+    # process the @tmpblock of aligned material
+    foreach $row (@tmpblock)
+    {
+
+        $linebreak='';
+        # check for line break \\
+        if($row =~ m/\\\\/)
+        {
+          # remove \\ and all characters that follow
+          # and put it back in later
+          $row =~ s/(\\\\.*$)//;
+          $linebreak = $1;
+        }
+
+        if($row =~ m/NOFORMATTING/)
+        {
+            $row =~ s/NOFORMATTING//;
+            $tmpstring=$row;
+
+            # tracing mode
+            print $logfile "\t\tLine $lineCounter\t maximum number of & NOT found- not aligning delimiters \n" if($tracingMode);
+        }
+        else
+        {
+          # remove trailing comments
+          $trailingcomments='';
+          if($row =~ m/((?<!\\)%.*$)/)
+          {
+               $row =~ s/((?<!\\)%.*)/%TC/;
+               $trailingcomments=$1;
+          }
+
+          $tmpstring = sprintf($fmtstring,split(/(?<!\\)&/,$row)).$linebreak."\n";
+
+          # put trailing comments back on
+          if($trailingcomments)
+          {
+               $tmpstring =~ s/%TC/$trailingcomments/;
+          }
+
+          # tracing mode
+          print $logfile "\t\tLine $lineCounter\t Found maximum number of & so aligning delimiters\n" if($tracingMode);
+        }
+
+        # remove trailing whitespace
+        if ($removeTrailingWhitespace)
+        {
+            print $logfile "Line $lineCounter\t removing trailing whitespace from delimiter aligned line\n" if ($tracingMode);
+            $tmpstring =~ s/\s+$/\n/;
+        }
+
+        push(@formattedblock,$tmpstring);
+
+        # increase the line counter
+        $lineCounter++;
+    }
+
+    # return the formatted block
+	@formattedblock;
+}
+
+sub increase_indent{
+       # PURPOSE: Adjust the indentation
+       #          of the current environment;
+       #          check that it's not an environment
+       #          that doesn't want indentation.
+
+       my $command = pop(@_);
+
+       # if the user has specified $indentRules{$command} and 
+       # $noAdditionalIndent{$command} then they are a bit confused-
+       # we remove the $indentRules{$command} and assume that they
+       # want $noAdditionalIndent{$command}
+       if(scalar($indentRules{$command}) and $noAdditionalIndent{$command})
+       {
+            print $logfile "WARNING\n\t Line $lineCounter\t $command is in *both* indentRules and noAdditionalIndent\n";
+            print $logfile "\t\t\t ignoring indentRules and prioritizing noAdditionalIndent\n";
+            print $logfile "\t\t\t Note that you only get this message once per command/environment\n";
+
+            # remove the key and value from %indentRules hash
+            # to avoid any further confusion
+            delete $indentRules{$command};
+       }
+
+       # if the command is in verbatimEnvironments and in indentRules then 
+       # remove it from %indentRules hash
+       # to avoid any further confusion
+       if($indentRules{$command} and $verbatimEnvironments{$command})
+       {
+            # remove the key and value from %indentRules hash
+            # to avoid any further confusion
+            print $logfile "WARNING\n\t Line $lineCounter\t $command is in *both* indentRules and verbatimEnvironments\n";
+            print $logfile "\t\t\t ignoring indentRules and prioritizing verbatimEnvironments\n";
+            print $logfile "\t\t\t Note that you only get this message once per environment\n";
+            delete $indentRules{$command};
+       }
+
+       if(scalar($indentRules{$command}))
+       {
+          # if there's a rule for indentation for this environment
+          push(@indent, $indentRules{$command});
+          # tracing mode
+          print $logfile "Line $lineCounter\t increasing indent using rule for $command (see indentRules)\n" if($tracingMode);
+          push(@indentNames,"$command (rule)");
+       }
+       else
+       {
+          # default indentation
+          if(!($noAdditionalIndent{$command} or $verbatimEnvironments{$command} or $inverbatim))
+          {
+            push(@indent, $defaultIndent);
+            push(@indentNames,"$command (default)");
+            # tracing mode
+            print $logfile "Line $lineCounter\t increasing indent using defaultIndent\n" if($tracingMode);
+          }
+          elsif($noAdditionalIndent{$command})
+          {
+            # tracing mode
+            print $logfile "Line $lineCounter\t no additional indent added for $command (see noAdditionalIndent)\n" if($tracingMode);
+          }
+       }
+}
+
+sub decrease_indent{
+       # PURPOSE: Adjust the indentation
+       #          of the current environment;
+       #          check that it's not an environment
+       #          that doesn't want indentation.
+
+       my $command = pop(@_);
+
+       if(!($noAdditionalIndent{$command} or $verbatimEnvironments{$command} or $inverbatim))
+       {
+            pop(@indent);
+            pop(@indentNames);
+            # tracing mode
+            print $logfile "Line $lineCounter\t decreasing indent to: ",join(", ",@indentNames),"\n" if($tracingMode);
+       }
+}
+