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diff --git a/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex b/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex
index bf62bb9ee1f..ae704bd4bc1 100644
--- a/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex
+++ b/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex
@@ -25,7 +25,7 @@
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}
@@ -94,18 +94,18 @@
\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
@@ -161,7 +161,7 @@
\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?
@@ -171,17 +171,17 @@
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
@@ -199,12 +199,12 @@
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
+ \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}
@@ -254,7 +254,7 @@
\label{poly:fig:evenpow}
\end{minipage}%
\end{figure}
-
+
%===================================
% Author: Hughes
% Date: March 2012
@@ -271,12 +271,12 @@
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
+ \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)
@@ -285,7 +285,7 @@
\begin{shortsolution}
The functions $f$, $g$, and $h$ have domain $(-\infty,\infty)$ and
are graphed below.
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -303,11 +303,11 @@
\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
+ \mathllap{\text{and }} f(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty
\end{align*}
The same is true for $g$ and $h$.
\end{shortsolution}
@@ -317,7 +317,7 @@
\begin{shortsolution}
The functions $F$, $G$, and $H$ have domain $(-\infty,\infty)$ and
are graphed below.
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -335,18 +335,18 @@
\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
+ \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
@@ -357,7 +357,7 @@
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
@@ -376,11 +376,11 @@
\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
@@ -416,7 +416,7 @@
\end{enumerate}
\end{pccsolution}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: March 2012
@@ -440,7 +440,7 @@
shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}.
\end{itemize}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -536,7 +536,7 @@
\caption{Graphs to illustrate typical curves of polynomial functions.}
\label{poly:fig:typical}
\end{figure}
-
+
%===================================
% Author: Hughes
% Date: March 2012
@@ -550,7 +550,7 @@
to guide you.
\begin{shortsolution}
$a_1<0$:
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -563,9 +563,9 @@
\addplot expression[domain=-10:8]{-(x+2)};
\end{axis}
\end{tikzpicture}
-
+
$a_2<0$
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -578,9 +578,9 @@
\addplot expression[domain=-4:4]{-(x^2-6)};
\end{axis}
\end{tikzpicture}
-
+
$a_3<0$
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -593,9 +593,9 @@
\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,
@@ -608,9 +608,9 @@
\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,
@@ -626,11 +626,11 @@
\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
@@ -642,24 +642,24 @@
\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
+ 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}
@@ -712,7 +712,7 @@
\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
@@ -724,7 +724,7 @@
\end{itemize}
If $n=1$, then we say that $p$ has a \emph{simple} zero at $a$.
\end{pccdefinition}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -786,7 +786,7 @@
Let's check if the formula we have written satisfies this requirement
\begin{align*}
p(1) & = (1)(4)(2)(-1) \\
- & = -8
+ & = -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
@@ -807,7 +807,7 @@
evaluate $p(2)$
\begin{align*}
p(2) & =k(4)^2(-1) \\
- & =-16k
+ & =-16k
\end{align*}
We solve the equation $4=-8k$ and obtain $k=-\frac{1}{4}$ and conclude that the
formula for $q(x)$ is
@@ -817,8 +817,8 @@
\end{enumerate}
\end{pccsolution}
\end{pccexample}
-
-
+
+
\fixthis{Chris: need sketching polynomial problems}
\begin{pccspecialcomment}[Steps to follow when sketching polynomial functions]
\begin{steps}
@@ -865,12 +865,12 @@
\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}
@@ -909,7 +909,7 @@
\end{figure}
\end{pccsolution}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: May 2012
@@ -934,7 +934,7 @@
\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}
@@ -973,7 +973,7 @@
\end{figure}
\end{pccsolution}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: May 2012
@@ -1000,7 +1000,7 @@
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}
@@ -1038,7 +1038,7 @@
\end{figure}
\end{pccsolution}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: March 2012
@@ -1058,7 +1058,7 @@
$x$ 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}
@@ -1080,21 +1080,21 @@
\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.
