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@@ -1,7 +1,8 @@
-% A sample chapter file- it contains a lot of
+% arara: indent: {overwrite: yes}
+% 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
+%
+% Don't try and compile this file using pdflatex etc, just
% compare the *format* of it to the format of the
% sampleAFTER.tex
%
@@ -9,5810 +10,5810 @@
% 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}
+ \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}
+ \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}
+ \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}
+ \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}