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diff --git a/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex b/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex deleted file mode 100644 index ae704bd4bc1..00000000000 --- a/Master/texmf-dist/doc/support/latexindent/success/sampleBEFORE.tex +++ /dev/null @@ -1,5819 +0,0 @@ -% arara: indent: {overwrite: true, trace: on} -% A sample chapter file- it contains a lot of -% environments, including tabulars, align, etc -% -% Don't try and compile this file using pdflatex etc, just -% compare the *format* of it to the format of the -% sampleAFTER.tex -% -% In particular, compare the tabular and align-type -% environments before and after running the script - -\section{Polynomial functions} - \reformatstepslist{P} % the steps list should be P1, P2, \ldots - In your previous mathematics classes you have studied \emph{linear} and - \emph{quadratic} functions. The most general forms of these types of - functions can be represented (respectively) by the functions $f$ - and $g$ that have formulas - \begin{equation}\label{poly:eq:linquad} - f(x)=mx+b, \qquad g(x)=ax^2+bx+c - \end{equation} - We know that $m$ is the slope of $f$, and that $a$ is the \emph{leading coefficient} - of $g$. We also know that the \emph{signs} of $m$ and $a$ completely - determine the behavior of the functions $f$ and $g$. For example, if $m>0$ - then $f$ is an \emph{increasing} function, and if $m<0$ then $f$ is - a \emph{decreasing} function. Similarly, if $a>0$ then $g$ is - \emph{concave up} and if $a<0$ then $g$ is \emph{concave down}. Graphical - representations of these statements are given in \cref{poly:fig:linquad}. - - \begin{figure}[!htb] - \setlength{\figurewidth}{.2\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-10:8]{(x+2)}; - \end{axis} - \end{tikzpicture} - \caption{$m>0$} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-10:8]{-(x+2)}; - \end{axis} - \end{tikzpicture} - \caption{$m<0$} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-4:4]{(x^2-6)}; - \end{axis} - \end{tikzpicture} - \caption{$a>0$} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-4:4]{-(x^2-6)}; - \end{axis} - \end{tikzpicture} - \caption{$a<0$} - \end{subfigure} - \caption{Typical graphs of linear and quadratic functions.} - \label{poly:fig:linquad} - \end{figure} - - Let's look a little more closely at the formulas for $f$ and $g$ in - \cref{poly:eq:linquad}. Note that the \emph{degree} - of $f$ is $1$ since the highest power of $x$ that is present in the - formula for $f(x)$ is $1$. Similarly, the degree of $g$ is $2$ since - the highest power of $x$ that is present in the formula for $g(x)$ - is $2$. - - In this section we will build upon our knowledge of these elementary - functions. In particular, we will generalize the functions $f$ and $g$ to a function $p$ that has - any degree that we wish. - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{essentialskills} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Quadratic functions] - Every quadratic function has the form $y=ax^2+bx+c$; state the value - of $a$ for each of the following functions, and hence decide if the - parabola that represents the function opens upward or downward. - \begin{multicols}{2} - \begin{subproblem} - $F(x)=x^2+3$ - \begin{shortsolution} - $a=1$; the parabola opens upward. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $G(t)=4-5t^2$ - \begin{shortsolution} - $a=-5$; the parabola opens downward. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $H(y)=4y^2-96y+8$ - \begin{shortsolution} - $a=4$; the parabola opens upward. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $K(z)=-19z^2$ - \begin{shortsolution} - $m=-19$; the parabola opens downward. - \end{shortsolution} - \end{subproblem} - \end{multicols} - Now let's generalize our findings for the most general quadratic function $g$ - that has formula $g(x)=a_2x^2+a_1x+a_0$. Complete the following sentences. - \begin{subproblem} - When $a_2>0$, the parabola that represents $y=g(x)$ opens $\ldots$ - \begin{shortsolution} - When $a_2>0$, the parabola that represents the function opens upward. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - When $a_2<0$, the parabola that represents $y=g(x)$ opens $\ldots$ - \begin{shortsolution} - When $a_2<0$, the parabola that represents the function opens downward. - \end{shortsolution} - \end{subproblem} - \end{problem} - \end{essentialskills} - - \subsection*{Power functions with positive exponents} - The study of polynomials will rely upon a good knowledge - of power functions| you may reasonably ask, what is a power function? - \begin{pccdefinition}[Power functions] - Power functions have the form - \[ - f(x) = a_n x^n - \] - where $n$ can be any real number. - - Note that for this section we will only be concerned with the - case when $n$ is a positive integer. - \end{pccdefinition} - - You may find assurance in the fact that you are already very comfortable - with power functions that have $n=1$ (linear) and $n=2$ (quadratic). Let's - explore some power functions that you might not be so familiar with. - As you read \cref{poly:ex:oddpow,poly:ex:evenpow}, try and spot - as many patterns and similarities as you can. - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{pccexample}[Power functions with odd positive exponents] - \label{poly:ex:oddpow} - Graph each of the following functions, state their domain, and their - long-run behavior as $x\rightarrow\pm\infty$ - \[ - f(x)=x^3, \qquad g(x)=x^5, \qquad h(x)=x^7 - \] - \begin{pccsolution} - The functions $f$, $g$, and $h$ are plotted in \cref{poly:fig:oddpow}. - The domain of each of the functions $f$, $g$, and $h$ is $(-\infty,\infty)$. Note that - the long-run behavior of each of the functions is the same, and in particular - \begin{align*} - f(x)\rightarrow\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty - \end{align*} - The same results hold for $g$ and $h$. - \end{pccsolution} - \end{pccexample} - - \begin{figure}[!htb] - \begin{minipage}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-1.5,xmax=1.5, - ymin=-5,ymax=5, - xtick={-1.0,-0.5,...,1.0}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\textwidth, - legend pos=north west, - ] - \addplot expression[domain=-1.5:1.5]{x^3}; - \addplot expression[domain=-1.379:1.379]{x^5}; - \addplot expression[domain=-1.258:1.258]{x^7}; - \addplot[soldot]coordinates{(-1,-1)} node[axisnode,anchor=north west]{$(-1,-1)$}; - \addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=south east]{$(1,1)$}; - \legend{$f$,$g$,$h$} - \end{axis} - \end{tikzpicture} - \caption{Odd power functions} - \label{poly:fig:oddpow} - \end{minipage}% - \hfill - \begin{minipage}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-2.5,xmax=2.5, - ymin=-5,ymax=5, - xtick={-2.0,-1.5,...,2.0}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\textwidth, - legend pos=south east, - ] - \addplot expression[domain=-2.236:2.236]{x^2}; - \addplot expression[domain=-1.495:1.495]{x^4}; - \addplot expression[domain=-1.307:1.307]{x^6}; - \addplot[soldot]coordinates{(-1,1)} node[axisnode,anchor=east]{$(-1,1)$}; - \addplot[soldot]coordinates{(1,1)} node[axisnode,anchor=west]{$(1,1)$}; - \legend{$F$,$G$,$H$} - \end{axis} - \end{tikzpicture} - \caption{Even power functions} - \label{poly:fig:evenpow} - \end{minipage}% - \end{figure} - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{pccexample}[Power functions with even positive exponents]\label{poly:ex:evenpow}% - Graph each of the following functions, state their domain, and their - long-run behavior as $x\rightarrow\pm\infty$ - \[ - F(x)=x^2, \qquad G(x)=x^4, \qquad H(x)=x^6 - \] - \begin{pccsolution} - The functions $F$, $G$, and $H$ are plotted in \cref{poly:fig:evenpow}. The domain - of each of the functions is $(-\infty,\infty)$. Note that the long-run behavior - of each of the functions is the same, and in particular - \begin{align*} - F(x)\rightarrow\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} F(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty - \end{align*} - The same result holds for $G$ and $H$. - \end{pccsolution} - \end{pccexample} - - \begin{doyouunderstand} - \begin{problem} - Repeat \cref{poly:ex:oddpow,poly:ex:evenpow} using (respectively) - \begin{subproblem} - $f(x)=-x^3, \qquad g(x)=-x^5, \qquad h(x)=-x^7$ - \begin{shortsolution} - The functions $f$, $g$, and $h$ have domain $(-\infty,\infty)$ and - are graphed below. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-1.5,xmax=1.5, - ymin=-5,ymax=5, - xtick={-1.0,-0.5,...,0.5}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\solutionfigurewidth, - legend pos=north east, - ] - \addplot expression[domain=-1.5:1.5]{-x^3}; - \addplot expression[domain=-1.379:1.379]{-x^5}; - \addplot expression[domain=-1.258:1.258]{-x^7}; - \legend{$f$,$g$,$h$} - \end{axis} - \end{tikzpicture} - - Note that - \begin{align*} - f(x)\rightarrow-\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow\infty & \text{ as } x\rightarrow-\infty - \end{align*} - The same is true for $g$ and $h$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $F(x)=-x^2, \qquad G(x)=-x^4, \qquad H(x)=-x^6$ - \begin{shortsolution} - The functions $F$, $G$, and $H$ have domain $(-\infty,\infty)$ and - are graphed below. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-2.5,xmax=2.5, - ymin=-5,ymax=5, - xtick={-1.0,-0.5,...,0.5}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\solutionfigurewidth, - legend pos=north east, - ] - \addplot expression[domain=-2.236:2.236]{-x^2}; - \addplot expression[domain=-1.495:1.495]{-x^4}; - \addplot expression[domain=-1.307:1.307]{-x^6}; - \legend{$F$,$G$,$H$} - \end{axis} - \end{tikzpicture} - - Note that - \begin{align*} - F(x)\rightarrow-\infty & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} F(x)\rightarrow-\infty & \text{ as } x\rightarrow-\infty - \end{align*} - The same is true for $G$ and $H$. - \end{shortsolution} - \end{subproblem} - \end{problem} - \end{doyouunderstand} - - \subsection*{Polynomial functions} - Now that we have a little more familiarity with power functions, - we can define polynomial functions. Provided that you were comfortable - with our opening discussion about linear and quadratic functions (see - $f$ and $g$ in \cref{poly:eq:linquad}) then there is every chance - that you'll be able to master polynomial functions as well; just remember - that polynomial functions are a natural generalization of linear - and quadratic functions. Once you've studied the examples and problems - in this section, you'll hopefully agree that polynomial functions - are remarkably predictable. - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccdefinition}[Polynomial functions] - Polynomial functions have the form - \[ - p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0 - \] - where $a_n$, $a_{n-1}$, $a_{n-2}$, \ldots, $a_0$ are real numbers. - \begin{itemize} - \item We call $n$ the degree of the polynomial, and require that $n$ - is a non-negative integer; - \item $a_n$, $a_{n-1}$, $a_{n-2}$, \ldots, $a_0$ are called the coefficients; - \item We typically write polynomial functions in descending powers of $x$. - \end{itemize} - In particular, we call $a_n$ the \emph{leading} coefficient, and $a_nx^n$ the - \emph{leading term}. - - Note that if a polynomial is given in factored form, then the degree can be found - by counting the number of linear factors. - \end{pccdefinition} - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{pccexample}[Polynomial or not] - Identify the following functions as polynomial or not; if the function - is a polynomial, state its degree. - \begin{multicols}{3} - \begin{enumerate} - \item $p(x)=x^2-3$ - \item $q(x)=-4x^{\nicefrac{1}{2}}+10$ - \item $r(x)=10x^5$ - \item $s(x)=x^{-2}+x^{23}$ - \item $f(x)=-8$ - \item $g(x)=3^x$ - \item $h(x)=\sqrt[3]{x^7}-x^2+x$ - \item $k(x)=4x(x+2)(x-3)$ - \item $j(x)=x^2(x-4)(5-x)$ - \end{enumerate} - \end{multicols} - \begin{pccsolution} - \begin{enumerate} - \item $p$ is a polynomial, and its degree is $2$. - \item $q$ is \emph{not} a polynomial, because $\frac{1}{2}$ is not an integer. - \item $r$ is a polynomial, and its degree is $5$. - \item $s$ is \emph{not} a polynomial, because $-2$ is not a positive integer. - \item $f$ is a polynomial, and its degree is $0$. - \item $g$ is \emph{not} a polynomial, because the independent - variable, $x$, is in the exponent. - \item $h$ is \emph{not} a polynomial, because $\frac{7}{3}$ is not an integer. - \item $k$ is a polynomial, and its degree is $3$. - \item $j$ is a polynomial, and its degree is $4$. - \end{enumerate} - \end{pccsolution} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{pccexample}[Typical graphs]\label{poly:ex:typical} - \Cref{poly:fig:typical} shows graphs of some polynomial functions; - the ticks have deliberately been left off the axis to allow us to concentrate - on the features of each graph. Note in particular that: - \begin{itemize} - \item \cref{poly:fig:typical1} shows a degree-$1$ polynomial (you might also - classify the function as linear) whose leading coefficient, $a_1$, is positive. - \item \cref{poly:fig:typical2} shows a degree-$2$ polynomial (you might also - classify the function as quadratic) whose leading coefficient, $a_2$, is positive. - \item \cref{poly:fig:typical3} shows a degree-$3$ polynomial whose leading coefficient, $a_3$, - is positive| compare its overall - shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}. - \item \cref{poly:fig:typical4} shows a degree-$4$ polynomial whose leading coefficient, $a_4$, - is positive|compare its overall shape and long-run behavior to the functions described in \cref{poly:ex:evenpow}. - \item \cref{poly:fig:typical5} shows a degree-$5$ polynomial whose leading coefficient, $a_5$, - is positive| compare its overall - shape and long-run behavior to the functions described in \cref{poly:ex:oddpow}. - \end{itemize} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{figure}[!htb] - \begin{widepage} - \setlength{\figurewidth}{\textwidth/6} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-10:8]{(x+2)}; - \end{axis} - \end{tikzpicture} - \caption{$a_1>0$} - \label{poly:fig:typical1} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-4:4]{(x^2-6)}; - \end{axis} - \end{tikzpicture} - \caption{$a_2>0$} - \label{poly:fig:typical2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-7.5:7.5]{0.05*(x+6)*x*(x-6)}; - \end{axis} - \end{tikzpicture} - \caption{$a_3>0$} - \label{poly:fig:typical3} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-2.35:5.35,samples=100]{0.2*(x-5)*x*(x-3)*(x+2)}; - \end{axis} - \end{tikzpicture} - \caption{$a_4>0$} - \label{poly:fig:typical4} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\textwidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-5.5:6.3,samples=100]{0.01*(x+2)*x*(x-3)*(x+5)*(x-6)}; - \end{axis} - \end{tikzpicture} - \caption{$a_5>0$} - \label{poly:fig:typical5} - \end{subfigure} - \end{widepage} - \caption{Graphs to illustrate typical curves of polynomial functions.} - \label{poly:fig:typical} - \end{figure} - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{doyouunderstand} - \begin{problem} - Use \cref{poly:ex:typical} and \cref{poly:fig:typical} to help you sketch - the graphs of polynomial functions that have negative leading coefficients| note - that there are many ways to do this! The intention with this problem - is to use your knowledge of transformations- in particular, \emph{reflections}- - to guide you. - \begin{shortsolution} - $a_1<0$: - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\solutionfigurewidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-10:8]{-(x+2)}; - \end{axis} - \end{tikzpicture} - - $a_2<0$ - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\solutionfigurewidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-4:4]{-(x^2-6)}; - \end{axis} - \end{tikzpicture} - - $a_3<0$ - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\solutionfigurewidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-7.5:7.5]{-0.05*(x+6)*x*(x-6)}; - \end{axis} - \end{tikzpicture} - - $a_4<0$ - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\solutionfigurewidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-2.35:5.35,samples=100]{-0.2*(x-5)*x*(x-3)*(x+2)}; - \end{axis} - \end{tikzpicture} - - $a_5<0$ - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - width=\solutionfigurewidth, - xtick={-11}, - ytick={-11}, - ] - \addplot expression[domain=-5.5:6.3,samples=100]{-0.01*(x+2)*x*(x-3)*(x+5)*(x-6)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{problem} - \end{doyouunderstand} - - \fixthis{poly: Need a more basic example here- it can have a similar - format to the multiple zeros example, but just keep it simple; it should - be halfway between the 2 examples surrounding it} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Multiple zeros] - Consider the polynomial functions $p$, $q$, and $r$ which are - graphed in \cref{poly:fig:moremultiple}. - The formulas for $p$, $q$, and $r$ are as follows - \begin{align*} - p(x) & =(x-3)^2(x+4)^2 \\ - q(x) & =x(x+2)^2(x-1)^2(x-3) \\ - r(x) & =x(x-3)^3(x+1)^2 - \end{align*} - Find the degree of $p$, $q$, and $r$, and decide if the functions bounce off or cut - through the horizontal axis at each of their zeros. - \begin{pccsolution} - The degree of $p$ is 4. Referring to \cref{poly:fig:bouncep}, - the curve bounces off the horizontal axis at both zeros, $3$ and $4$. - - The degree of $q$ is 6. Referring to \cref{poly:fig:bounceq}, - the curve bounces off the horizontal axis at $-2$ and $1$, and cuts - through the horizontal axis at $0$ and $3$. - - The degree of $r$ is 6. Referring to \cref{poly:fig:bouncer}, - the curve bounces off the horizontal axis at $-1$, and cuts through - the horizontal axis at $0$ and at $3$, although is flattened immediately to the left and right of $3$. - \end{pccsolution} - \end{pccexample} - - \setlength{\figurewidth}{0.25\textwidth} - \begin{figure}[!htb] - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-6,xmax=5, - ymin=-30,ymax=200, - xtick={-4,-2,...,4}, - width=\textwidth, - ] - \addplot