@@ -1113,7 +1113,7 @@
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}
@@ -1143,7 +1143,7 @@
We begin by factoring $p$
\begin{align*}
p(x) & =x^4-2x^3+5x^2 \\
- & =x^2(x^2-2x+5)
+ & =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
@@ -1154,7 +1154,7 @@
\begin{align*}
x & =\frac{2\pm\sqrt{(-2)^2}-20}{2(1)} \\
& =\frac{2\pm\sqrt{-16}}{2} \\
- & =1\pm 2i
+ & =1\pm 2i
\end{align*}
We conclude that $p$ has $4$ zeros: $0$ (multiplicity $2$), and $1\pm 2i$ (simple).
\end{pccsolution}
@@ -1169,13 +1169,13 @@
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
+ & =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
@@ -1235,8 +1235,8 @@
\end{enumerate}
\end{shortsolution}
\end{problem}
-
-
+
+
\begin{figure}[!htb]
\setlength{\figurewidth}{0.3\textwidth}
\begin{subfigure}{\figurewidth}
@@ -1287,10 +1287,10 @@
\caption{}
\label{poly:fig:findformula}
\end{figure}
-
-
-
-
+
+
+
+
\begin{exercises}
%===================================
% Author: Hughes
@@ -1459,14 +1459,14 @@
\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)
+ 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,
@@ -1477,9 +1477,9 @@
\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,
@@ -1490,9 +1490,9 @@
\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,
@@ -1503,7 +1503,7 @@
\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}
@@ -1520,7 +1520,7 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
+
\begin{figure}[!htb]
\begin{widepage}
\setlength{\figurewidth}{0.3\textwidth}
@@ -1773,7 +1773,7 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -1792,7 +1792,7 @@
$\dd\lim_{x\rightarrow\infty}s(x)=\infty$,
\end{shortsolution}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -1917,7 +1917,7 @@
is positive.
\begin{shortsolution}
Assuming that $a_3>0$:
-
+
\begin{tikzpicture}
\begin{axis}[
xmin=-10,xmax=10,
@@ -1937,7 +1937,7 @@
is negative.
\begin{shortsolution}
Assuming that $a_3<0$:
-
+
\begin{tikzpicture}
\begin{axis}[
xmin=-10,xmax=10,
@@ -1961,7 +1961,7 @@
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,
@@ -1982,7 +1982,7 @@
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,
@@ -2194,7 +2194,7 @@
\end{subproblem}
\end{multicols}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: July 2012
@@ -2238,7 +2238,7 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: July 2012
@@ -2290,7 +2290,7 @@
\end{subproblem}
\end{multicols}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -2298,7 +2298,7 @@
\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}
@@ -2382,7 +2382,7 @@
\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}
@@ -2433,7 +2433,7 @@
\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
@@ -2455,21 +2455,21 @@
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
+ \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^+
+ \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}
@@ -2525,8 +2525,8 @@
\label{rat:fig:evenpow}
\end{minipage}%
\end{figure}
-
-
+
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -2543,7 +2543,7 @@
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
+ \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 $y=0$.
@@ -2551,7 +2551,7 @@
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^+
+ \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.
@@ -2569,7 +2569,7 @@
\begin{shortsolution}
The functions $k$, $m$, and $n$ have domain $(-\infty,0)\cup (0,\infty)$, and
are graphed below.
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -2590,14 +2590,14 @@
\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^+
+ \mathllap{\text{and }} k(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+
\end{align*}
The same are true for $m$ and $n$.
\end{shortsolution}
@@ -2607,7 +2607,7 @@
\begin{shortsolution}
The functions $K$, $M$, and $N$ have domain $(-\infty,0)\cup (0,\infty)$, and
are graphed below.
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -2628,21 +2628,21 @@
\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^+
+ \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
@@ -2650,7 +2650,7 @@
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
@@ -2658,13 +2658,13 @@
\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
@@ -2699,7 +2699,7 @@
\end{enumerate}
\end{pccsolution}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -2779,7 +2779,7 @@
\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
@@ -2787,18 +2787,18 @@
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
@@ -2809,7 +2809,7 @@
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
@@ -2827,7 +2827,7 @@
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:
@@ -2910,7 +2910,7 @@
\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.
