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authorKarl Berry <karl@freefriends.org>2013-11-15 00:51:07 +0000
committerKarl Berry <karl@freefriends.org>2013-11-15 00:51:07 +0000
commitbc1b5e3250223eb944aa85d3e420f400b63e0254 (patch)
tree74ca0858368ead9c5ae0af04bc8597ee1f0a8170 /Master/texmf-dist
parent4909c173ce26f3b8abbb198131d4c2846918c8b0 (diff)
latexindent (13nov13)
git-svn-id: svn://tug.org/texlive/trunk@32150 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist')
-rw-r--r--Master/texmf-dist/doc/support/latexindent/README47
-rw-r--r--Master/texmf-dist/doc/support/latexindent/documentation/manual.pdfbin0 -> 231794 bytes
-rw-r--r--Master/texmf-dist/doc/support/latexindent/documentation/manual.tex1042
-rw-r--r--Master/texmf-dist/doc/support/latexindent/indent.yaml38
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/bigTest.tex293
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/braceTest.tex57
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/braceTestsmall.tex31
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/environments.tex38
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/filecontents.tex73
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/matrix.tex14
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/nestedalignment.tex33
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/nestedalignment1.tex25
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/outputfile.tex57
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/preamble.tex29
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/pstricks1.tex25
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/pstricks2.tex68
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/pstricks3.tex49
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/sampleAFTER.tex5818
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex5819
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/stylefile.tex132
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/table1.tex22
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/table2.tex26
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/table3.tex26
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/testHeadings.tex59
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/testHeadings1.tex25
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/theorem.tex47
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/tikz1.tex88
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/tikz2.tex61
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/tikz3.tex53
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/tikz4.tex27
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/tikz5.tex34
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/torusPGF.tex20
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/torusPSTricks.tex18
-rw-r--r--Master/texmf-dist/doc/support/latexindent/success/trailingComments.tex7
-rwxr-xr-xMaster/texmf-dist/scripts/latexindent/defaultSettings.yaml181
-rwxr-xr-xMaster/texmf-dist/scripts/latexindent/latexindent.pl1784
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
new file mode 100644
index 00000000000..606460409a4
--- /dev/null
+++ b/Master/texmf-dist/doc/support/latexindent/README
@@ -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
Binary files differ
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);
+ }
+}
+