expression[domain=-5.63733:4.63733,samples=50]{(x-3)^2*(x+4)^2}; - \addplot[soldot]coordinates{(3,0)(-4,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=p(x)$} - \label{poly:fig:bouncep} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-3,xmax=4, - xtick={-2,...,3}, - ymin=-60,ymax=40, - width=\textwidth, - ] - \addplot+[samples=50] expression[domain=-2.49011:3.11054]{x*(x+2)^2*(x-1)^2*(x-3)}; - \addplot[soldot]coordinates{(-2,0)(0,0)(1,0)(3,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=q(x)$} - \label{poly:fig:bounceq} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-2,xmax=4, - xtick={-1,...,3}, - ymin=-40,ymax=40, - width=\textwidth, - ] - \addplot expression[domain=-1.53024:3.77464,samples=50]{x*(x-3)^3*(x+1)^2}; - \addplot[soldot]coordinates{(-1,0)(0,0)(3,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=r(x)$} - \label{poly:fig:bouncer} - \end{subfigure} - \caption{} - \label{poly:fig:moremultiple} - \end{figure} - - \begin{pccdefinition}[Multiple zeros]\label{poly:def:multzero} - Let $p$ be a polynomial that has a repeated linear factor $(x-a)^n$. Then we say - that $p$ has a multiple zero at $a$ of multiplicity $n$ and - \begin{itemize} - \item if the factor $(x-a)$ is repeated an even number of times, the graph of $y=p(x)$ does not - cross the $x$ axis at $a$, but `bounces' off the horizontal axis at $a$. - \item if the factor $(x-a)$ is repeated an odd number of times, the graph of $y=p(x)$ crosses the - horizontal axis at $a$, but it looks `flattened' there - \end{itemize} - If $n=1$, then we say that $p$ has a \emph{simple} zero at $a$. - \end{pccdefinition} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Find a formula] - Find formulas for the polynomial functions, $p$ and $q$, graphed in \cref{poly:fig:findformulademoboth}. - \begin{figure}[!htb] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[framed, - xmin=-5,xmax=5, - ymin=-10,ymax=10, - xtick={-4,-2,...,4}, - minor xtick={-3,-1,...,3}, - ytick={-8,-6,...,8}, - width=\textwidth, - grid=both] - \addplot expression[domain=-3.25842:2.25842,samples=50]{-x*(x-2)*(x+3)*(x+1)}; - \addplot[soldot]coordinates{(1,8)}node[axisnode,inner sep=.35cm,anchor=west]{$(1,8)$}; - \addplot[soldot]coordinates{(-3,0)(-1,0)(0,0)(2,0)}; - \end{axis} - \end{tikzpicture} - \caption{$p$} - \label{poly:fig:findformulademo} - \end{subfigure} - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[framed, - xmin=-5,xmax=5, - ymin=-10,ymax=10, - xtick={-4,-2,...,4}, - minor xtick={-3,-1,...,3}, - ytick={-8,-6,...,8}, - width=\textwidth, - grid=both] - \addplot expression[domain=-4.33:4.08152]{-.25*(x+2)^2*(x-3)}; - \addplot[soldot]coordinates{(2,4)}node[axisnode,anchor=south west]{$(2,4)$}; - \addplot[soldot]coordinates{(-2,0)(3,0)}; - \end{axis} - \end{tikzpicture} - \caption{$q$} - \label{poly:fig:findformulademo1} - \end{subfigure} - \caption{} - \label{poly:fig:findformulademoboth} - \end{figure} - \begin{pccsolution} - \begin{enumerate} - \item We begin by noting that the horizontal intercepts of $p$ are $(-3,0)$, $(-1,0)$, $(0,0)$ and $(2,0)$. - We also note that each zero is simple (multiplicity $1$). - If we assume that $p$ has no other zeros, then we can start by writing - \begin{align*} - p(x) & =(x+3)(x+1)(x-0)(x-2) \\ - & =x(x+3)(x+1)(x-2) \\ - \end{align*} - According to \cref{poly:fig:findformulademo}, the point $(1,8)$ lies - on the curve $y=p(x)$. - Let's check if the formula we have written satisfies this requirement - \begin{align*} - p(1) & = (1)(4)(2)(-1) \\ - & = -8 - \end{align*} - which is clearly not correct| it is close though. We can correct this by - multiplying $p$ by a constant $k$; so let's assume that - \[ - p(x)=kx(x+3)(x+1)(x-2) - \] - Then $p(1)=-8k$, and if this is to equal $8$, then $k=-1$. Therefore - the formula for $p(x)$ is - \[ - p(x)=-x(x+3)(x+1)(x-2) - \] - \item The function $q$ has a zero at $-2$ of multiplicity $2$, and zero of - multiplicity $1$ at $3$ (so $3$ is a simple zero of $q$); we can therefore assume that $q$ has the form - \[ - q(x)=k(x+2)^2(x-3) - \] - where $k$ is some real number. In order to find $k$, we use the given ordered pair, $(2,4)$, and - evaluate $p(2)$ - \begin{align*} - p(2) & =k(4)^2(-1) \\ - & =-16k - \end{align*} - We solve the equation $4=-8k$ and obtain $k=-\frac{1}{4}$ and conclude that the - formula for $q(x)$ is - \[ - q(x)=-\frac{1}{4}(x+2)^2(x-3) - \] - \end{enumerate} - \end{pccsolution} - \end{pccexample} - - - \fixthis{Chris: need sketching polynomial problems} - \begin{pccspecialcomment}[Steps to follow when sketching polynomial functions] - \begin{steps} - \item \label{poly:step:first} Determine the degree of the polynomial, - its leading term and leading coefficient, and hence determine - the long-run behavior of the polynomial| does it behave like $\pm x^2$ or $\pm x^3$ - as $x\rightarrow\pm\infty$? - \item Determine the zeros and their multiplicity. Mark all zeros - and the vertical intercept on the graph using solid circles $\bullet$. - \item \label{poly:step:last} Deduce the overall shape of the curve, and sketch it. If there isn't - enough information from the previous steps, then construct a table of values. - \end{steps} - Remember that until we have the tools of calculus, we won't be able to - find the exact coordinates of local minimums, local maximums, and points - of inflection. - \end{pccspecialcomment} - Before we demonstrate some examples, it is important to remember the following: - \begin{itemize} - \item our sketches will give a good representation of the overall - shape of the graph, but until we have the tools of calculus (from MTH 251) - we can not find local minimums, local maximums, and inflection points algebraically. This - means that we will make our best guess as to where these points are. - \item we will not concern ourselves too much with the vertical scale (because of - our previous point)| we will, however, mark the vertical intercept (assuming there is one), - and any horizontal asymptotes. - \end{itemize} - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{poly:ex:simplecubic} - Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $p$ - that has formula - \[ - p(x)=\frac{1}{2}(x-4)(x-1)(x+3) - \] - \begin{pccsolution} - \begin{steps} - \item $p$ has degree $3$. The leading term of $p$ is $\frac{1}{2}x^3$, so the leading coefficient of $p$ - is $\frac{1}{2}$. The long-run behavior of $p$ is therefore similar to that of $x^3$. - \item The zeros of $p$ are $-3$, $1$, and $4$; each zero is simple (i.e, it has multiplicity $1$). - This means that the curve of $p$ cuts the horizontal axis at each zero. The vertical - intercept of $p$ is $(0,6)$. - \item We draw the details we have obtained so far on \cref{poly:fig:simplecubicp1}. Given - that the curve of $p$ looks like the curve of $x^3$ in the long-run, we are able to complete a sketch of the - graph of $p$ in \cref{poly:fig:simplecubicp2}. - - Note that we can not find the coordinates of the local minimums, local maximums, and inflection - points| for the moment we make reasonable guesses as to where these points are (you'll find how - to do this in calculus). - \end{steps} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=15, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:simplecubicp1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=15, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-3,0)(1,0)(4,0)(0,6)}node[axisnode,anchor=south west]{$(0,6)$}; - \addplot[pccplot] expression[domain=-3.57675:4.95392,samples=100]{.5*(x-4)*(x-1)*(x+3)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:simplecubicp2} - \end{subfigure}% - \caption{$y=\dfrac{1}{2}(x-4)(x-1)(x+3)$} - \label{poly:fig:simplecubic} - \end{figure} - \end{pccsolution} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{poly:ex:degree5} - Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $q$ - that has formula - \[ - q(x)=\frac{1}{200}(x+7)^2(2-x)(x-6)^2 - \] - \begin{pccsolution} - \begin{steps} - \item $q$ has degree $4$. The leading term of $q$ is - \[ - -\frac{1}{200}x^5 - \] - so the leading coefficient of $q$ is $-\frac{1}{200}$. The long-run behavior of $q$ - is therefore similar to that of $-x^5$. - \item The zeros of $q$ are $-7$ (multiplicity 2), $2$ (simple), and $6$ (multiplicity $2$). - The curve of $q$ bounces off the horizontal axis at the zeros with multiplicity $2$ and - cuts the horizontal axis at the simple zeros. The vertical intercept of $q$ is $\left( 0,\frac{441}{25} \right)$. - \item We mark the details we have found so far on \cref{poly:fig:degree5p1}. Given that - the curve of $q$ looks like the curve of $-x^5$ in the long-run, we can complete \cref{poly:fig:degree5p2}. - \end{steps} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=40, - xtick={-8,-6,...,8}, - ytick={-5,0,...,35}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\left( 0, \frac{441}{25} \right)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:degree5p1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=40, - xtick={-8,-6,...,8}, - ytick={-5,0,...,35}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-7,0)(2,0)(6,0)(0,441/25)}node[axisnode,anchor=south west]{$\left( 0, \frac{441}{25} \right)$}; - \addplot[pccplot] expression[domain=-8.83223:7.34784,samples=50]{1/200*(x+7)^2*(2-x)*(x-6)^2}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:degree5p2} - \end{subfigure}% - \caption{$y=\dfrac{1}{200}(x+7)^2(2-x)(x-6)^2$} - \label{poly:fig:degree5} - \end{figure} - \end{pccsolution} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample} - Use \crefrange{poly:step:first}{poly:step:last} to sketch a graph of the function $r$ - that has formula - \[ - r(x)=\frac{1}{100}x^3(x+4)(x-4)(x-6) - \] - \begin{pccsolution} - \begin{steps} - \item $r$ has degree $6$. The leading term of $r$ is - \[ - \frac{1}{100}x^6 - \] - so the leading coefficient of $r$ is $\frac{1}{100}$. The long-run behavior of $r$ - is therefore similar to that of $x^6$. - \item The zeros of $r$ are $-4$ (simple), $0$ (multiplicity $3$), $4$ (simple), - and $6$ (simple). The vertical intercept of $r$ is $(0,0)$. The curve of $r$ - cuts the horizontal axis at the simple zeros, and goes through the axis - at $(0,0)$, but does so in a flattened way. - \item We mark the zeros and vertical intercept on \cref{poly:fig:degree6p1}. Given that - the curve of $r$ looks like the curve of $x^6$ in the long-run, we complete the graph - of $r$ in \cref{poly:fig:degree6p2}. - \end{steps} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=10, - ymin=-20,ymax=10, - xtick={-4,-2,...,8}, - ytick={-15,-10,...,5}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:degree6p1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=10, - ymin=-20,ymax=10, - xtick={-4,-2,...,8}, - ytick={-15,-10,...,5}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-4,0)(0,0)(4,0)(6,0)}; - \addplot[pccplot] expression[domain=-4.16652:6.18911,samples=100]{1/100*(x+4)*x^3*(x-4)*(x-6)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:degree6p2} - \end{subfigure}% - \caption{$y=\dfrac{1}{100}(x+4)x^3(x-4)(x-6)$} - \end{figure} - \end{pccsolution} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{pccexample}[An open-topped box] - A cardboard company makes open-topped boxes for their clients. The specifications - dictate that the box must have a square base, and that it must be open-topped. - The company uses sheets of cardboard that are $\unit[1200]{cm^2}$. Assuming that - the base of each box has side $x$ (measured in cm), it can be shown that the volume of each box, $V(x)$, - has formula - \[ - V(x)=\frac{x}{4}(1200-x^2) - \] - Find the dimensions of the box that maximize the volume. - \begin{pccsolution} - We graph $y=V(x)$ in \cref{poly:fig:opentoppedbox}. Note that because - $x$ represents the length of a side, and $V(x)$ represents the volume - of the box, we necessarily require both values to be positive; we illustrate - the part of the curve that applies to this problem using a solid line. - - \begin{figure}[!htb] - \centering - \begin{tikzpicture} - \begin{axis}[framed, - xmin=-50,xmax=50, - ymin=-5000,ymax=5000, - xtick={-40,-30,...,40}, - minor xtick={-45,-35,...,45}, - minor ytick={-3000,-1000,1000,3000}, - width=.75\textwidth, - height=.5\textwidth, - grid=both] - \addplot[pccplot,dashed,<-] expression[domain=-40:0,samples=50]{x/4*(1200-x^2)}; - \addplot[pccplot,-] expression[domain=0:34.64,samples=50]{x/4*(1200-x^2)}; - \addplot[pccplot,dashed,->] expression[domain=34.64:40,samples=50]{x/4*(1200-x^2)}; - \addplot[soldot] coordinates{(20,4000)}; - \end{axis} - \end{tikzpicture} - \caption{$y=V(x)$} - \label{poly:fig:opentoppedbox} - \end{figure} - - According to \cref{poly:fig:opentoppedbox}, the maximum volume of such a box is - approximately $\unit[4000]{cm^2}$, and we achieve it using a base of length - approximately $\unit[20]{cm}$. Since the base is square and each sheet of cardboard - is $\unit[1200]{cm^2}$, we conclude that the dimensions of each box are $\unit[20]{cm}\times\unit[20]{cm}\times\unit[30]{cm}$. - \end{pccsolution} - \end{pccexample} - - \subsection*{Complex zeros} - There has been a pattern to all of the examples that we have seen so far| - the degree of the polynomial has dictated the number of \emph{real} zeros that the - polynomial has. For example, the function $p$ in \cref{poly:ex:simplecubic} - has degree $3$, and $p$ has $3$ real zeros; the function $q$ in \cref{poly:ex:degree5} - has degree $5$ and $q$ has $5$ real zeros. - - You may wonder if this result can be generalized| does every polynomial that - has degree $n$ have $n$ real zeros? Before we tackle the general result, - let's consider an example that may help motivate it. - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{pccexample}\label{poly:ex:complx} - Consider the polynomial function $c$ that has formula - \[ - c(x)=x(x^2+1) - \] - It is clear that $c$ has degree $3$, and that $c$ has a (simple) zero at $0$. Does - $c$ have any other zeros, i.e, can we find any values of $x$ that satisfy the equation - \begin{equation}\label{poly:eq:complx} - x^2+1=0 - \end{equation} - The solutions to \cref{poly:eq:complx} are $\pm i$. - - We conclude that $c$ has $3$ zeros: $0$ and $\pm i$; we note that \emph{not - all of them are real}. - \end{pccexample} - \Cref{poly:ex:complx} shows that not every degree-$3$ polynomial has $3$ - \emph{real} zeros; however, if we are prepared to venture into the complex numbers, - then we can state the following theorem. - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{pccspecialcomment}[The fundamental theorem of algebra] - Every polynomial function of degree $n$ has $n$ roots, some of which may - be complex, and some may be repeated. - \end{pccspecialcomment} - \fixthis{Fundamental theorem of algebra: is this wording ok? do we want - it as a theorem?} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{pccexample} - Find all the zeros of the polynomial function $p$ that has formula - \[ - p(x)=x^4-2x^3+5x^2 - \] - \begin{pccsolution} - We begin by factoring $p$ - \begin{align*} - p(x) & =x^4-2x^3+5x^2 \\ - & =x^2(x^2-2x+5) - \end{align*} - We note that $0$ is a zero of $p$ with multiplicity $2$. The other zeros of $p$ - can be found by solving the equation - \[ - x^2-2x+5=0 - \] - This equation can not be factored, so we use the quadratic formula - \begin{align*} - x & =\frac{2\pm\sqrt{(-2)^2}-20}{2(1)} \\ - & =\frac{2\pm\sqrt{-16}}{2} \\ - & =1\pm 2i - \end{align*} - We conclude that $p$ has $4$ zeros: $0$ (multiplicity $2$), and $1\pm 2i$ (simple). - \end{pccsolution} - \end{pccexample} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{pccexample} - Find a polynomial that has zeros at $2\pm i\sqrt{2}$. - \begin{pccsolution} - We know that the zeros of a polynomial can be found by analyzing the linear - factors. We are given the zeros, and have to work backwards to find the - linear factors. - - We begin by assuming that $p$ has the form - \begin{align*} - p(x) & =(x-(2-i\sqrt{2}))(x-(2+i\sqrt{2})) \\ - & =x^2-x(2+i\sqrt{2})-x(2-i\sqrt{2})+(2-i\sqrt{2})(2+i\sqrt{2}) \\ - & =x^2-4x+(4-2i^2) \\ - & =x^2-4x+6 - \end{align*} - We conclude that a possible formula for a polynomial function, $p$, - that has zeros at $2\pm i\sqrt{2}$ is - \[ - p(x)=x^2-4x+6 - \] - Note that we could multiply $p$ by any real number and still ensure - that $p$ has the same zeros. - \end{pccsolution} - \end{pccexample} - \investigation*{} - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Find a formula from a graph] - For each of the polynomials in \cref{poly:fig:findformula} - \begin{enumerate} - \item count the number of times the curve turns round, and cuts/bounces off the $x$ axis; - \item approximate the degree of the polynomial; - \item use your information to find the linear factors of each polynomial, and therefore write a possible formula for each; - \item make sure your polynomial goes through the given ordered pair. - \end{enumerate} - \begin{shortsolution} - \Vref{poly:fig:findformdeg2}: - \begin{enumerate} - \item the curve turns round once; - \item the degree could be 2; - \item based on the zeros, the linear factors are $(x+5)$ and $(x-3)$; since the - graph opens downwards, we will assume the leading coefficient is negative: $p(x)=-k(x+5)(x-3)$; - \item $p$ goes through $(2,2)$, so we need to solve $2=-k(7)(-1)$ and therefore $k=\nicefrac{2}{7}$, so - \[ - p(x)=-\frac{2}{7}(x+5)(x-3) - \] - \end{enumerate} - \Vref{poly:fig:findformdeg3}: - \begin{enumerate} - \item the curve turns around twice; - \item the degree could be 3; - \item based on the zeros, the linear factors are $(x+2)^2$, and $(x-1)$; - based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+2)^2(x-1)$; - \item $p$ goes through $(0,-2)$, so we need to solve $-2=k(4)(-1)$ and therefore $k=\nicefrac{1}{2}$, so - \[ - p(x)=\frac{1}{2}(x+2)^2(x-1) - \] - \end{enumerate} - \Vref{poly:fig:findformdeg5}: - \begin{enumerate} - \item the curve turns around 4 times; - \item the degree could be 5; - \item based on the zeros, the linear factors are $(x+5)^2$, $(x+1)$, $(x-2)$, $(x-3)$; - based on the behavior of $p$, we assume that the leading coefficient is positive, and try $p(x)=k(x+5)^2(x+1)(x-2)(x-3)$; - \item $p$ goes through $(-3,-50)$, so we need to solve $-50=k(64)(-2)(-5)(-6)$ and therefore $k=\nicefrac{5}{384}$, so - \[ - p(x)=\frac{5}{384}(x+5)^2(x+1)(x-2)(x-3) - \] - \end{enumerate} - \end{shortsolution} - \end{problem} - - - \begin{figure}[!htb] - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-2,ymax=5, - width=\textwidth, - ] - \addplot expression[domain=-4.5:3.75]{-1/3*(x+4)*(x-3)}; - \addplot[soldot] coordinates{(-4,0)(3,0)(2,2)} node[axisnode,above right]{$(2,2)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:findformdeg2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-3,xmax=2, - ymin=-2,ymax=4, - xtick={-2,...,1}, - width=\textwidth, - ] - \addplot expression[domain=-2.95:1.75]{1/3*(x+2)^2*(x-1)}; - \addplot[soldot]coordinates{(-2,0)(1,0)(0,-1.33)}node[axisnode,anchor=north west]{$(0,-2)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:findformdeg3} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-100,ymax=150, - width=\textwidth, - ] - \addplot expression[domain=-4.5:3.4,samples=50]{(x+4)^2*(x+1)*(x-2)*(x-3)}; - \addplot[soldot]coordinates{(-4,0)(-1,0)(2,0)(3,0)(-3,-60)}node[axisnode,anchor=north]{$(-3,-50)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:findformdeg5} - \end{subfigure} - \caption{} - \label{poly:fig:findformula} - \end{figure} - - - - - \begin{exercises} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Prerequisite classifacation skills] - Decide if each of the following functions are linear or quadratic. - \begin{multicols}{3} - \begin{subproblem} - $f(x)=2x+3$ - \begin{shortsolution} - $f$ is linear. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $g(x)=10-7x$ - \begin{shortsolution} - $g$ is linear - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $h(x)=-x^2+3x-9$ - \begin{shortsolution} - $h$ is quadratic. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $k(x)=-17$ - \begin{shortsolution} - $k$ is linear. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $l(x)=-82x^2-4$ - \begin{shortsolution} - $l$ is quadratic - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $m(x)=6^2x-8$ - \begin{shortsolution} - $m$ is linear. - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Prerequisite slope identification] - State the slope of each of the following linear functions, and - hence decide if each function is increasing or decreasing. - \begin{multicols}{4} - \begin{subproblem} - $\alpha(x)=4x+1$ - \begin{shortsolution} - $m=4$; $\alpha$ is increasing. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $\beta(x)=-9x$ - \begin{shortsolution} - $m=-9$; $\beta$ is decreasing. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $\gamma(t)=18t+100$ - \begin{shortsolution} - $m=18$; $\gamma$ is increasing. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $\delta(y)=23-y$ - \begin{shortsolution} - $m=-1$; $\delta$ is decreasing. - \end{shortsolution} - \end{subproblem} - \end{multicols} - Now let's generalize our findings for the most general linear function $f$ - that has formula $f(x)=mx+b$. Complete the following sentences. - \begin{subproblem} - When $m>0$, the function $f$ is $\ldots$ - \begin{shortsolution} - When $m>0$, the function $f$ is $\ldots$ \emph{increasing}. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - When $m<0$, the function $f$ is $\ldots$ - \begin{shortsolution} - When $m<0$, the function $f$ is $\ldots$ \emph{decreasing}. - \end{shortsolution} - \end{subproblem} - \end{problem} - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Polynomial or not?] - Identify whether each of the following functions is a polynomial or not. - If the function is a polynomial, state its degree. - \begin{multicols}{3} - \begin{subproblem} - $p(x)=2x+1$ - \begin{shortsolution} - $p$ is a polynomial (you might also describe $p$ as linear). The degree of $p$ is 1. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=7x^2+4x$ - \begin{shortsolution} - $p$ is a polynomial (you might also describe $p$ as quadratic). The degree of $p$ is 2. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=\sqrt{x}+2x+1$ - \begin{shortsolution} - $p$ is not a polynomial; we require the powers of $x$ to be integer values. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=2^x-45$ - \begin{shortsolution} - $p$ is not a polynomial; the $2^x$ term is exponential. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=6x^4-5x^3+9$ - \begin{shortsolution} - $p$ is a polynomial, and the degree of $p$ is $6$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=-5x^{17}+9x+2$ - \begin{shortsolution} - $p$ is a polynomial, and the degree of $p$ is 17. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=4x(x+7)^2(x-3)^3$ - \begin{shortsolution} - $p$ is a polynomial, and the degree of $p$ is $6$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=4x^{-5}-x^2+x$ - \begin{shortsolution} - $p$ is not a polynomial because $-5$ is not a positive integer. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=-x^6(x^2+1)(x^3-2)$ - \begin{shortsolution} - $p$ is a polynomial, and the degree of $p$ is $11$. - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Polynomial graphs] - Three polynomial functions $p$, $m$, and $n$ are shown in \crefrange{poly:fig:functionp}{poly:fig:functionn}. - The functions have the following formulas - \begin{align*} - p(x) & = (x-1)(x+2)(x-3) \\ - m(x) & = -(x-1)(x+2)(x-3) \\ - n(x) & = (x-1)(x+2)(x-3)(x+1)(x+4) - \end{align*} - Note that for our present purposes we are not concerned with the vertical scale of the graphs. - \begin{subproblem} - Identify both on the graph {\em and} algebraically, the zeros of each polynomial. - \begin{shortsolution} - $y=p(x)$ is shown below. - - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-10,ymax=10, - width=\solutionfigurewidth, - ] - \addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)}; - \addplot[soldot] coordinates{(-2,0)(1,0)(3,0)}; - \end{axis} - \end{tikzpicture} - - $y=m(x)$ is shown below. - - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-10,ymax=10, - width=\solutionfigurewidth, - ] - \addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)}; - \addplot[soldot] coordinates{(-2,0)(1,0)(3,0)}; - \end{axis} - \end{tikzpicture} - - $y=n(x)$ is shown below. - - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-90,ymax=70, - width=\solutionfigurewidth, - ] - \addplot expression[domain=-4.15:3.15,samples=50]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)}; - \addplot[soldot] coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)}; - \end{axis} - \end{tikzpicture} - - The zeros of $p$ are $-2$, $1$, and $3$; the zeros of $m$ are $-2$, $1$, and $3$; the zeros of $n$ are - $-4$, $-2$, $-1$, and $3$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Write down the degree, how many times the curve of each function `turns around', - and how many zeros it has - \begin{shortsolution} - \begin{itemize} - \item The degree of $p$ is 3, and the curve $y=p(x)$ turns around twice. - \item The degree of $q$ is also 3, and the curve $y=q(x)$ turns around twice. - \item The degree of $n$ is $5$, and the curve $y=n(x)$ turns around 4 times. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \end{problem} - - \begin{figure}[!htb] - \begin{widepage} - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-10,ymax=10, - ytick={-5,5}, - width=\textwidth, - ] - \addplot expression[domain=-2.5:3.5,samples=50]{(x-1)*(x+2)*(x-3)}; - \addplot[soldot]coordinates{(-2,0)(1,0)(3,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=p(x)$} - \label{poly:fig:functionp} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-10,ymax=10, - ytick={-5,5}, - width=\textwidth, - ] - \addplot expression[domain=-2.5:3.5,samples=50]{-1*(x-1)*(x+2)*(x-3)}; - \addplot[soldot]coordinates{(-2,0)(1,0)(3,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=m(x)$} - \label{poly:fig:functionm} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-90,ymax=70, - width=\textwidth, - ] - \addplot expression[domain=-4.15:3.15,samples=100]{(x-1)*(x+2)*(x-3)*(x+1)*(x+4)}; - \addplot[soldot]coordinates{(-4,0)(-2,0)(-1,0)(1,0)(3,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=n(x)$} - \label{poly:fig:functionn} - \end{subfigure} - \caption{} - \end{widepage} - \end{figure} - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Horizontal intercepts]\label{poly:prob:matchpolys}% - State the horizontal intercepts (as ordered pairs) of the following polynomials. - \begin{multicols}{2} - \begin{subproblem}\label{poly:prob:degree5} - $p(x)=(x-1)(x+2)(x-3)(x+1)(x+4)$ - \begin{shortsolution} - $(-4,0)$, $(-2,0)$, $(-1,0)$, $(1,0)$, $(3,0)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $q(x)=-(x-1)(x+2)(x-3)$ - \begin{shortsolution} - $(-2,0)$, $(1,0)$, $(3,0)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(x)=(x-1)(x+2)(x-3)$ - \begin{shortsolution} - $(-2,0)$, $(1,0)$, $(3,0)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem}\label{poly:prob:degree2} - $s(x)=(x-2)(x+2)$ - \begin{shortsolution} - $(-2,0)$, $(2,0)$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Minimums, maximums, and concavity]\label{poly:prob:incdec} - Four polynomial functions are graphed in \cref{poly:fig:incdec}. The formulas - for these functions are (not respectively) - \begin{gather*} - p(x)=\frac{x^3}{6}-\frac{x^2}{4}-3x, \qquad q(x)=\frac{x^4}{20}+\frac{x^3}{15}-\frac{6}{5}x^2+1\\ - r(x)=-\frac{x^5}{50}-\frac{x^4}{40}+\frac{2x^3}{5}+6, \qquad s(x)=-\frac{x^6}{6000}-\frac{x^5}{2500}+\frac{67x^4}{4000}+\frac{17x^3}{750}-\frac{42x^2}{125} - \end{gather*} - \begin{figure}[!htb] - \begin{widepage} - \setlength{\figurewidth}{.23\textwidth} - \centering - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - width=\textwidth, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-8,-6,...,8}, - grid=major, - ] - \addplot expression[domain=-5.28:4.68,samples=50]{-x^5/50-x^4/40+2*x^3/5+6}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:incdec3} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - width=\textwidth, - xmin=-10,xmax=10,ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-8,-6,...,8}, - grid=major, - ] - \addplot expression[domain=-6.08:4.967,samples=50]{x^4/20+x^3/15-6/5*x^2+1}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:incdec2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - width=\textwidth, - xmin=-6,xmax=8,ymin=-10,ymax=10, - xtick={-4,-2,...,6}, - ytick={-8,-4,4,8}, - minor ytick={-6,-2,...,6}, - grid=both, - ] - \addplot expression[domain=-4.818:6.081,samples=50]{x^3/6-x^2/4-3*x}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:incdec1} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - width=\textwidth, - xmin=-10,xmax=10,ymin=-10,ymax=10, - xtick={-8,-4,4,8}, - ytick={-8,-4,4,8}, - minor xtick={-6,-2,...,6}, - minor ytick={-6,-2,...,6}, - grid=both, - ] - \addplot expression[domain=-9.77:8.866,samples=50]{-x^6/6000-x^5/2500+67*x^4/4000+17/750*x^3-42/125*x^2}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:incdec4} - \end{subfigure} - \caption{Graphs for \cref{poly:prob:incdec}.} - \label{poly:fig:incdec} - \end{widepage} - \end{figure} - \begin{subproblem} - Match each of the formulas with one of the given graphs. - \begin{shortsolution} - \begin{itemize} - \item $p$ is graphed in \vref{poly:fig:incdec1}; - \item $q$ is graphed in \vref{poly:fig:incdec2}; - \item $r$ is graphed in \vref{poly:fig:incdec3}; - \item $s$ is graphed in \vref{poly:fig:incdec4}. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Approximate the zeros of each function using the appropriate graph. - \begin{shortsolution} - \begin{itemize} - \item $p$ has simple zeros at about $-3.8$, $0$, and $5$. - \item $q$ has simple zeros at about $-5.9$, $-1$, $1$, and $4$. - \item $r$ has simple zeros at about $-5$, $-2.9$, and $4.1$. - \item $s$ has simple zeros at about $-9$, $-6$, $4.2$, $8.1$, and a zero of multiplicity $2$ at $0$. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Approximate the local maximums and minimums of each of the functions. - \begin{shortsolution} - \begin{itemize} - \item $p$ has a local maximum of approximately $3.9$ at $-2$, and a local minimum of approximately $-6.5$ at $3$. - \item $q$ has a local minimum of approximately $-10$ at $-4$, and $-4$ at $3$; $q$ has a local maximum of approximately $1$ at $0$. - \item $r$ has a local minimum of approximately $-5.5$ at $-4$, and a local maximum of approximately $10$ at $3$. - \item $s$ has a local maximum of approximately $5$ at $-8$, $0$ at $0$, and $5$ at $7$; $s$ has local minimums - of approximately $-3$ at $-4$, and $-1$ at $3$. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Approximate the global maximums and minimums of each of the functions. - \begin{shortsolution} - \begin{itemize} - \item $p$ does not have a global maximum, nor a global minimum. - \item $q$ has a global minimum of approximately $-10$; it does not have a global maximum. - \item $r$ does not have a global maximum, nor a global minimum. - \item $s$ has a global maximum of approximately $5$; it does not have a global minimum. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Approximate the intervals on which each function is increasing and decreasing. - \begin{shortsolution} - \begin{itemize} - \item $p$ is increasing on $(-\infty,-2)\cup (3,\infty)$, and decreasing on $(-2,3)$. - \item $q$ is increasing on $(-4,0)\cup (3,\infty)$, and decreasing on $(-\infty,-4)\cup (0,3)$. - \item $r$ is increasing on $(-4,3)$, and decreasing on $(-\infty,-4)\cup (3,\infty)$. - \item $s$ is increasing on $(-\infty,-8)\cup (-4,0)\cup (3,5)$, and decreasing on $(-8,-4)\cup (0,3)\cup (5,\infty)$. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Approximate the intervals on which each function is concave up and concave down. - \begin{shortsolution} - \begin{itemize} - \item $p$ is concave up on $(1,\infty)$, and concave down on $(-\infty,1)$. - \item $q$ is concave up on $(-\infty,-1)\cup (1,\infty)$, and concave down on $(-1,1)$. - \item $r$ is concave up on $(-\infty,-3)\cup (0,2)$, and concave down on $(-3,0)\cup (2,\infty)$. - \item $s$ is concave up on $(-6,-2)\cup (2,5)$, and concave down on $(-\infty,-6)\cup (-2,2)\cup (5,\infty)$. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - The degree of $q$ is $5$. Assuming that all of the real zeros of $q$ are - shown in its graph, how many complex zeros does $q$ have? - \begin{shortsolution} - \Vref{poly:fig:incdec2} shows that $q$ has $3$ real zeros - since the curve of $q$ cuts the horizontal axis $3$ times. - Since $q$ has degree $5$, $q$ must have $2$ complex zeros. - \end{shortsolution} - \end{subproblem} - \end{problem} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Long-run behaviour of polynomials] - Describe the long-run behavior of each of polynomial functions in - \crefrange{poly:prob:degree5}{poly:prob:degree2}. - \begin{shortsolution} - $\dd\lim_{x\rightarrow-\infty}p(x)=-\infty$, - $\dd\lim_{x\rightarrow\infty}p(x)=\infty$, - $\dd\lim_{x\rightarrow-\infty}q(x)=\infty$, - $\dd\lim_{x\rightarrow\infty}q(x)=-\infty$, - $\dd\lim_{x\rightarrow-\infty}r(x)=-\infty$, - $\dd\lim_{x\rightarrow\infty}r(x)=\infty$, - $\dd\lim_{x\rightarrow-\infty}s(x)=\infty$, - $\dd\lim_{x\rightarrow\infty}s(x)=\infty$, - \end{shortsolution} - \end{problem} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[True of false?] - Let $p$ be a polynomial function. - Label each of the following statements as true (T) or false (F); if they are false, - provide an example that supports your answer. - \begin{subproblem} - If $p$ has degree $3$, then $p$ has $3$ distinct zeros. - \begin{shortsolution} - False. Consider $p(x)=x^2(x+1)$ which has only 2 distinct zeros. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - If $p$ has degree $4$, then $\dd\lim_{x\rightarrow-\infty}p(x)=\infty$ and $\dd\lim_{x\rightarrow\infty}p(x)=\infty$. - \begin{shortsolution} - False. Consider $p(x)=-x^4$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - If $p$ has even degree, then it is possible that $p$ can have no real zeros. - \begin{shortsolution} - True. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - If $p$ has odd degree, then it is possible that $p$ can have no real zeros. - \begin{shortsolution} - False. All odd degree polynomials will cut the horizontal axis at least once. - \end{shortsolution} - \end{subproblem} - \end{problem} - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Find a formula from a description] - In each of the following problems, give a possible formula for a polynomial - function that has the specified properties. - \begin{subproblem} - Degree 2 and has zeros at $4$ and $5$. - \begin{shortsolution} - Possible option: $p(x)=(x-4)(x-5)$. Note we could multiply $p$ by any real number, and still meet the requirements. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Degree 3 and has zeros at $4$,$5$ and $-3$. - \begin{shortsolution} - Possible option: $p(x)=(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Degree 4 and has zeros at $0$, $4$, $5$, $-3$. - \begin{shortsolution} - Possible option: $p(x)=x(x-4)(x-5)(x+3)$. Note we could multiply $p$ by any real number, and still meet the requirements. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Degree 4, with zeros that make the graph cut at $2$, $-5$, and a zero that makes the graph touch at $-2$; - \begin{shortsolution} - Possible option: $p(x)=(x-2)(x+5)(x+2)^2$. Note we could multiply $p$ by any real number, and still meet the requirements. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Degree 3, with only one zero at $-1$. - \begin{shortsolution} - Possible option: $p(x)=(x+1)^3$. Note we could multiply $p$ by any real number, and still meet the requirements. - \end{shortsolution} - \end{subproblem} - \end{problem} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{problem}[\Cref{poly:step:last}] - \pccname{Saheed} is graphing a polynomial function, $p$. - He is following \crefrange{poly:step:first}{poly:step:last} and has so far - marked the zeros of $p$ on \cref{poly:fig:optionsp1}. Saheed tells you that - $p$ has degree $3$, but does \emph{not} say if the leading coefficient - of $p$ is positive or negative. - \begin{figure}[!htbp] - \begin{widepage} - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-15}, - width=\textwidth, - height=.5\textwidth, - ] - \addplot[soldot] coordinates{(-5,0)(2,0)(6,0)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:optionsp1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-15}, - width=\textwidth, - height=.5\textwidth, - ] - \addplot[soldot] coordinates{(-5,0)(6,0)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{poly:fig:optionsp2} - \end{subfigure}% - \caption{} - \end{widepage} - \end{figure} - \begin{subproblem} - Use the information in \cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient - is positive. - \begin{shortsolution} - Assuming that $a_3>0$: - - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-15}, - width=\solutionfigurewidth, - ] - \addplot expression[domain=-6.78179:8.35598,samples=50]{1/20*(x+5)*(x-2)*(x-6)}; - \addplot[soldot] coordinates{(-5,0)(2,0)(6,0)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Use the information in \cref{poly:fig:optionsp1} to help sketch $p$, assuming that the leading coefficient - is negative. - \begin{shortsolution} - Assuming that $a_3<0$: - - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-15}, - width=\solutionfigurewidth, - ] - \addplot expression[domain=-6.78179:8.35598,samples=50]{-1/20*(x+5)*(x-2)*(x-6)}; - \addplot[soldot] coordinates{(-5,0)(2,0)(6,0)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - Saheed now turns his attention to another polynomial function, $q$. He finds - the zeros of $q$ (there are only $2$) and marks them on \cref{poly:fig:optionsp2}. - Saheed knows that $q$ has degree $3$, but doesn't know if the leading - coefficient is positive or negative. - \begin{subproblem} - Use the information in \cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading - coefficient of $q$ is positive. Hint: only one of the zeros is simple. - \begin{shortsolution} - Assuming that $a_4>0$ there are $2$ different options: - - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-15}, - width=\solutionfigurewidth, - ] - \addplot expression[domain=-8.68983:7.31809,samples=50]{1/20*(x+5)^2*(x-6)}; - \addplot expression[domain=-6.31809:9.68893,samples=50]{1/20*(x+5)*(x-6)^2}; - \addplot[soldot] coordinates{(-5,0)(6,0)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Use the information in \cref{poly:fig:optionsp2} to help sketch $q$, assuming that the leading - coefficient of $q$ is negative. - \begin{shortsolution} - Assuming that $a_4<0$ there are $2$ different options: - - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-15}, - width=\solutionfigurewidth, - ] - \addplot expression[domain=-8.68983:7.31809,samples=50]{-1/20*(x+5)^2*(x-6)}; - \addplot expression[domain=-6.31809:9.68893,samples=50]{-1/20*(x+5)*(x-6)^2}; - \addplot[soldot] coordinates{(-5,0)(6,0)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \end{problem} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{problem}[Zeros] - Find all zeros of each of the following polynomial functions, making - sure to detail their multiplicity. Note that - you may need to use factoring, or the quadratic formula, or both! Also note - that some zeros may be repeated, and some may be complex. - \begin{multicols}{3} - \begin{subproblem} - $p(x)=x^2+1$ - \begin{shortsolution} - $\pm i$ (simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $q(y)=(y^2-9)(y^2-7)$ - \begin{shortsolution} - $\pm 3$, $\pm \sqrt{7}$ (all are simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(z)=-4z^3(z^2+3)(z^2+64)$ - \begin{shortsolution} - $0$ (multiplicity $3$), $\pm\sqrt{3}$ (simple), $\pm\sqrt{8}$ (simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $a(x)=x^4-81$ - \begin{shortsolution} - $\pm 3$, $\pm 3i$ (all are simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $b(y)=y^3-8$ - \begin{shortsolution} - $2$, $-1\pm i\sqrt{3}$ (all are simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $c(m)=m^3-m^2$ - \begin{shortsolution} - $0$ (multiplicity $2$), $1$ (simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $h(n)=(n+1)(n^2+4)$ - \begin{shortsolution} - $-1$, $\pm 2i$ (all are simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $f(\alpha)=(\alpha^2-16)(\alpha^2-5\alpha+4)$ - \begin{shortsolution} - $-4$ (simple), $4$ (multiplicity $2$), $1$ (simple). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $g(\beta)=(\beta^2-25)(\beta^2-5\beta-4)$ - \begin{shortsolution} - $\pm 5$, $\dfrac{5\pm\sqrt{41}}{2}$ (all are simple). - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{problem}[Given zeros, find a formula] - In each of the following problems you are given the zeros of a polynomial. - Write a possible formula for each polynomial| you may leave your - answer in factored form, but it may not contain complex numbers. Unless - otherwise stated, assume that the zeros are simple. - \begin{multicols}{3} - \begin{subproblem} - $1$, $2$ - \begin{shortsolution} - $p(x)=(x-1)(x-2)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $0$, $5$, $13$ - \begin{shortsolution} - $p(x)=x(x-5)(x-13)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $-7$, $2$ (multiplicity $3$), $5$ - \begin{shortsolution} - $p(x)=(x+7)(x-2)^3(x-5)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $0$, $\pm i$ - \begin{shortsolution} - $p(x)=x(x^2+1)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $\pm 2i$, $\pm 7$ - \begin{shortsolution} - $p(x)=(x^2+4)(x^2-49)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $-2\pm i\sqrt{6}$ - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{problem}[Composition of polynomials] - Let $p$ and $q$ be polynomial functions that have formulas - \[ - p(x)=(x+1)(x+2)(x+5), \qquad q(x)=3-x^4 - \] - Evaluate each of the following. - \begin{multicols}{4} - \begin{subproblem} - $(p\circ q)(0)$ - \begin{shortsolution} - $160$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(q\circ p)(0)$ - \begin{shortsolution} - $-9997$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(p\circ q)(1)$ - \begin{shortsolution} - $84$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(p\circ p)(0)$ - \begin{shortsolution} - $1980$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{problem}[Piecewise polynomial functions] - Let $P$ be the piecewise-defined function with formula - \[ - P(x)=\begin{cases} - (1-x)(2x+5)(x^2+1), & x\leq -3\\ - 4-x^2, & -3<x < 4\\ - x^3 & x\geq 4 - \end{cases} - \] - Evaluate each of the following - \begin{multicols}{5} - \begin{subproblem} - $P(-4)$ - \begin{shortsolution} - $-255$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $P(0)$ - \begin{shortsolution} - $4$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $P(4)$ - \begin{shortsolution} - $64$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $P(-3)$ - \begin{shortsolution} - $-40$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(P\circ P)(0)$ - \begin{shortsolution} - $64$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - - %=================================== - % Author: Hughes - % Date: July 2012 - %=================================== - \begin{problem}[Function algebra] - Let $p$ and $q$ be the polynomial functions that have formulas - \[ - p(x)=x(x+1)(x-3)^2, \qquad q(x)=7-x^2 - \] - Evaluate each of the following (if possible). - \begin{multicols}{4} - \begin{subproblem} - $(p+q)(1)$ - \begin{shortsolution} - $14$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(p-q)(0)$ - \begin{shortsolution} - $7$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(p\cdot q)(\sqrt{7})$ - \begin{shortsolution} - $0$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $\left( \frac{q}{p} \right)(1)$ - \begin{shortsolution} - $\frac{3}{4}$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \begin{subproblem} - What is the domain of the function $\frac{q}{p}$? - \begin{shortsolution} - $(-\infty,-1)\cup (-1,0)\cup (0,3)\cup (3,\infty)$ - \end{shortsolution} - \end{subproblem} - \end{problem} - - %=================================== - % Author: Hughes - % Date: July 2012 - %=================================== - \begin{problem}[Transformations: given the transformation, find the formula] - Let $p$ be the polynomial function that has formula. - \[ - p(x)=4x(x^2-1)(x+3) - \] - In each of the following - problems apply the given transformation to the function $p$ and - write a formula for the transformed version of $p$. - \begin{multicols}{2} - \begin{subproblem} - Shift $p$ to the right by $5$ units. - \begin{shortsolution} - $p(x-5)=4(x-5)(x-2)(x^2-10x+24)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Shift $p$ to the left by $6$ units. - \begin{shortsolution} - $p(x+6)=4(x+6)(x+9)(x^2+12x+35)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Shift $p$ up by $12$ units. - \begin{shortsolution} - $p(x)+12=4x(x^2-1)(x+3)+12$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Shift $p$ down by $2$ units. - \begin{shortsolution} - $p(x)-2=4x(x^2-1)(x+3)-2$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Reflect $p$ over the horizontal axis. - \begin{shortsolution} - $-p(x)=-4x(x^2-1)(x+3)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Reflect $p$ over the vertical axis. - \begin{shortsolution} - $p(-x)=-4x(x^2-1)(3-x)$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Find a formula from a table]\label{poly:prob:findformula} - \Crefrange{poly:tab:findformulap}{poly:tab:findformulas} show values of polynomial functions, $p$, $q$, - $r$, and $s$. - - \begin{table}[!htb] - \centering - \begin{widepage} - \caption{Tables for \cref{poly:prob:findformula}} - \label{poly:tab:findformula} - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=p(x)$} - \label{poly:tab:findformulap} - \begin{tabular}{rr} - \beforeheading - \heading{$x$} & \heading{$y$} \\ - \afterheading - $-4$ & $-56$ \\\normalline - $-3$ & $-18$ \\\normalline - $-2$ & $0$ \\\normalline - $-1$ & $4$ \\\normalline - $0$ & $0$ \\\normalline - $1$ & $-6$ \\\normalline - $2$ & $-8$ \\\normalline - $3$ & $0$ \\\normalline - $4$ & $24$ \\\lastline - \end{tabular} - \end{subtable} - \hfill - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=q(x)$} - \label{poly:tab:findformulaq} - \begin{tabular}{rr} - \beforeheading - \heading{$x$} & \heading{$y$} \\ \afterheading - $-4$ & $-16$ \\\normalline - $-3$ & $-3$ \\\normalline - $-2$ & $0$ \\\normalline - $-1$ & $-1$ \\\normalline - $0$ & $0$ \\\normalline - $1$ & $9$ \\\normalline - $2$ & $32$ \\\normalline - $3$ & $75$ \\\normalline - $4$ & $144$ \\\lastline - \end{tabular} - \end{subtable} - \hfill - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=r(x)$} - \label{poly:tab:findformular} - \begin{tabular}{rr} - \beforeheading - \heading{$x$} & \heading{$y$} \\ \afterheading - $-4$ & $105$ \\\normalline - $-3$ & $0$ \\\normalline - $-2$ & $-15$ \\\normalline - $-1$ & $0$ \\\normalline - $0$ & $9$ \\\normalline - $1$ & $0$ \\\normalline - $2$ & $-15$ \\\normalline - $3$ & $0$ \\\normalline - $4$ & $105$ \\\lastline - \end{tabular} - \end{subtable} - \hfill - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=s(x)$} - \label{poly:tab:findformulas} - \begin{tabular}{rr} - \beforeheading - \heading{$x$} & \heading{$y$} \\ \afterheading - $-4$ & $75$ \\\normalline - $-3$ & $0$ \\\normalline - $-2$ & $-9$ \\\normalline - $-1$ & $0$ \\\normalline - $0$ & $3$ \\\normalline - $1$ & $0$ \\\normalline - $2$ & $15$ \\\normalline - $3$ & $96$ \\\normalline - $4$ & $760$ \\\lastline - \end{tabular} - \end{subtable} - \end{widepage} - \end{table} - - \begin{subproblem} - Assuming that all of the zeros of $p$ are shown (in \cref{poly:tab:findformulap}), how many zeros does $p$ have? - \begin{shortsolution} - $p$ has 3 zeros. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - What is the degree of $p$? - \begin{shortsolution} - $p$ is degree 3. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Write a formula for $p(x)$. - \begin{shortsolution} - $p(x)=x(x+2)(x-3)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Assuming that all of the zeros of $q$ are shown (in \cref{poly:tab:findformulaq}), how many zeros does $q$ have? - \begin{shortsolution} - $q$ has 2 zeros. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Describe the difference in behavior of $p$ and $q$ at $-2$. - \begin{shortsolution} - $p$ changes sign at $-2$, and $q$ does not change sign at $-2$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Given that $q$ is a degree-$3$ polynomial, write a formula for $q(x)$. - \begin{shortsolution} - $q(x)=x(x+2)^2$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Assuming that all of the zeros of $r$ are shown (in \cref{poly:tab:findformular}), find a formula for $r(x)$. - \begin{shortsolution} - $r(x)=(x+3)(x+1)(x-1)(x-3)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Assuming that all of the zeros of $s$ are shown (in \cref{poly:tab:findformulas}), find a formula for $s(x)$. - \begin{shortsolution} - $s(x)=(x+3)(x+1)(x-1)^2$ - \end{shortsolution} - \end{subproblem} - \end{problem} - \end{exercises} - -\section{Rational functions} - \subsection*{Power functions with negative exponents} - The study of rational functions will rely upon a good knowledge - of power functions with negative exponents. \Cref{rat:ex:oddpow,rat:ex:evenpow} are - simple but fundamental to understanding the behavior of rational functions. - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Power functions with odd negative exponents]\label{rat:ex:oddpow} - Graph each of the following functions on your calculator, state their domain in interval notation, and their - behavior as $x\rightarrow 0^-$ and $x\rightarrow 0^+$. - \[ - f(x)=\frac{1}{x},\qquad g(x)=\dfrac{1}{x^3},\qquad h(x)=\dfrac{1}{x^5} - \] - \begin{pccsolution} - The functions $f$, $g$, and $k$ are plotted in \cref{rat:fig:oddpow}. - The domain of each of the functions $f$, $g$, and $h$ is $(-\infty,0)\cup (0,\infty)$. Note that - the long-run behavior of each of the functions is the same, and in particular - \begin{align*} - f(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty - \end{align*} - The same results hold for $g$ and $h$. Note also that each of the functions - has a \emph{vertical asymptote} at $0$. We see that - \begin{align*} - f(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} f(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ - \end{align*} - The same results hold for $g$ and $h$. - - The curve of a function that has a vertical asymptote is necessarily separated - into \emph{branches}| each of the functions $f$, $g$, and $h$ have $2$ branches. - \end{pccsolution} - \end{pccexample} - - \begin{figure}[!htb] - \begin{minipage}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-3,xmax=3, - ymin=-5,ymax=5, - xtick={-2,-1,...,2}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\textwidth, - legend pos=north west, - ] - \addplot expression[domain=-3:-0.2]{1/x}; - \addplot expression[domain=-3:-0.584]{1/x^3}; - \addplot expression[domain=-3:-0.724]{1/x^5}; - \addplot expression[domain=0.2:3]{1/x}; - \addplot expression[domain=0.584:3]{1/x^3}; - \addplot expression[domain=0.724:3]{1/x^5}; - \addplot[soldot]coordinates{(-1,-1)}node[axisnode,anchor=north east]{$(-1,-1)$}; - \addplot[soldot]coordinates{(1,1)}node[axisnode,anchor=south west]{$(1,1)$}; - \legend{$f$,$g$,$h$} - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:oddpow} - \end{minipage}% - \hfill - \begin{minipage}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-3,xmax=3, - ymin=-5,ymax=5, - xtick={-2,-1,...,2}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\textwidth, - legend pos=south east, - ] - \addplot expression[domain=-3:-0.447]{1/x^2}; - \addplot expression[domain=-3:-0.668]{1/x^4}; - \addplot expression[domain=-3:-0.764]{1/x^6}; - \addplot expression[domain=0.447:3]{1/x^2}; - \addplot expression[domain=0.668:3]{1/x^4}; - \addplot expression[domain=0.764:3]{1/x^6}; - \addplot[soldot]coordinates{(-1,1)}node[axisnode,anchor=south east]{$(-1,1)$}; - \addplot[soldot]coordinates{(1,1)}node[axisnode,anchor=south west]{$(1,1)$}; - \legend{$F$,$G$,$H$} - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:evenpow} - \end{minipage}% - \end{figure} - - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Power functions with even negative exponents]\label{rat:ex:evenpow}% - Graph each of the following functions, state their domain, and their - behavior as $x\rightarrow 0^-$ and $x\rightarrow 0^+$. - \[ - f(x)=\frac{1}{x^2},\qquad g(x)=\frac{1}{x^4},\qquad h(x)=\frac{1}{x^6} - \] - \begin{pccsolution} - The functions $F$, $G$, and $H$ are plotted in \cref{rat:fig:evenpow}. - The domain of each of the functions $F$, $G$, and $H$ is $(-\infty,0)\cup (0,\infty)$. Note that - the long-run behavior of each of the functions is the same, and in particular - \begin{align*} - F(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} f(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty - \end{align*} - As in \cref{rat:ex:oddpow}, $F$ has a horizontal asymptote that - has equation $y=0$. - The same results hold for $G$ and $H$. Note also that each of the functions - has a \emph{vertical asymptote} at $0$. We see that - \begin{align*} - F(x)\rightarrow \infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} F(x)\rightarrow \infty & \text{ as } x\rightarrow 0^+ - \end{align*} - The same results hold for $G$ and $H$. Each of the functions $F$, $G$, and $H$ - have $2$ branches. - \end{pccsolution} - \end{pccexample} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{doyouunderstand} - \begin{problem} - Repeat \cref{rat:ex:oddpow,rat:ex:evenpow} using (respectively) - \begin{subproblem} - $k(x)=-\dfrac{1}{x}$, $ m(x)=-\dfrac{1}{x^3}$, $ n(x)=-\dfrac{1}{x^5}$ - \begin{shortsolution} - The functions $k$, $m$, and $n$ have domain $(-\infty,0)\cup (0,\infty)$, and - are graphed below. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-3,xmax=3, - ymin=-5,ymax=5, - xtick={-2,-1,...,2}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\solutionfigurewidth, - legend pos=north east, - ] - \addplot expression[domain=-3:-0.2]{-1/x}; - \addplot expression[domain=-3:-0.584]{-1/x^3}; - \addplot expression[domain=-3:-0.724]{-1/x^5}; - \addplot expression[domain=0.2:3]{-1/x}; - \addplot expression[domain=0.584:3]{-1/x^3}; - \addplot expression[domain=0.724:3]{-1/x^5}; - \legend{$k$,$m$,$n$} - \end{axis} - \end{tikzpicture} - - Note that - \begin{align*} - k(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} k(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty \\ - \intertext{and also} - k(x)\rightarrow \infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} k(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+ - \end{align*} - The same are true for $m$ and $n$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $ K(x)=-\dfrac{1}{x^2}$, $ M(x)=-\dfrac{1}{x^4}$, $ N(x)=-\dfrac{1}{x^6}$ - \begin{shortsolution} - The functions $K$, $M$, and $N$ have domain $(-\infty,0)\cup (0,\infty)$, and - are graphed below. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-3,xmax=3, - ymin=-5,ymax=5, - xtick={-2,-1,...,2}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\solutionfigurewidth, - legend pos=north east, - ] - \addplot expression[domain=-3:-0.447]{-1/x^2}; - \addplot expression[domain=-3:-0.668]{-1/x^4}; - \addplot expression[domain=-3:-0.764]{-1/x^6}; - \addplot expression[domain=0.447:3]{-1/x^2}; - \addplot expression[domain=0.668:3]{-1/x^4}; - \addplot expression[domain=0.764:3]{-1/x^6}; - \legend{$K$,$M$,$N$} - \end{axis} - \end{tikzpicture} - - Note that - \begin{align*} - K(x)\rightarrow 0 & \text{ as } x\rightarrow\infty \\ - \mathllap{\text{and }} K(x)\rightarrow 0 & \text{ as } x\rightarrow-\infty \\ - \intertext{and also} - K(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^- \\ - \mathllap{\text{and }} K(x)\rightarrow -\infty & \text{ as } x\rightarrow 0^+ - \end{align*} - The same are true for $M$ and $N$. - \end{shortsolution} - \end{subproblem} - \end{problem} - \end{doyouunderstand} - - \subsection*{Rational functions} - \begin{pccdefinition}[Rational functions]\label{rat:def:function} - Rational functions have the form - \[ - r(x) = \frac{p(x)}{q(x)} - \] - where both $p$ and $q$ are polynomials. - - Note that - \begin{itemize} - \item the domain or $r$ will be all real numbers, except those that - make the \emph{denominator}, $q(x)$, equal to $0$; - \item the zeros of $r$ are the zeros of $p$, i.e the real numbers - that make the \emph{numerator}, $p(x)$, equal to $0$. - \end{itemize} - - \Cref{rat:ex:oddpow,rat:ex:evenpow} are particularly important because $r$ - will behave like $\frac{1}{x}$, or $\frac{1}{x^2}$ around its vertical asymptotes, - depending on the power that the relevant term is raised to| we will demonstrate - this in what follows. - \end{pccdefinition} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Rational or not] - Identify whether each of the following functions is a rational or not. If - the function is rational, state the domain. - \begin{multicols}{3} - \begin{enumerate} - \item $r(x)=\dfrac{1}{x}$ - \item $f(x)=2^x+3$ - \item $g(x)=19$ - \item $h(x)=\dfrac{3+x}{4-x}$ - \item $k(x)=\dfrac{x^3+2x}{x-15}$ - \item $l(x)=9-4x$ - \item $m(x)=\dfrac{x+5}{(x-7)(x+9)}$ - \item $n(x)=x^2+6x+7$ - \item $q(x)=1-\dfrac{3}{x+1}$ - \end{enumerate} - \end{multicols} - \begin{pccsolution} - \begin{enumerate} - \item $r$ is rational; the domain of $r$ is $(-\infty,0)\cup(0,\infty)$. - \item $f$ is not rational. - \item $g$ is not rational; $g$ is constant. - \item $h$ is rational; the domain of $h$ is $(-\infty,4)\cup(4,\infty)$. - \item $k$ is rational; the domain of $k$ is $(-\infty,15)\cup(15,\infty)$. - \item $l$ is not rational; $l$ is linear. - \item $m$ is rational; the domain of $m$ is $(-\infty,-9)\cup(-9,7)\cup(7,\infty)$. - \item $n$ is not rational; $n$ is quadratic (or you might describe $n$ as a polynomial). - \item $q$ is rational; the domain of $q$ is $(-\infty,-1)\cup (-1,\infty)$. - \end{enumerate} - \end{pccsolution} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Match formula to graph] - Each of the following functions is graphed in \cref{rat:fig:whichiswhich}. - Which is which? - \[ - r(x)=\frac{1}{x-3}, \qquad q(x)=\frac{x-2}{x+5}, \qquad k(x)=\frac{1}{(x+2)(x-3)} - \] - \begin{figure}[!htb] - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-2)/(x+5);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-6,ymax=6, - xtick={-8,-6,...,8}, - minor ytick={-4,-3,...,4}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:-6.37]{f}; - \addplot[pccplot] expression[domain=-3.97:10]{f}; - \addplot[soldot] coordinates{(2,0)}; - \addplot[asymptote,domain=-6:6]({-5},{x}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:which1} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=1/(x-3);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-5,ymax=6, - xtick={-8,-6,...,8}, - ytick={-4,4}, - minor ytick={-3,...,5}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:2.8]{f}; - \addplot[pccplot] expression[domain=3.17:10]{f}; - \addplot[asymptote,domain=-6:6]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:which2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=1/((x-3)*(x+2));}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-5,ymax=5, - xtick={-8,-6,...,8}, - ytick={-4,4}, - minor ytick={-3,...,3}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:-2.03969]{f}; - \addplot[pccplot] expression[domain=-1.95967:2.95967]{f}; - \addplot[pccplot] expression[domain=3.03969:10]{f}; - \addplot[asymptote,domain=-5:5]({-2},{x}); - \addplot[asymptote,domain=-5:5]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:which3} - \end{subfigure} - \caption{} - \label{rat:fig:whichiswhich} - \end{figure} - - \begin{pccsolution} - Let's start with the function $r$. Note that domain of $r$ is $(-\infty,3)\cup(0,3)$, so - we search for a function that has a vertical asymptote at $3$. There - are two possible choices: the functions graphed in \cref{rat:fig:which2,rat:fig:which3}, - but note that the function in \cref{rat:fig:which3} also has a vertical asymptote at $-2$ - which is not consistent with the formula for $r(x)$. Therefore, $y=r(x)$ - is graphed in \cref{rat:fig:which2}. - - The function $q$ has domain $(-\infty,-5)\cup(-5,\infty)$, so we search - for a function that has a vertical asymptote at $-5$. The only candidate - is the curve shown in \cref{rat:fig:which1}; note that the curve also goes through $(2,0)$, - which is consistent with the formula for $q(x)$, since $q(2)=0$, i.e $q$ - has a zero at $2$. - - The function $k$ has domain $(-\infty,-2)\cup(-2,3)\cup(3,\infty)$, and - has vertical asymptotes at $-2$ and $3$. This is consistent with - the graph in \cref{rat:fig:which3} (and is the only curve that - has $3$ branches). - - We note that each function behaves like $\frac{1}{x}$ around its vertical asymptotes, - because each linear factor in each denominator is raised to the power $1$; if (for example) - the definition of $r$ was instead - \[ - r(x)=\frac{1}{(x-3)^2} - \] - then we would see that $r$ behaves like $\frac{1}{x^2}$ around its vertical asymptote, and - the graph of $r$ would be very different. We will deal with these cases in the examples that follow. - \end{pccsolution} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Repeated factors in the denominator] - Consider the functions $f$, $g$, and $h$ that have formulas - \[ - f(x)=\frac{x-2}{(x-3)(x+2)}, \qquad g(x)=\frac{x-2}{(x-3)^2(x+2)}, \qquad h(x)=\frac{x-2}{(x-3)(x+2)^2} - \] - which are graphed in \cref{rat:fig:repfactd}. Note that each function has $2$ - vertical asymptotes, and the domain of each function is - \[ - (-\infty,-2)\cup(-2,3)\cup(3,\infty) - \] - so we are not surprised to see that each curve has $3$ branches. We also note that - the numerator of each function is the same, which tells us that each function has - only $1$ zero at $2$. - - The functions $g$ and $h$ are different from those that we have considered previously, - because they have a repeated factor in the denominator. Notice in particular - the way that the functions behave around their asymptotes: - \begin{itemize} - \item $f$ behaves like $\frac{1}{x}$ around both of its asymptotes; - \item $g$ behaves like $\frac{1}{x}$ around $-2$, and like $\frac{1}{x^2}$ around $3$; - \item $h$ behaves like $\frac{1}{x^2}$ around $-2$, and like $\frac{1}{x}$ around $3$. - \end{itemize} - \end{pccexample} - \begin{figure}[!htb] - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)*(x-3));}] - \begin{axis}[ - % framed, - xmin=-5,xmax=5, - ymin=-4,ymax=4, - xtick={-4,-2,...,4}, - ytick={-2,2}, - % grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-5:-2.201]{f}; - \addplot[pccplot] expression[domain=-1.802:2.951]{f}; - \addplot[pccplot] expression[domain=3.052:5]{f}; - \addplot[soldot] coordinates{(2,0)}; - % \addplot[asymptote,domain=-6:6]({-2},{x}); - % \addplot[asymptote,domain=-6:6]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{$y=\dfrac{x-2}{(x+2)(x-3)}$} - \label{rat:fig:repfactd1} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)*(x-3)^2);}] - \begin{axis}[ - % framed, - xmin=-5,xmax=5, - ymin=-4,ymax=4, - xtick={-4,-2,...,4}, - ytick={-2,2}, - % grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-5:-2.039]{f}; - \addplot[pccplot] expression[domain=-1.959:2.796]{f}; - \addplot[pccplot] expression[domain=3.243:5]{f}; - \addplot[soldot] coordinates{(2,0)}; - % \addplot[asymptote,domain=-4:4]({-2},{x}); - % \addplot[asymptote,domain=-4:4]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{$y=\dfrac{x-2}{(x+2)(x-3)^2}$} - \label{rat:fig:repfactd2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+2)^2*(x-3));}] - \begin{axis}[ - % framed, - xmin=-5,xmax=5, - ymin=-4,ymax=4, - xtick={-4,-2,...,2}, - ytick={-2,2}, - % grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-5:-2.451]{f}; - \addplot[pccplot] expression[domain=-1.558:2.990]{f}; - \addplot[pccplot] expression[domain=3.010:6]{f}; - \addplot[soldot] coordinates{(2,0)}; - % \addplot[asymptote,domain=-4:4]({-2},{x}); - % \addplot[asymptote,domain=-4:4]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{$y=\dfrac{x-2}{(x+2)^2(x-3)}$} - \label{rat:fig:repfactd3} - \end{subfigure} - \caption{} - \label{rat:fig:repfactd} - \end{figure} - - \Cref{rat:def:function} says that the zeros of - the rational function $r$ that has formula $r(x)=\frac{p(x)}{q(x)}$ are - the zeros of $p$. Let's explore this a little more. - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}[Zeros] Find the zeros of each of the following functions - \[ - \alpha(x)=\frac{x+5}{3x-7}, \qquad \beta(x)=\frac{9-x}{x+1}, \qquad \gamma(x)=\frac{17x^2-10}{2x+1} - \] - \begin{pccsolution} - We find the zeros of each function in turn by setting the numerator equal to $0$. The zeros of - $\alpha$ are found by solving - \[ - x+5=0 - \] - The zero of $\alpha$ is $-5$. - - Similarly, we may solve $9-x=0$ to find the zero of $\beta$, which is clearly $9$. - - The zeros of $\gamma$ satisfy the equation - \[ - 17x^2-10=0 - \] - which we can solve using the square root property to obtain - \[ - x=\pm\frac{10}{17} - \] - The zeros of $\gamma$ are $\pm\frac{10}{17}$. - \end{pccsolution} - \end{pccexample} - - \subsection*{Long-run behavior} - Our focus so far has been on the behavior of rational functions around - their \emph{vertical} asymptotes. In fact, rational functions also - have interesting long-run behavior around their \emph{horizontal} or - \emph{oblique} asymptotes. A rational function will always have either - a horizontal or an oblique asymptote| the case is determined by the degree - of the numerator and the degree of the denominator. - \begin{pccdefinition}[Long-run behavior]\label{rat:def:longrun} - Let $r$ be the rational function that has formula - \[ - r(x) = \frac{a_n x^n + a_{n-1}x^{n-1}+\ldots + a_0}{b_m x^m + b_{m-1}x^{m-1}+\ldots+b_0} - \] - We can classify the long-run behavior of the rational function $r$ - according to the following criteria: - \begin{itemize} - \item if $n<m$ then $r$ has a horizontal asymptote with equation $y=0$; - \item if $n=m$ then $r$ has a horizontal asymptote with equation $y=\dfrac{a_n}{b_m}$; - \item if $n>m$ then $r$ will have an oblique asymptote as $x\rightarrow\pm\infty$ (more on this in \cref{rat:sec:oblique}) - \end{itemize} - \end{pccdefinition} - We will concentrate on functions that have horizontal asymptotes until - we reach \cref{rat:sec:oblique}. - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}[Long-run behavior graphically]\label{rat:ex:horizasymp} - \pccname{Kebede} has graphed the following functions in his graphing calculator - \[ - r(x)=\frac{x+1}{x-3}, \qquad s(x)=\frac{2(x+1)}{x-3}, \qquad t(x)=\frac{3(x+1)}{x-3} - \] - and obtained the curves shown in \cref{rat:fig:horizasymp}. Kebede decides - to test his knowledgeable friend \pccname{Oscar}, and asks him - to match the formulas to the graphs. - - \begin{figure}[!htb] - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=2*(x+1)/(x-3);}] - \begin{axis}[ - framed, - xmin=-15,xmax=15, - ymin=-6,ymax=6, - xtick={-12,-8,...,12}, - minor ytick={-4,-3,...,4}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-15:2]{f}; - \addplot[pccplot] expression[domain=5:15]{f}; - \addplot[soldot] coordinates{(-1,0)}; - \addplot[asymptote,domain=-6:6]({3},{x}); - \addplot[asymptote,domain=-15:15]({x},{2}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:horizasymp1} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x+1)/(x-3);}] - \begin{axis}[ - framed, - xmin=-15,xmax=15, - ymin=-6,ymax=6, - xtick={-12,-8,...,12}, - minor ytick={-4,-3,...,4}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-15:2.42857,samples=50]{f}; - \addplot[pccplot] expression[domain=3.8:15,samples=50]{f}; - \addplot[soldot] coordinates{(-1,0)}; - \addplot[asymptote,domain=-6:6]({3},{x}); - \addplot[asymptote,domain=-15:15]({x},{1}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:horizasymp2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=3*(x+1)/(x-3);}] - \begin{axis}[ - framed, - xmin=-15,xmax=15, - ymin=-6,ymax=6, - xtick={-12,-8,...,12}, - minor ytick={-4,-3,...,4}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-15:1.6666,samples=50]{f}; - \addplot[pccplot] expression[domain=7:15]{f}; - \addplot[soldot] coordinates{(-1,0)}; - \addplot[asymptote,domain=-6:6]({3},{x}); - \addplot[asymptote,domain=-15:15]({x},{3}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:horizasymp3} - \end{subfigure} - \caption{Horizontal asymptotes} - \label{rat:fig:horizasymp} - \end{figure} - - Oscar notices that each function has a vertical asymptote at $3$ and a zero at $-1$. - The main thing that catches Oscar's eye is that each function has a different - coefficient in the numerator, and that each curve has a different horizontal asymptote. - In particular, Oscar notes that - \begin{itemize} - \item the curve shown in \cref{rat:fig:horizasymp1} has a horizontal asymptote with equation $y=2$; - \item the curve shown in \cref{rat:fig:horizasymp2} has a horizontal asymptote with equation $y=1$; - \item the curve shown in \cref{rat:fig:horizasymp3} has a horizontal asymptote with equation $y=3$. - \end{itemize} - Oscar is able to tie it all together for Kebede by referencing \cref{rat:def:longrun}. He says - that since the degree of the numerator and the degree of the denominator is the same - for each of the functions $r$, $s$, and $t$, the horizontal asymptote will be determined - by evaluating the ratio of their leading coefficients. - - Oscar therefore says that $r$ should have a horizontal asymptote $y=\frac{1}{1}=1$, $s$ should - have a horizontal asymptote $y=\frac{2}{1}=2$, and $t$ should have a horizontal asymptote - $y=\frac{3}{1}=3$. Kebede is able to finish the problem from here, and says that $r$ is - shown in \cref{rat:fig:horizasymp2}, $s$ is shown in \cref{rat:fig:horizasymp1}, and - $t$ is shown in \cref{rat:fig:horizasymp3}. - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}[Long-run behavior numerically] - \pccname{Xiao} and \pccname{Dwayne} saw \cref{rat:ex:horizasymp} but are a little confused - about horizontal asymptotes. What does it mean to say that a function $r$ has a horizontal - asymptote? - - They decide to explore the concept by - constructing a table of values for the rational functions $R$ and $S$ that have formulas - \[ - R(x)=\frac{-5(x+1)}{x-3}, \qquad S(x)=\frac{7(x-5)}{2(x+1)} - \] - In \cref{rat:tab:plusinfty} they model the behavior of $R$ and $S$ as $x\rightarrow\infty$, - and in \cref{rat:tab:minusinfty} they model the behavior of $R$ and $S$ as $x\rightarrow-\infty$ - by substituting very large values of $|x|$ into each function. - \begin{table}[!htb] - \begin{minipage}{.5\textwidth} - \centering - \caption{$R$ and $S$ as $x\rightarrow\infty$} - \label{rat:tab:plusinfty} - \begin{tabular}{crr} - \beforeheading - $x$ & $R(x)$ & $S(x)$ \\ \afterheading - $1\times 10^2$ & $-5.20619$ & $3.29208$ \\\normalline - $1\times 10^3$ & $-5.02006$ & $3.47902$ \\\normalline - $1\times 10^4$ & $-5.00200$ & $3.49790$ \\\normalline - $1\times 10^5$ & $-5.00020$ & $3.49979$ \\\normalline - $1\times 10^6$ & $-5.00002$ & $3.49998$ \\\lastline - \end{tabular} - \end{minipage}% - \begin{minipage}{.5\textwidth} - \centering - \caption{$R$ and $S$ as $x\rightarrow-\infty$} - \label{rat:tab:minusinfty} - \begin{tabular}{crr} - \beforeheading - $x$ & $R(x)$ & $S(x)$ \\ \afterheading - $-1\times 10^2$ & $-4.80583$ & $3.71212$ \\\normalline - $-1\times 10^3$ & $-4.98006$ & $3.52102$ \\\normalline - $-1\times 10^4$ & $-4.99800$ & $3.50210$ \\\normalline - $-1\times 10^5$ & $-4.99980$ & $3.50021$ \\\normalline - $-1\times 10^6$ & $-4.99998$ & $3.50002$ \\\lastline - \end{tabular} - \end{minipage} - \end{table} - - Xiao and Dwayne study \cref{rat:tab:plusinfty,rat:tab:minusinfty} and