@@ -2929,9 +2929,9 @@
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
@@ -2943,7 +2943,7 @@
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
@@ -2966,7 +2966,7 @@
\end{pccdefinition}
We will concentrate on functions that have horizontal asymptotes until
we reach \cref{rat:sec:oblique}.
-
+
%===================================
% Author: Hughes
% Date: May 2012
@@ -2979,7 +2979,7 @@
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}
@@ -3050,7 +3050,7 @@
\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.
@@ -3064,14 +3064,14 @@
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
@@ -3080,7 +3080,7 @@
\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
\[
@@ -3119,13 +3119,13 @@
\end{tabular}
\end{minipage}
\end{table}
-
+
Xiao and Dwayne study \cref{rat:tab:plusinfty,rat:tab:minusinfty} and decide that
the functions $R$ and $S$ never actually touch their horizontal asymptotes, but they
do get infinitely close. They also feel as if they have a better understanding of
what it means to study the behavior of a function as $x\rightarrow\pm\infty$.
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -3142,7 +3142,7 @@
\]
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
@@ -3152,7 +3152,7 @@
\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}
@@ -3162,7 +3162,7 @@
study this more in \cref{rat:sec:oblique}).
\end{itemize}
\end{pccexample}
-
+
\begin{figure}[!htb]
\setlength{\figurewidth}{0.3\textwidth}
\begin{subfigure}{\figurewidth}
@@ -3235,7 +3235,7 @@
\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
@@ -3250,7 +3250,7 @@
$(a,r(a))$ on the curve $y=r(x)$ by
using a hollow circle, $\circ$.
\end{pccdefinition}
-
+
%===================================
% Author: Hughes
% Date: March 2012
@@ -3262,12 +3262,12 @@
\]
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
+ & = 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
@@ -3275,7 +3275,7 @@
(-\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]
@@ -3319,7 +3319,7 @@
\end{minipage}%
\end{figure}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -3333,12 +3333,12 @@
make the denominator equal to $0$. Notice that
\begin{align*}
f(x) & = \frac{x(x+3)}{x(x-4)} \\
- & = \frac{x+3}{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
@@ -3362,9 +3362,9 @@
\label{rat:fig:holeex}
\end{figure}
\end{pccexample}
-
-
-
+
+
+
%===================================
% Author: Hughes
% Date: March 2012
@@ -3374,7 +3374,7 @@
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
\[
@@ -3383,7 +3383,7 @@
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);}]
@@ -3427,19 +3427,19 @@
\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
@@ -3448,12 +3448,12 @@
\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}
@@ -3537,7 +3537,7 @@
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,
@@ -3567,9 +3567,9 @@
\end{subproblem}
The Queen looks forward to more food-related investigations from her jester.
\end{problem}
-
-
-
+
+
+
%===================================
% Author: Adams (Hughes)
% Date: March 2012
@@ -3593,17 +3593,17 @@
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
+ & \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
+ & \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.
@@ -3619,20 +3619,20 @@
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
+ & \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
+ & \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}
@@ -3662,7 +3662,7 @@
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
+ & \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.
@@ -3670,7 +3670,7 @@
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
+ & \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
@@ -3681,7 +3681,7 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
+
\begin{exercises}
%===================================
% Author: Hughes
@@ -3783,7 +3783,7 @@
$\begin{aligned}[t]
r(0) & =\frac{(0-2)(0+3)}{(0+5)(0-7)} \\
& =\frac{-6}{-35} \\
- & =\frac{6}{35}
+ & =\frac{6}{35}
\end{aligned}$
\end{shortsolution}
\end{subproblem}
@@ -3793,7 +3793,7 @@
$\begin{aligned}[t]
r(1) & =\frac{(1-2)(1+3)}{(1+5)(1-7)} \\
& =\frac{-4}{-36} \\
- & =\frac{1}{9}
+ & =\frac{1}{9}
\end{aligned}$
\end{shortsolution}
\end{subproblem}
@@ -3803,7 +3803,7 @@
$\begin{aligned}[t]
r(2) & =\frac{(2-2)(2+3)}{(2+5)(2-7)} \\
& = \frac{0}{-50} \\
- & =0
+ & =0
\end{aligned}$
\end{shortsolution}
\end{subproblem}
@@ -3813,7 +3813,7 @@
$\begin{aligned}[t]
r(4) & =\frac{(4-2)(4+3)}{(4+5)(4-7)} \\
& =\frac{14}{-27} \\
- & =-\frac{14}{27}
+ & =-\frac{14}{27}
\end{aligned}$
\end{shortsolution}
\end{subproblem}
@@ -3822,9 +3822,9 @@
\begin{shortsolution}
$\begin{aligned}[t]
r(7) & =\frac{(7-2)(7+3)}{(7+5)(7-7)} \\
- & =\frac{50}{0}
+ & =\frac{50}{0}
\end{aligned}$
-
+
$r(7)$ is undefined.
\end{shortsolution}
\end{subproblem}
@@ -3834,7 +3834,7 @@
$\begin{aligned}[t]
r(-3) & =\frac{(-3-2)(-3+3)}{(-3+5)(-3-7)} \\
& =\frac{0}{-20} \\
- & =0
+ & =0
\end{aligned}$
\end{shortsolution}
\end{subproblem}
@@ -3843,9 +3843,9 @@
\begin{shortsolution}
$\begin{aligned}[t]
r(-5) & =\frac{(-5-2)(-5+3)}{(-5+5)(-5-7)} \\
- & =\frac{14}{0}
+ & =\frac{14}{0}
\end{aligned}$
-
+
$r(-5)$ is undefined.
\end{shortsolution}
\end{subproblem}
@@ -3856,7 +3856,7 @@
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}
+ & =\frac{37}{143}
\end{aligned}$
\end{shortsolution}
\end{subproblem}
@@ -3908,7 +3908,7 @@
\end{subproblem}
\end{multicols}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -3925,7 +3925,7 @@
\end{itemize}
\end{shortsolution}
\end{problem}
-
+
\begin{figure}[!htb]
\begin{widepage}
\setlength{\figurewidth}{0.3\textwidth}
@@ -3997,7 +3997,7 @@
\label{rat:fig:findformula}
\end{widepage}
\end{figure}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -4034,7 +4034,7 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -4099,7 +4099,7 @@
\end{subproblem}
\end{multicols}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: May 2012
@@ -4181,7 +4181,7 @@
\end{subproblem}
\end{multicols}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -4212,7 +4212,7 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: Feb 2011
@@ -4265,8 +4265,8 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
-
+
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -4309,7 +4309,7 @@
Sketch a graph of $r$.
\begin{shortsolution}
A graph of $r$ is shown below.
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -4329,7 +4329,7 @@
\end{shortsolution}
\end{subproblem}
\end{problem}
-
+
%===================================
% Author: Hughes
% Date: July 2012
@@ -4367,8 +4367,8 @@
\end{subproblem}
\end{multicols}
\end{problem}
-
-
+
+
%===================================
% Author: Hughes
% Date: July 2012
@@ -4419,8 +4419,8 @@
\end{subproblem}
\end{multicols}
\end{problem}
-
-
+
+
%===================================
% Author: Hughes
% Date: May 2011
@@ -4428,7 +4428,7 @@
\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
@@ -4525,7 +4525,7 @@
r(-4) & = \frac{-4-3}{-4+2} \\
& = \frac{7}{2} \\
\end{aligned}$
-
+
$r(-3)=\ldots$ etc
\end{shortsolution}
\end{subproblem}
@@ -4541,9 +4541,9 @@
\begin{shortsolution}
$\begin{aligned}[t]
s(-4) & =\frac{-4+2}{(-4-3)(-4+1)} \\
- & =-\frac{2}{21}
+ & =-\frac{2}{21}
\end{aligned}$
-
+
$s(-3)=\ldots$ etc
\end{shortsolution}
\end{subproblem}
@@ -4563,13 +4563,13 @@
\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
\[
@@ -4583,7 +4583,7 @@
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
@@ -4612,10 +4612,10 @@
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
@@ -4641,7 +4641,7 @@
\end{steps}
\end{pccsolution}
\end{pccexample}
-
+
\begin{figure}[!htbp]
\begin{subfigure}{.45\textwidth}
\begin{tikzpicture}
@@ -4678,14 +4678,14 @@
\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
@@ -4711,12 +4711,12 @@
is, because we know what the overall shape will be. Let's compute $v(2)$
\begin{align*}
v(2) & =\dfrac{10}{2} \\
- & = 5
+ & = 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}
@@ -4760,7 +4760,7 @@
\end{figure}
\end{pccsolution}
\end{pccexample}
-
+
%===================================
% Author: Hughes
% Date: May 2012
@@ -4778,7 +4778,7 @@
\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}
+ & =\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.
@@ -4788,7 +4788,7 @@
\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}
@@ -4833,12 +4833,12 @@
\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
@@ -4862,20 +4862,20 @@
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}
@@ -4921,12 +4921,12 @@
\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
@@ -4949,17 +4949,17 @@
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
@@ -4967,7 +4967,7 @@
\end{steps}
\end{pccsolution}
\end{pccexample}
-
+
\begin{figure}[!htbp]
\begin{subfigure}{.45\textwidth}
\begin{tikzpicture}
@@ -5011,7 +5011,7 @@
\end{subfigure}%
\caption{$y=\dfrac{100}{(x+5)(x-4)^2}$}
\end{figure}
-
+
%===================================
% Author: Hughes
% Date: May 2012
@@ -5037,19 +5037,19 @@
\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}
@@ -5093,14 +5093,14 @@
\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
@@ -5127,20 +5127,20 @@
\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}
@@ -5199,7 +5199,7 @@
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
@@ -5207,7 +5207,7 @@
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,
@@ -5222,9 +5222,9 @@
\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,
@@ -5239,9 +5239,9 @@
\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,
@@ -5258,7 +5258,7 @@
\end{tikzpicture}
\end{shortsolution}
\end{problem}
-
+
\begin{figure}[!htb]
\begin{widepage}
\setlength{\figurewidth}{0.3\textwidth}
@@ -5316,7 +5316,7 @@
\label{rat:fig:deducecurve}
\end{widepage}
\end{figure}
-
+
%===================================
% Author: Hughes
% Date: June 2012
@@ -5331,13 +5331,13 @@
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,
@@ -5354,9 +5354,9 @@
\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,
@@ -5373,9 +5373,9 @@
\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,
@@ -5392,10 +5392,10 @@
\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}
@@ -5470,7 +5470,7 @@
$y=\dfrac{4}{x+2}$
\begin{shortsolution}
Vertical intercept: $(0,2)$; vertical asymptote: $x=-2$, horizontal asymptote: $y=0$.
-
+
\begin{tikzpicture}
\begin{axis}[
framed,
@@ -5494,7 +5494,7 @@
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,
@@ -5519,7 +5519,7 @@
\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,
@@ -5544,7 +5544,7 @@
\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,
@@ -5568,7 +5568,7 @@
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,
@@ -5594,7 +5594,7 @@
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,
@@ -5724,7 +5724,7 @@
R(x)=
\begin{dcases}
\frac{2}{x+3}, & x<-5 \\
- \frac{x-4}{x-10}, & x\geq -5
+ \frac{x-4}{x-10}, & x\geq -5
\end{dcases}
\]
Evaluate each of the following.
@@ -5762,7 +5762,7 @@
\end{subproblem}
\end{problem}
\end{exercises}
-
+
\section{Graphing rational functions (oblique asymptotes)}\label{rat:sec:oblique}
\begin{subproblem}
$y=\dfrac{x^2+1}{x-4}$
@@ -5771,7 +5771,7 @@
\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,
@@ -5798,7 +5798,7 @@
\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,