decide that - the functions $R$ and $S$ never actually touch their horizontal asymptotes, but they - do get infinitely close. They also feel as if they have a better understanding of - what it means to study the behavior of a function as $x\rightarrow\pm\infty$. - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample}[Repeated factors in the numerator] - Consider the functions $f$, $g$, and $h$ that have formulas - \[ - f(x)=\frac{(x-2)^2}{(x-3)(x+1)}, \qquad g(x)=\frac{x-2}{(x-3)(x+1)}, \qquad h(x)=\frac{(x-2)^3}{(x-3)(x+1)} - \] - which are graphed in \cref{rat:fig:repfactn}. We note that each function has vertical - asymptotes at $-1$ and $3$, and so the domain of each function is - \[ - (-\infty,-1)\cup(-1,3)\cup(3,\infty) - \] - We also notice that the numerators of each function are quite similar| indeed, each - function has a zero at $2$, but how does each function behave around their zero? - - Using \cref{rat:fig:repfactn} to guide us, we note that - \begin{itemize} - \item $f$ has a horizontal intercept $(2,0)$, but the curve of - $f$ does not cut the horizontal axis| it bounces off it; - \item $g$ also has a horizontal intercept $(2,0)$, and the curve - of $g$ \emph{does} cut the horizontal axis; - \item $h$ has a horizontal intercept $(2,0)$, and the curve of $h$ - also cuts the axis, but appears flattened as it does so. - \end{itemize} - - We can further enrich our study by discussing the long-run behavior of each function. - Using the tools of \cref{rat:def:longrun}, we can deduce that - \begin{itemize} - \item $f$ has a horizontal asymptote with equation $y=1$; - \item $g$ has a horizontal asymptote with equation $y=0$; - \item $h$ does \emph{not} have a horizontal asymptote| it has an oblique asymptote (we'll - study this more in \cref{rat:sec:oblique}). - \end{itemize} - \end{pccexample} - - \begin{figure}[!htb] - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-2)^2/((x+1)*(x-3));}] - \begin{axis}[ - % framed, - xmin=-5,xmax=5, - ymin=-10,ymax=10, - xtick={-4,-2,...,4}, - ytick={-8,-4,...,8}, - % grid=both, - width=\figurewidth, - ] - \addplot[pccplot] expression[domain=-5:-1.248,samples=50]{f}; - \addplot[pccplot] expression[domain=-0.794:2.976,samples=50]{f}; - \addplot[pccplot] expression[domain=3.026:5,samples=50]{f}; - \addplot[soldot] coordinates{(2,0)}; - % \addplot[asymptote,domain=-6:6]({-1},{x}); - % \addplot[asymptote,domain=-6:6]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{$y=\dfrac{(x-2)^2}{(x+1)(x-3)}$} - \label{rat:fig:repfactn1} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-2)/((x+1)*(x-3));}] - \begin{axis}[ - % framed, - xmin=-5,xmax=5, - ymin=-10,ymax=10, - xtick={-4,-2,...,4}, - ytick={-8,-4,...,8}, - % grid=both, - width=\figurewidth, - ] - \addplot[pccplot] expression[domain=-5:-1.075]{f}; - \addplot[pccplot] expression[domain=-0.925:2.975]{f}; - \addplot[pccplot] expression[domain=3.025:5]{f}; - \addplot[soldot] coordinates{(2,0)}; - % \addplot[asymptote,domain=-6:6]({-1},{x}); - % \addplot[asymptote,domain=-6:6]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{$y=\dfrac{x-2}{(x+1)(x-3)}$} - \label{rat:fig:repfactn2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-2)^3/((x+1)*(x-3));}] - \begin{axis}[ - % framed, - xmin=-5,xmax=5, - xtick={-8,-6,...,8}, - % grid=both, - ymin=-30,ymax=30, - width=\figurewidth, - ] - \addplot[pccplot] expression[domain=-5:-1.27]{f}; - \addplot[pccplot] expression[domain=-0.806:2.99185]{f}; - \addplot[pccplot] expression[domain=3.0085:5]{f}; - \addplot[soldot] coordinates{(2,0)}; - % \addplot[asymptote,domain=-30:30]({-1},{x}); - % \addplot[asymptote,domain=-30:30]({3},{x}); - \end{axis} - \end{tikzpicture} - \caption{$y=\dfrac{(x-2)^3}{(x+1)(x-3)}$} - \label{rat:fig:repfactn3} - \end{subfigure} - \caption{} - \label{rat:fig:repfactn} - \end{figure} - - \subsection*{Holes} - Rational functions have a vertical asymptote at $a$ if the denominator is $0$ at $a$. - What happens if the numerator is $0$ at the same place? In this case, we say that the rational - function has a \emph{hole} at $a$. - \begin{pccdefinition}[Holes] - The rational function - \[ - r(x)=\frac{p(x)}{q(x)} - \] - has a hole at $a$ if $p(a)=q(a)=0$. Note that holes are different from - a vertical asymptotes. We represent that $r$ has a hole at the point - $(a,r(a))$ on the curve $y=r(x)$ by - using a hollow circle, $\circ$. - \end{pccdefinition} - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{pccexample} - \pccname{Mohammed} and \pccname{Sue} have graphed the function $r$ that has formula - \[ - r(x)=\frac{x^2+x-6}{(x-2)} - \] - in their calculators, and can not decide if the correct graph - is \cref{rat:fig:hole} or \cref{rat:fig:hole1}. - - Luckily for them, Oscar is nearby, and can help them settle the debate. - Oscar demonstrates that - \begin{align*} - r(x) & =\frac{(x+3)(x-2)}{(x-2)} \\ - & = x+3 - \end{align*} - but only when $x\ne 2$, because the function is undefined at $2$. Oscar - says that this necessarily means that the domain or $r$ is - \[ - (-\infty,2)\cup(2,\infty) - \] - and that $r$ must have a hole at $2$. - - Mohammed and Sue are very grateful for the clarification, and conclude that - the graph of $r$ is shown in \cref{rat:fig:hole1}. - \begin{figure}[!htb] - \begin{minipage}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-4,...,8}, - ytick={-8,-4,...,8}, - grid=both, - width=\textwidth, - ] - \addplot expression[domain=-10:7]{x+3}; - \addplot[soldot] coordinates{(-3,0)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:hole} - \end{minipage}% - \hfill - \begin{minipage}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-4,...,8}, - ytick={-8,-4,...,8}, - grid=both, - width=\textwidth, - ] - \addplot expression[domain=-10:7]{x+3}; - \addplot[holdot] coordinates{(2,5)}; - \addplot[soldot] coordinates{(-3,0)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:hole1} - \end{minipage}% - \end{figure} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{pccexample} - Consider the function $f$ that has formula - \[ - f(x)=\frac{x(x+3)}{x^2-4x} - \] - The domain of $f$ is $(-\infty,0)\cup(0,4)\cup(4,\infty)$ because both $0$ and $4$ - make the denominator equal to $0$. Notice that - \begin{align*} - f(x) & = \frac{x(x+3)}{x(x-4)} \\ - & = \frac{x+3}{x-4} - \end{align*} - provided that $x\ne 0$. Since $0$ makes the numerator - and the denominator 0 at the same time, we say that $f$ has a hole at $(0,-\nicefrac{3}{4})$. - Note that this necessarily means that $f$ does not have a vertical intercept. - - We also note $f$ has a vertical asymptote at $4$; the function is graphed in \cref{rat:fig:holeex}. - \begin{figure}[!htb] - \centering - \begin{tikzpicture}[/pgf/declare function={f=(x+3)/(x-4);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-8,-6,...,8}, - grid=both, - ] - \addplot[pccplot] expression[domain=-10:3.36364,samples=50]{f}; - \addplot[pccplot] expression[domain=4.77:10]{f}; - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[holdot]coordinates{(0,-0.75)}; - \addplot[soldot] coordinates{(-3,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=\dfrac{x(x+3)}{x^2-4x}$} - \label{rat:fig:holeex} - \end{figure} - \end{pccexample} - - - - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{pccexample}[Minimums and maximums] - \pccname{Seamus} and \pccname{Trang} are discussing rational functions. Seamus says that - if a rational function has a vertical asymptote, then it can - not possibly have local minimums and maximums, nor can it have - global minimums and maximums. - - Trang says this statement is not always true. She plots the functions - $f$ and $g$ that have formulas - \[ - f(x)=-\frac{32(x-1)(x+1)}{(x-2)^2(x+2)^2}, \qquad g(x)=\frac{32(x-1)(x+1)}{(x-2)^2(x+2)^2} - \] - in \cref{rat:fig:minmax1,rat:fig:minmax2} and shows them to Seamus. On seeing the graphs, - Seamus quickly corrects himself, and says that $f$ has a local (and global) - maximum of $2$ at $0$, and that $g$ has a local (and global) minimum of $-2$ at $0$. - - \begin{figure}[!htb] - \begin{minipage}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=-32*(x-1)*(x+1)/(( x-2)^2*(x+2)^2);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-8,-6,...,8}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:-3.01]{f}; - \addplot[pccplot] expression[domain=-1.45:1.45]{f}; - \addplot[pccplot] expression[domain=3.01:10]{f}; - \addplot[soldot] coordinates{(-1,0)(1,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=f(x)$} - \label{rat:fig:minmax1} - \end{minipage}% - \hfill - \begin{minipage}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=32*(x-1)*(x+1)/(( x-2)^2*(x+2)^2);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-8,-6,...,8}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:-3.01]{f}; - \addplot[pccplot] expression[domain=-1.45:1.45]{f}; - \addplot[pccplot] expression[domain=3.01:10]{f}; - \addplot[soldot] coordinates{(-1,0)(1,0)}; - \end{axis} - \end{tikzpicture} - \caption{$y=g(x)$} - \label{rat:fig:minmax2} - \end{minipage}% - \end{figure} - - Seamus also notes that (in its domain) the function $f$ is always concave down, and - that (in its domain) the function $g$ is always concave up. Furthermore, Trang - observes that each function behaves like $\frac{1}{x^2}$ around each of its vertical - asymptotes, because each linear factor in the denominator is raised to the power $2$. - - \pccname{Oscar} stops by and reminds both students about the long-run behavior; according - to \cref{rat:def:longrun} since the degree of the denominator is greater than the - degree of the numerator (in both functions), each function has a horizontal asymptote - at $y=0$. - \end{pccexample} - - - \investigation*{} - %=================================== - % Author: Pettit/Hughes - % Date: March 2012 - %=================================== - \begin{problem}[The spaghetti incident] - The same Queen from \vref{exp:prob:queenschessboard} has recovered from - the rice experiments, and has called her loyal jester for another challenge. - - The jester has an $11-$inch piece of uncooked spaghetti that he puts on a table; - he uses a book to cover $\unit[1]{inch}$ of it so that - $\unit[10]{inches}$ hang over the edge. The jester then produces a box of $\unit{mg}$ - weights that can be hung from the spaghetti. - - The jester says it will take $\unit[y]{mg}$ to break the spaghetti when hung - $\unit[x]{inches}$ from the edge, according to the rule $y=\frac{100}{x}$. - \begin{margintable} - \centering - \captionof{table}{} - \label{rat:tab:spaghetti} - \begin{tabular}{cc} - \beforeheading - \heading{$x$} & \heading{$y$} \\ - \afterheading - $1$ & \\\normalline - $2$ & \\\normalline - $3$ & \\\normalline - $4$ & \\\normalline - $5$ & \\\normalline - $6$ & \\\normalline - $7$ & \\\normalline - $8$ & \\\normalline - $9$ & \\\normalline - $10$ & \\\lastline - \end{tabular} - \end{margintable} - \begin{subproblem}\label{rat:prob:spaggt1} - Help the Queen complete \cref{rat:tab:spaghetti}, and use $2$ digits after the decimal - where appropriate. - \begin{shortsolution} - \begin{tabular}[t]{ld{2}} - \beforeheading - \heading{$x$} & \heading{$y$} \\ - \afterheading - $1$ & 100 \\\normalline - $2$ & 50 \\\normalline - $3$ & 33.33 \\\normalline - $4$ & 25 \\\normalline - $5$ & 20 \\\normalline - $6$ & 16.67 \\\normalline - $7$ & 14.29 \\\normalline - $8$ & 12.50 \\\normalline - $9$ & 11.11 \\\normalline - $10$ & 10 \\\lastline - \end{tabular} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - What do you notice about the number of $\unit{mg}$ that it takes to break - the spaghetti as $x$ increases? - \begin{shortsolution} - It seems that the number of $\unit{mg}$ that it takes to break the spaghetti decreases - as $x$ increases. - \end{shortsolution} - \end{subproblem} - \begin{subproblem}\label{rat:prob:spaglt1} - The Queen wonders what happens when $x$ gets very small| help the Queen construct - a table of values for $x$ and $y$ when $x=0.0001, 0.001, 0.01, 0.1, 0.5, 1$. - \begin{shortsolution} - \begin{tabular}[t]{d{2}l} - \beforeheading - \heading{$x$} & \heading{$y$} \\ - \afterheading - 0.0001 & $1000000$ \\\normalline - 0.001 & $100000$ \\\normalline - 0.01 & $10000$ \\\normalline - 0.1 & $1000$ \\\normalline - 0.5 & $200$ \\\normalline - 1 & $100$ \\\lastline - \end{tabular} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - What do you notice about the number of $\unit{mg}$ that it takes to break the spaghetti - as $x\rightarrow 0$? Would it ever make sense to let $x=0$? - \begin{shortsolution} - The number of $\unit{mg}$ required to break the spaghetti increases as $x\rightarrow 0$. - We can not allow $x$ to be $0$, as we can not divide by $0$, and we can not - be $0$ inches from the edge of the table. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Plot your results from \cref{rat:prob:spaggt1,rat:prob:spaglt1} on the same graph, - and join the points using a smooth curve| set the maximum value of $y$ as $200$, and - note that this necessarily means that you will not be able to plot all of the points. - \begin{shortsolution} - The graph of $y=\frac{100}{x}$ is shown below. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-2,xmax=11, - ymin=-20,ymax=200, - xtick={2,4,...,10}, - ytick={20,40,...,180}, - grid=major, - width=\solutionfigurewidth, - ] - \addplot+[-] expression[domain=0.5:10]{100/x}; - \addplot[soldot] coordinates{(0.5,200)(1,100)(2,50)(3,33.33) - (4,25)(5,20)(16.67)(7,14.29)(8,12.50)(9,11.11)(10,10)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Using your graph, observe what happens to $y$ as $x$ increases. If we could somehow - construct a piece of uncooked spaghetti that was $\unit[101]{inches}$ long, how many - $\unit{mg}$ would it take to break the spaghetti? - \begin{shortsolution} - As $x$ increases, $y\rightarrow 0$. If we could construct a piece of spaghetti - $\unit[101]{inches}$ long, it would only take $\unit[1]{mg}$ to break it $\left(\frac{100}{100}=1\right)$. Of course, - the weight of spaghetti would probably cause it to break without the weight. - \end{shortsolution} - \end{subproblem} - The Queen looks forward to more food-related investigations from her jester. - \end{problem} - - - - %=================================== - % Author: Adams (Hughes) - % Date: March 2012 - %=================================== - \begin{problem}[Debt Amortization] - To amortize a debt means to pay it off in a given length of time using - equal periodic payments. The payments include interest on the unpaid - balance. The following formula gives the monthly payment, $M$, in dollars - that is necessary to amortize a debt of $P$ dollars in $n$ months - at a monthly interest rate of $i$ - \[ - M=\frac{P\cdot i}{1-(1+i)^{-n}} - \] - Use this formula in each of the following problems. - \begin{subproblem} - What monthly payments are necessary on a credit card debt of \$2000 at - $\unit[1.5]{\%}$ monthly if you want to pay off the debt in $2$ years? - In one year? How much money will you save by paying off the debt in the - shorter amount of time? - \begin{shortsolution} - Paying off the debt in $2$ years, we use - \begin{align*} - M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-24}} \\ - & \approx 99.85 - \end{align*} - The monthly payments are \$99.85. - - Paying off the debt in $1$ year, we use - \begin{align*} - M & = \frac{2000\cdot 0.015}{1-(1+0.015)^{-12}} \\ - & \approx 183.36 - \end{align*} - The monthly payments are \$183.36 - - In the $2$-year model we would pay a total of $\$99.85\cdot 12=\$2396.40$. In the - $1$-year model we would pay a total of $\$183.36\cdot 12=\$2200.32$. We would therefore - save $\$196.08$ if we went with the $1$-year model instead of the $2$-year model. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - To purchase a home, a family needs a loan of \$300,000 at $\unit[5.2]{\%}$ - annual interest. Compare a $20$ year loan to a $30$ year loan and make - a recommendation for the family. - (Note: when given an annual interest rate, it is a common business practice to divide by - $12$ to get a monthly rate.) - \begin{shortsolution} - For the $20$-year loan we use - \begin{align*} - M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 20}} \\ - & \approx 2013.16 - \end{align*} - The monthly payments are \$2013.16. - - For the $30$-year loan we use - \begin{align*} - M & = \frac{300000\cdot \frac{0.052}{12}}{1-\left( 1+\frac{0.052}{12} \right)^{-12\cdot 30}} \\ - & \approx 1647.33 - \end{align*} - The monthly payments are \$1647.33. - - The total amount paid during the $20$-year loan is $\$2013.16\cdot 12\cdot 20=\$483,158.40$. - The total amount paid during the $30$-year loan is $\$1647.33\cdot 12\cdot 30=\$593,038.80$. - - Recommendation: if you can afford the payments, choose the $20$-year loan. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - \pccname{Ellen} wants to make monthly payments of \$100 to pay off a debt of \$3000 - at \unit[12]{\%} annual interest. How long will it take her to pay off the - debt? - \begin{shortsolution} - We are given $M=100$, $P=3000$, $i=0.01$, and we need to find $n$ - in the equation - \[ - 100 = \frac{3000\cdot 0.01}{1-(1+0.01)^{-n}} - \] - Using logarithms, we find that $n\approx 36$. It will take - Ellen about $3$ years to pay off the debt. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - \pccname{Jake} is going to buy a new car. He puts \$2000 down and wants to finance the - remaining \$14,000. The dealer will offer him \unit[4]{\%} annual interest for - $5$ years, or a \$2000 - rebate which he can use to reduce the amount of the loan and \unit[8]{\%} - annual interest for 5 years. Which should he choose? - \begin{shortsolution} - \begin{description} - \item[Option 1:] $\unit[4]{\%}$ annual interest for $5$ years on \$14,000. - This means that the monthly payments will be calculated using - \begin{align*} - M & = \frac{14000\cdot \frac{0.04}{12}}{1-\left( 1+\frac{0.04}{12} \right)^{-12\cdot 5}} \\ - & \approx 257.83 - \end{align*} - The monthly payments will be $\$257.83$. The total amount paid will be - $\$257.83\cdot 5\cdot 12=\$15,469.80$, of which $\$1469.80$ is interest. - \item[Option 2:] $\unit[8]{\%}$ annual interest for $5$ years on \$12,000. - This means that the monthly payments will be calculated using - \begin{align*} - M & = \frac{12000\cdot \frac{0.08}{12}}{1-\left( 1+\frac{0.08}{12} \right)^{-12\cdot 5}} \\ - & \approx 243.32 - \end{align*} - The monthly payments will be $\$243.32$. The total amount paid - will be $\$243.32\cdot 5\cdot 12 =\$14,599.20$, of which $\$2599.2$ is - interest. - \end{description} - Jake should choose option 1 to minimize the amount of interest - he has to pay. - \end{shortsolution} - \end{subproblem} - \end{problem} - - \begin{exercises} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Rational or not] - Decide if each of the following functions are rational or not. If - they are rational, state their domain. - \begin{multicols}{3} - \begin{subproblem} - $r(x)=\dfrac{3}{x}$ - \begin{shortsolution} - $r$ is rational; the domain of $r$ is $(-\infty,0)\cup (0,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $s(y)=\dfrac{y}{6}$ - \begin{shortsolution} - $s$ is not rational ($s$ is linear). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $t(z)=\dfrac{4-x}{7-8z}$ - \begin{shortsolution} - $t$ is rational; the domain of $t$ is $\left( -\infty,\dfrac{7}{8} \right)\cup \left( \dfrac{7}{8},\infty \right)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $u(w)=\dfrac{w^2}{(w-3)(w+4)}$ - \begin{shortsolution} - $u$ is rational; the domain of $w$ is $(-\infty,-4)\cup(-4,3)\cup(3,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $v(x)=\dfrac{4}{(x-2)^2}$ - \begin{shortsolution} - $v$ is rational; the domain of $v$ is $(-\infty,2)\cup(2,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $w(x)=\dfrac{9-x}{x+17}$ - \begin{shortsolution} - $w$ is rational; the domain of $w$ is $(-\infty,-17)\cup(-17,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $a(x)=x^2+4$ - \begin{shortsolution} - $a$ is not rational ($a$ is quadratic, or a polynomial of degree $2$). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $b(y)=3^y$ - \begin{shortsolution} - $b$ is not rational ($b$ is exponential). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $c(z)=\dfrac{z^2}{z^3}$ - \begin{shortsolution} - $c$ is rational; the domain of $c$ is $(-\infty,0)\cup (0,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $d(x)=x^2(x+3)(5x-7)$ - \begin{shortsolution} - $d$ is not rational ($d$ is a polynomial). - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $e(\alpha)=\dfrac{\alpha^2}{\alpha^2-1}$ - \begin{shortsolution} - $e$ is rational; the domain of $e$ is $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $f(\beta)=\dfrac{3}{4}$ - \begin{shortsolution} - $f$ is not rational ($f$ is constant). - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Function evaluation] - Let $r$ be the function that has formula - \[ - r(x)=\frac{(x-2)(x+3)}{(x+5)(x-7)} - \] - Evaluate each of the following (if possible); if the value is undefined, - then state so. - \begin{multicols}{4} - \begin{subproblem} - $r(0)$ - \begin{shortsolution} - $\begin{aligned}[t] - r(0) & =\frac{(0-2)(0+3)}{(0+5)(0-7)} \\ - & =\frac{-6}{-35} \\ - & =\frac{6}{35} - \end{aligned}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(1)$ - \begin{shortsolution} - $\begin{aligned}[t] - r(1) & =\frac{(1-2)(1+3)}{(1+5)(1-7)} \\ - & =\frac{-4}{-36} \\ - & =\frac{1}{9} - \end{aligned}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(2)$ - \begin{shortsolution} - $\begin{aligned}[t] - r(2) & =\frac{(2-2)(2+3)}{(2+5)(2-7)} \\ - & = \frac{0}{-50} \\ - & =0 - \end{aligned}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(4)$ - \begin{shortsolution} - $\begin{aligned}[t] - r(4) & =\frac{(4-2)(4+3)}{(4+5)(4-7)} \\ - & =\frac{14}{-27} \\ - & =-\frac{14}{27} - \end{aligned}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(7)$ - \begin{shortsolution} - $\begin{aligned}[t] - r(7) & =\frac{(7-2)(7+3)}{(7+5)(7-7)} \\ - & =\frac{50}{0} - \end{aligned}$ - - $r(7)$ is undefined. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(-3)$ - \begin{shortsolution} - $\begin{aligned}[t] - r(-3) & =\frac{(-3-2)(-3+3)}{(-3+5)(-3-7)} \\ - & =\frac{0}{-20} \\ - & =0 - \end{aligned}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(-5)$ - \begin{shortsolution} - $\begin{aligned}[t] - r(-5) & =\frac{(-5-2)(-5+3)}{(-5+5)(-5-7)} \\ - & =\frac{14}{0} - \end{aligned}$ - - $r(-5)$ is undefined. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r\left( \frac{1}{2} \right)$ - \begin{shortsolution} - $\begin{aligned}[t] - r\left( \frac{1}{2} \right) & = \frac{\left( \frac{1}{2}-2 \right)\left( \frac{1}{2}+3 \right)}{\left( \frac{1}{2}+5 \right)\left( \frac{1}{2}-7 \right)} \\ - & =\frac{-\frac{3}{2}\cdot\frac{7}{2}}{\frac{11}{2}\left( -\frac{13}{2} \right)} \\ - & =\frac{-\frac{21}{4}}{-\frac{143}{4}} \\ - & =\frac{37}{143} - \end{aligned}$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Holes or asymptotes?] - State the domain of each of the following rational functions. Identify - any holes or asymptotes. - \begin{multicols}{3} - \begin{subproblem} - $f(x)=\dfrac{12}{x-2}$ - \begin{shortsolution} - $f$ has a vertical asymptote at $2$; the domain of $f$ is $(-\infty,2)\cup (2,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $g(x)=\dfrac{x^2+x}{(x+1)(x-2)}$ - \begin{shortsolution} - $g$ has a vertical asymptote at $2$, and a hole at $-1$; the domain of $g$ is $(-\infty,-1)\cup(-1,2)\cup(2,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $h(x)=\dfrac{x^2+5x+4}{x^2+x-12}$ - \begin{shortsolution} - $h$ has a vertical asymptote at $3$, and a whole at $-4$; the domain of $h$ is $(-\infty,-4)\cup(-4,3)\cup(3,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $k(z)=\dfrac{z+2}{2z-3}$ - \begin{shortsolution} - $k$ has a vertical asymptote at $\dfrac{3}{2}$; the domain of $k$ is $\left( -\infty,\dfrac{3}{2} \right)\cup\left( \dfrac{3}{2},\infty \right)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $l(w)=\dfrac{w}{w^2+1}$ - \begin{shortsolution} - $l$ does not have any vertical asymptotes nor holes; the domain of $w$ is $(-\infty,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $m(t)=\dfrac{14}{13-t^2}$ - \begin{shortsolution} - $m$ has vertical asymptotes at $\pm\sqrt{13}$; the domain of $m$ is $(-\infty,\sqrt{13})\cup(-\sqrt{13},\sqrt{13})\cup(\sqrt{13},\infty)$. - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Find a formula from a graph] - Consider the rational functions graphed in \cref{rat:fig:findformula}. Find - the vertical asymptotes for each function, together with any zeros, and - give a possible formula for each. - \begin{shortsolution} - \begin{itemize} - \item \Vref{rat:fig:formula1}: possible formula is $r(x)=\dfrac{1}{x+5}$ - \item \Vref{rat:fig:formula2}: possible formula is $r(x)=\dfrac{(x+3)}{(x-5)}$ - \item \Vref{rat:fig:formula3}: possible formula is $r(x)=\dfrac{1}{(x-4)(x+3)}$. - \end{itemize} - \end{shortsolution} - \end{problem} - - \begin{figure}[!htb] - \begin{widepage} - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=1/(x+4);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-6,ymax=6, - xtick={-8,-6,...,8}, - minor ytick={-4,-3,...,4}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:-4.16667,samples=50]{f}; - \addplot[pccplot] expression[domain=-3.83333:10,samples=50]{f}; - \addplot[asymptote,domain=-6:6]({-4},{x}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:formula1} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=(x+3)/(x-5);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-6,ymax=6, - xtick={-8,-6,...,8}, - minor ytick={-4,-3,...,4}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:3.85714]{f}; - \addplot[pccplot] expression[domain=6.6:10]{f}; - \addplot[soldot] coordinates{(-3,0)}; - \addplot[asymptote,domain=-6:6]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{1}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:formula2} - \end{subfigure} - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture}[/pgf/declare function={f=1/((x-4)*(x+3));}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-3,ymax=3, - xtick={-8,-6,...,8}, - minor ytick={-4,-3,...,4}, - grid=both, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:-3.0473]{f}; - \addplot[pccplot] expression[domain=-2.95205:3.95205]{f}; - \addplot[pccplot] expression[domain=4.0473:10]{f}; - \addplot[asymptote,domain=-3:3]({-3},{x}); - \addplot[asymptote,domain=-3:3]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{0}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:formula3} - \end{subfigure} - \caption{} - \label{rat:fig:findformula} - \end{widepage} - \end{figure} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Find a formula from a description] - In each of the following problems, give a formula of a rational - function that has the listed properties. - \begin{subproblem} - Vertical asymptote at $2$. - \begin{shortsolution} - Possible option: $r(x)=\dfrac{1}{x-2}$. Note that we could multiply the - numerator or denominator by any real number and still have the desired properties. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Vertical asymptote at $5$. - \begin{shortsolution} - Possible option: $r(x)=\dfrac{1}{x-5}$. Note that we could multiply the - numerator or denominator by any real number and still have the desired properties. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Vertical asymptote at $-2$, and zero at $6$. - \begin{shortsolution} - Possible option: $r(x)=\dfrac{x-6}{x+2}$. Note that we could multiply the - numerator or denominator by any real number and still have the desired properties. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Zeros at $2$ and $-5$ and vertical asymptotes at $1$ and $-7$. - \begin{shortsolution} - Possible option: $r(x)=\dfrac{(x-2)(x+5)}{(x-1)(x+7)}$. Note that we could multiply the - numerator or denominator by any real number and still have the desired properties. - \end{shortsolution} - \end{subproblem} - \end{problem} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Given formula, find horizontal asymptotes] - Each of the following functions has a horizontal asymptote. Write the equation - of the horizontal asymptote for each function. - \begin{multicols}{3} - \begin{subproblem} - $f(x) = \dfrac{1}{x}$ - \begin{shortsolution} - $y=0$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $g(x) = \dfrac{2x+3}{x}$ - \begin{shortsolution} - $y=2$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $h(x) = \dfrac{x^2+2x}{x^2+3}$ - \begin{shortsolution} - $y=1$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $k(x) = \dfrac{x^2+7}{x}$ - \begin{shortsolution} - $y=1$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $l(x)=\dfrac{3x-2}{5x+8}$ - \begin{shortsolution} - $y=\dfrac{3}{5}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $m(x)=\dfrac{3x-2}{5x^2+8}$ - \begin{shortsolution} - $y=0$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $n(x)=\dfrac{(6x+1)(x-7)}{(11x-8)(x-5)}$ - \begin{shortsolution} - $y=\dfrac{6}{11}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $p(x)=\dfrac{19x^3}{5-x^4}$ - \begin{shortsolution} - $y=0$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $q(x)=\dfrac{14x^2+x}{1-7x^2}$ - \begin{shortsolution} - $y=-2$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{problem}[Given horizontal asymptotes, find formula] - In each of the following problems, give a formula for a function that - has the given horizontal asymptote. Note that there may be more than one option. - \begin{multicols}{4} - \begin{subproblem} - $y=7$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{7(x-2)}{x+1}$. Note that there - are other options, provided that the degree of the numerator is the same as the degree - of the denominator, and that the ratio of the leading - coefficients is $7$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=-1$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{5-x^2}{x^2+10}$. Note that there - are other options, provided that the degree of the numerator is the same as the degree - of the denominator, and that the ratio of the leading - coefficients is $10$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=53$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{53x^3}{x^3+4x^2-7}$. Note that there - are other options, provided that the degree of the numerator is the same as the degree - of the denominator, and that the ratio of the leading - coefficients is $53$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=-17$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{34(x+2)}{7-2x}$. Note that there - are other options, provided that the degree of the numerator is the same as the degree - of the denominator, and that the ratio of the leading - coefficients is $-17$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=\dfrac{3}{2}$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{3x+4}{2(x+1)}$. Note that there - are other options, provided that the degree of the numerator is the same as the degree - of the denominator, and that the ratio of the leading - coefficients is $\dfrac{3}{2}$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=0$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{4}{x}$. Note that there - are other options, provided that the degree of the numerator is less than the degree - of the denominator. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=-1$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{10x}{5-10x}$. Note that there - are other options, provided that the degree of the numerator is the same as the degree - of the denominator, and that the ratio of the leading - coefficients is $-1$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=2$ - \begin{shortsolution} - Possible option: $f(x)=\dfrac{8x-3}{4x+1}$. Note that there - are other options, provided that the degree of the numerator is the same as the degree - of the denominator, and that the ratio of the leading - coefficients is $2$. - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Find a formula from a description] - In each of the following problems, give a formula for a function that - has the prescribed properties. Note that there may be more than one option. - \begin{subproblem} - $f(x)\rightarrow 3$ as $x\rightarrow\pm\infty$. - \begin{shortsolution} - Possible option: $f(x)=\dfrac{3(x-2)}{x+7}$. Note that - the zero and asymptote of $f$ could be changed, and $f$ would still have the desired properties. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $r(x)\rightarrow -4$ as $x\rightarrow\pm\infty$. - \begin{shortsolution} - Possible option: $r(x)=\dfrac{-4(x-2)}{x+7}$. Note that - the zero and asymptote of $r$ could be changed, and $r$ would still have the desired properties. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $k(x)\rightarrow 2$ as $x\rightarrow\pm\infty$, and $k$ has vertical asymptotes at $-3$ and $5$. - \begin{shortsolution} - Possible option: $k(x)=\dfrac{2x^2}{(x+3)(x-5)}$. Note that the denominator - must have the given factors; the numerator could be any degree $2$ polynomial, provided the - leading coefficient is $2$. - \end{shortsolution} - \end{subproblem} - \end{problem} - - %=================================== - % Author: Hughes - % Date: Feb 2011 - %=================================== - \begin{problem} - Let $r$ be the rational function that has - \[ - r(x) = \frac{(x+2)(x-1)}{(x+3)(x-4)} - \] - Each of the following questions are in relation to this function. - \begin{subproblem} - What is the vertical intercept of this function? State your answer as an - ordered pair. \index{rational functions!vertical intercept} - \begin{shortsolution} - $\left(0,\frac{1}{6}\right)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem}\label{rat:prob:rational} - What values of $x$ make the denominator equal to $0$? - \begin{shortsolution} - $-3,4$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Use your answer to \cref{rat:prob:rational} to write the domain of the function in - both interval, and set builder notation. %\index{rational functions!domain}\index{domain!rational functions} - \begin{shortsolution} - Interval notation: $(-\infty,-3)\cup (-3,4)\cup (4,\infty)$. - Set builder: $\{x|x\ne -3, \mathrm{and}\, x\ne 4\}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - What are the vertical asymptotes of the function? State your answers in - the form $x=$ - \begin{shortsolution} - $x=-3$ and $x=4$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem}\label{rat:prob:zeroes} - What values of $x$ make the numerator equal to $0$? - \begin{shortsolution} - $-2,1$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Use your answer to \cref{rat:prob:zeroes} to write the horizontal intercepts of - $r$ as ordered pairs. - \begin{shortsolution} - $(-2,0)$ and $(1,0)$ - \end{shortsolution} - \end{subproblem} - \end{problem} - - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Holes] - \pccname{Josh} and \pccname{Pedro} are discussing the function - \[ - r(x)=\frac{x^2-1}{(x+3)(x-1)} - \] - \begin{subproblem} - What is the domain of $r$? - \begin{shortsolution} - The domain of $r$ is $(-\infty,-3)\cup(-3,1)\cup(1,\infty)$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Josh notices that the numerator can be factored- can you see how? - \begin{shortsolution} - $(x^2-1)=(x-1)(x+1)$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Pedro asks, `Doesn't that just mean that - \[ - r(x)=\frac{x+1}{x+3} - \] - for all values of $x$?' Josh says, `Nearly\ldots but not for all values of $x$'. - What does Josh mean? - \begin{shortsolution} - $r(x)=\dfrac{x+1}{x+3}$ provided that $x\ne -1$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Where does $r$ have vertical asymptotes, and where does it have holes? - \begin{shortsolution} - The function $r$ has a vertical asymptote at $-3$, and a hole at $1$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Sketch a graph of $r$. - \begin{shortsolution} - A graph of $r$ is shown below. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-8,-6,...,8}, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot] expression[domain=-10:-3.25]{(x+1)/(x+3)}; - \addplot[pccplot] expression[domain=-2.75:10]{(x+1)/(x+3)}; - \addplot[asymptote,domain=-10:10]({-3},{x}); - \addplot[holdot]coordinates{(1,0.5)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \end{problem} - - %=================================== - % Author: Hughes - % Date: July 2012 - %=================================== - \begin{problem}[Function algebra] - Let $r$ and $s$ be the rational functions that have formulas - \[ - r(x)=\frac{2-x}{x+3}, \qquad s(x)=\frac{x^2}{x-4} - \] - Evaluate each of the following (if possible). - \begin{multicols}{4} - \begin{subproblem} - $(r+s)(5)$ - \begin{shortsolution} - $\frac{197}{8}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(r-s)(3)$ - \begin{shortsolution} - $\frac{53}{6}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(r\cdot s)(4)$ - \begin{shortsolution} - Undefined. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $\left( \frac{r}{s} \right)(1)$ - \begin{shortsolution} - $-\frac{3}{4}$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - - - %=================================== - % Author: Hughes - % Date: July 2012 - %=================================== - \begin{problem}[Transformations: given the transformation, find the formula] - Let $r$ be the rational function that has formula. - \[ - r(x)=\frac{x+5}{2x-3} - \] - In each of the following problems apply the given transformation to the function $r$ and - write a formula for the transformed version of $r$. - \begin{multicols}{2} - \begin{subproblem} - Shift $r$ to the right by $3$ units. - \begin{shortsolution} - $r(x-3)=\frac{x+2}{2x-9}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Shift $r$ to the left by $4$ units. - \begin{shortsolution} - $r(x+4)=\frac{x+9}{2x+5}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Shift $r$ up by $\pi$ units. - \begin{shortsolution} - $r(x)+\pi=\frac{x+5}{2x-3}+\pi$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Shift $r$ down by $17$ units. - \begin{shortsolution} - $r(x)-17=\frac{x+5}{2x-3}-17$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Reflect $r$ over the horizontal axis. - \begin{shortsolution} - $-r(x)=-\frac{x+5}{2x-3}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Reflect $r$ over the vertical axis. - \begin{shortsolution} - $r(-x)=\frac{x-5}{2x+3}$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - - - %=================================== - % Author: Hughes - % Date: May 2011 - %=================================== - \begin{problem}[Find a formula from a table]\label{rat:prob:findformula} - \Crefrange{rat:tab:findformular}{rat:tab:findformulau} show values of rational functions $r$, $q$, $s$, - and $t$. Assume that any values marked with an X are undefined. - - \begin{table}[!htb] - \begin{widepage} - \centering - \caption{Tables for \cref{rat:prob:findformula}} - \label{rat:tab:findformula} - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=r(x)$} - \label{rat:tab:findformular} - \begin{tabular}{rr} - \beforeheading - $x$ & $y$ \\ \afterheading - $-4$ & $\nicefrac{7}{2}$ \\\normalline - $-3$ & $-18$ \\\normalline - $-2$ & X \\\normalline - $-1$ & $-4$ \\\normalline - $0$ & $\nicefrac{-3}{2}$ \\\normalline - $1$ & $\nicefrac{-2}{3}$ \\\normalline - $2$ & $\nicefrac{-1}{4}$ \\\normalline - $3$ & $0$ \\\normalline - $4$ & $\nicefrac{1}{6}$ \\\lastline - \end{tabular} - \end{subtable} - \hfill - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=s(x)$} - \label{rat:tab:findformulas} - \begin{tabular}{rr} - \beforeheading - $x$ & $y$ \\ \afterheading - $-4$ & $\nicefrac{-2}{21}$ \\\normalline - $-3$ & $\nicefrac{-1}{12}$ \\\normalline - $-2$ & $0$ \\\normalline - $-1$ & X \\\normalline - $0$ & $\nicefrac{-2}{3}$ \\\normalline - $1$ & $\nicefrac{-3}{4}$ \\\normalline - $2$ & $\nicefrac{-4}{3}$ \\\normalline - $3$ & X \\\normalline - $4$ & $\nicefrac{6}{5}$ \\\lastline - \end{tabular} - \end{subtable} - \hfill - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=t(x)$} - \label{rat:tab:findformulat} - \begin{tabular}{rr} - \beforeheading - $x$ & $y$ \\ \afterheading - $-4$ & $\nicefrac{3}{5}$ \\\normalline - $-3$ & $0$ \\\normalline - $-2$ & X \\\normalline - $-1$ & $3$ \\\normalline - $0$ & $3$ \\\normalline - $1$ & X \\\normalline - $2$ & $0$ \\\normalline - $3$ & $\nicefrac{3}{5}$ \\\normalline - $4$ & $\nicefrac{7}{9}$ \\\lastline - \end{tabular} - \end{subtable} - \hfill - \begin{subtable}{.2\textwidth} - \centering - \caption{$y=u(x)$} - \label{rat:tab:findformulau} - \begin{tabular}{rr} - \beforeheading - $x$ & $y$ \\ \afterheading - $-4$ & $\nicefrac{16}{7}$ \\\normalline - $-3$ & X \\\normalline - $-2$ & $-\nicefrac{4}{5}$ \\\normalline - $-1$ & $-\nicefrac{1}{8}$ \\\normalline - $0$ & $0$ \\\normalline - $1$ & $-\nicefrac{1}{8}$ \\\normalline - $2$ & $-\nicefrac{4}{5}$ \\\normalline - $3$ & X \\\normalline - $4$ & $\nicefrac{16}{7}$ \\\lastline - \end{tabular} - \end{subtable} - \end{widepage} - \end{table} - \begin{subproblem} - Given that the formula for $r(x)$ has the form $r(x)=\dfrac{x-A}{x-B}$, use \cref{rat:tab:findformular} - to find values of $A$ and $B$. - \begin{shortsolution} - $A=3$ and $B=-2$, so $r(x)=\dfrac{x-3}{x+2}$. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Check your formula by computing $r(x)$ at the values specified in the table. - \begin{shortsolution} - $\begin{aligned}[t] - r(-4) & = \frac{-4-3}{-4+2} \\ - & = \frac{7}{2} \\ - \end{aligned}$ - - $r(-3)=\ldots$ etc - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - The function $s$ in \cref{rat:tab:findformulas} has two vertical asymptotes and one zero. - Can you find a formula for $s(x)$? - \begin{shortsolution} - $s(x)=\dfrac{x+2}{(x-3)(x+1)}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Check your formula by computing $s(x)$ at the values specified in the table. - \begin{shortsolution} - $\begin{aligned}[t] - s(-4) & =\frac{-4+2}{(-4-3)(-4+1)} \\ - & =-\frac{2}{21} - \end{aligned}$ - - $s(-3)=\ldots$ etc - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Given that the formula for $t(x)$ has the form $t(x)=\dfrac{(x-A)(x-B)}{(x-C)(x-D)}$, use \cref{rat:tab:findformulat} to find the - values of $A$, $B$, $C$, and $D$; hence write a formula for $t(x)$. - \begin{shortsolution} - $t(x)=\dfrac{(x+3)(x-2)}{(x+2)(x+1)}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Given that the formula for $u(x)$ has the form $u(x)=\dfrac{(x-A)^2}{(x-B)(x-C)}$, use \cref{rat:tab:findformulau} to find the - values of $A$, $B$, and $C$; hence write a formula for $u(x)$. - \begin{shortsolution} - $u(x)=\dfrac{x^2}{(x+3)(x-3)}$ - \end{shortsolution} - \end{subproblem} - \end{problem} - \end{exercises} - -\section{Graphing rational functions (horizontal asymptotes)} - \reformatstepslist{R} % the steps list should be R1, R2, \ldots - We studied rational functions in the previous section, but were - not asked to graph them; in this section we will demonstrate the - steps to be followed in order to sketch graphs of the functions. - - Remember from \vref{rat:def:function} that rational functions have - the form - \[ - r(x)=\frac{p(x)}{q(x)} - \] - In this section we will restrict attention to the case when - \[ - \text{degree of }p\leq \text{degree of }q - \] - Note that this necessarily means that each function that we consider - in this section \emph{will have a horizontal asymptote} (see \vref{rat:def:longrun}). - The cases in which the degree of $p$ is greater than the degree of $q$ - is covered in the next section. - - Before we begin, it is important to remember the following: - \begin{itemize} - \item Our sketches will give a good representation of the overall - shape of the graph, but until we have the tools of calculus (from MTH 251) - we can not find local minimums, local maximums, and inflection points algebraically. This - means that we will make our best guess as to where these points are. - \item We will not concern ourselves too much with the vertical scale (because of - our previous point)| we will, however, mark the vertical intercept (assuming there is one), - and any horizontal asymptotes. - \end{itemize} - \begin{pccspecialcomment}[Steps to follow when sketching rational functions]\label{rat:def:stepsforsketch} - \begin{steps} - \item \label{rat:step:first} Find all vertical asymptotes and holes, and mark them on the - graph using dashed vertical lines and open circles $\circ$ respectively. - \item Find any intercepts, and mark them using solid circles $\bullet$; - determine if the curve cuts the axis, or bounces off it at each zero. - \item Determine the behavior of the function around each asymptote| does - it behave like $\frac{1}{x}$ or $\frac{1}{x^2}$? - \item \label{rat:step:penultimate} Determine the long-run behavior of the function, and mark the horizontal - asymptote using a dashed horizontal line. - \item \label{rat:step:last} Deduce the overall shape of the curve, and sketch it. If there isn't - enough information from the previous steps, then construct a table of values - including sample points from each branch. - \end{steps} - Remember that until we have the tools of calculus, we won't be able to - find the exact coordinates of local minimums, local maximums, and points - of inflection. - \end{pccspecialcomment} - - The examples that follow show how \crefrange{rat:step:first}{rat:step:last} can be - applied to a variety of different rational functions. - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{rat:ex:1overxminus2p2} - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $r$ - that has formula - \[ - r(x)=\frac{1}{x-2} - \] - \begin{pccsolution} - \begin{steps} - \item $r$ has a vertical asymptote at $2$; $r$ does not have any holes. The curve of - $r$ will have $2$ branches. - \item $r$ does not have any zeros since the numerator is never equal to $0$. The - vertical intercept of $r$ is $\left( 0,-\frac{1}{2} \right)$. - \item $r$ behaves like $\frac{1}{x}$ around its vertical asymptote since $(x-2)$ - is raised to the power $1$. - \item Since the degree of the numerator is less than the degree of the denominator, - according to \vref{rat:def:longrun} the horizontal asymptote of $r$ has equation $y=0$. - \item We put the details we have obtained so far on \cref{rat:fig:1overxminus2p1}. Notice - that there is only one way to complete the graph, which we have done in \cref{rat:fig:1overxminus2p2}. - \end{steps} - \end{pccsolution} - \end{pccexample} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-5,ymax=5, - width=\textwidth, - ] - \addplot[asymptote,domain=-5:5]({2},{x}); - \addplot[asymptote,domain=-5:5]({x},{0}); - \addplot[soldot] coordinates{(0,-0.5)}node[axisnode,anchor=north east]{$\left( 0,-\frac{1}{2} \right)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:1overxminus2p1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=1/(x-2);}] - \begin{axis}[ - xmin=-5,xmax=5, - ymin=-5,ymax=5, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-5:1.8,samples=50]{f}; - \addplot[pccplot] expression[domain=2.2:5]{f}; - \addplot[asymptote,domain=-5:5]({2},{x}); - \addplot[asymptote,domain=-5:5]({x},{0}); - \addplot[soldot] coordinates{(0,-0.5)}node[axisnode,anchor=north east]{$\left( 0,-\frac{1}{2} \right)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:1overxminus2p2} - \end{subfigure}% - \caption{$y=\dfrac{1}{x-2}$} - \end{figure} - - The function $r$ in \cref{rat:ex:1overxminus2p2} has a horizontal asymptote which has equation $y=0$. - This asymptote lies on the horizontal axis, and you might (understandably) find it hard - to distinguish between the two lines (\cref{rat:fig:1overxminus2p2}). When faced - with such a situation, it is perfectly acceptable to draw the horizontal axis - as a dashed line| just make sure to label it correctly. We will demonstrate this - in the next example. - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{rat:ex:1overxp1} - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $v$ - that has formula - \[ - v(x)=\frac{10}{x} - \] - \begin{pccsolution} - \begin{steps} - \item $v$ has a vertical asymptote at $0$. $v$ does not have - any holes. The curve of $v$ will have $2$ branches. - \item $v$ does not have any zeros (since $10\ne 0$). Furthermore, $v$ - does not have a vertical intercept since $v(0)$ is undefined. - \item $v$ behaves like $\frac{1}{x}$ around its vertical asymptote. - \item $v$ has a horizontal asymptote with equation $y=0$. - \item We put the details we have obtained so far in \cref{rat:fig:1overxp1}. - We do not have enough information to sketch $v$ yet (because $v$ does - not have any intercepts), so let's pick a sample - point in either of the $2$ branches| it doesn't matter where our sample point - is, because we know what the overall shape will be. Let's compute $v(2)$ - \begin{align*} - v(2) & =\dfrac{10}{2} \\ - & = 5 - \end{align*} - We therefore mark the point $(2,5)$ on \cref{rat:fig:1overxp2}, and then complete the sketch using - the details we found in the previous steps. - \end{steps} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-5,5}, - ytick={-5,5}, - axis line style={color=white}, - width=\textwidth, - ] - \addplot[asymptote,<->,domain=-10:10]({0},{x}); - \addplot[asymptote,<->,domain=-10:10]({x},{0}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:1overxp1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=10/x;}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-5,5}, - ytick={-5,5}, - axis line style={color=white}, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:-1]{f}; - \addplot[pccplot] expression[domain=1:10]{f}; - \addplot[soldot] coordinates{(2,5)}node[axisnode,anchor=south west]{$(2,5)$}; - \addplot[asymptote,<->,domain=-10:10]({0},{x}); - \addplot[asymptote,<->,domain=-10:10]({x},{0}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:1overxp2} - \end{subfigure}% - \caption{$y=\dfrac{10}{x}$} - \end{figure} - \end{pccsolution} - \end{pccexample} - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{rat:ex:asympandholep1} - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $u$ - that has formula - \[ - u(x)=\frac{-4(x^2-9)}{x^2-8x+15} - \] - \begin{pccsolution} - \begin{steps} - \item We begin by factoring both the numerator and denominator of $u$ to help - us find any vertical asymptotes or holes - \begin{align*} - u(x) & =\frac{-4(x^2-9)}{x^2-8x+15} \\ - & =\frac{-4(x+3)(x-3)}{(x-5)(x-3)} \\ - & =\frac{-4(x+3)}{x-5} - \end{align*} - provided that $x\ne 3$. Therefore $u$ has a vertical asymptote at $5$ and - a hole at $3$. The curve of $u$ has $2$ branches. - \item $u$ has a simple zero at $-3$. The vertical intercept of $u$ is $\left( 0,\frac{12}{5} \right)$. - \item $u$ behaves like $\frac{1}{x}$ around its vertical asymptote at $4$. - \item Using \vref{rat:def:longrun} the equation of the horizontal asymptote of $u$ is $y=-4$. - \item We put the details we have obtained so far on \cref{rat:fig:1overxminus2p1}. Notice - that there is only one way to complete the graph, which we have done in \cref{rat:fig:1overxminus2p2}. - \end{steps} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-20,ymax=20, - xtick={-8,-6,...,8}, - ytick={-10,10}, - width=\textwidth, - ] - \addplot[asymptote,domain=-20:20]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{-4}); - \addplot[soldot] coordinates{(-3,0)(0,2.4)}node[axisnode,anchor=south east]{$\left( 0,\frac{12}{5} \right)$}; - \addplot[holdot] coordinates{(3,12)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:asympandholep1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=-4*(x+3)/(x-5);}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-20,ymax=20, - xtick={-8,-6,...,8}, - ytick={-10,10}, - width=\textwidth, - ] - \addplot[pccplot] expression[domain=-10:3.6666,samples=50]{f}; - \addplot[pccplot] expression[domain=7:10]{f}; - \addplot[asymptote,domain=-20:20]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{-4}); - \addplot[soldot] coordinates{(-3,0)(0,2.4)}node[axisnode,anchor=south east]{$\left( 0,\frac{12}{5} \right)$}; - \addplot[holdot] coordinates{(3,12)}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:asympandholep2} - \end{subfigure}% - \caption{$y=\dfrac{-4(x+3)}{x-5}$} - \end{figure} - \end{pccsolution} - \end{pccexample} - - \Cref{rat:ex:1overxminus2p2,rat:ex:1overxp1,rat:ex:asympandholep1} have focused on functions - that only have one vertical asymptote; the remaining examples in this section - concern functions that have more than one vertical asymptote. We will demonstrate - that \crefrange{rat:step:first}{rat:step:last} still apply. - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{rat:ex:sketchtwoasymp} - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $w$ - that has formula - \[ - w(x)=\frac{2(x+3)(x-5)}{(x+5)(x-4)} - \] - \begin{pccsolution} - \begin{steps} - \item $w$ has vertical asymptotes at $-5$ and $4$. $w$ does not have - any holes. The curve of $w$ will have $3$ branches. - \item $w$ has simple zeros at $-3$ and $5$. The vertical intercept of $w$ - is $\left( 0,\frac{3}{2} \right)$. - \item $w$ behaves like $\frac{1}{x}$ around both of its vertical - asymptotes. - \item The degree of the numerator of $w$ is $2$ and the degree of the - denominator of $w$ is also $2$. Using the ratio of the leading coefficients - of the numerator and denominator, we say that $w$ has a horizontal - asymptote with equation $y=\frac{2}{1}=2$. - \item We put the details we have obtained so far on \cref{rat:fig:sketchtwoasymptp1}. - - The function $w$ is a little more complicated than the functions that - we have considered in the previous examples because the curve has $3$ - branches. When graphing such functions, it is generally a good idea to start with the branch - for which you have the most information| in this case, that is the \emph{middle} branch - on the interval $(-5,4)$. - - Once we have drawn the middle branch, there is only one way to complete the graph - (because of our observations about the behavior of $w$ around its vertical asymptotes), - which we have done in \cref{rat:fig:sketchtwoasymptp2}. - \end{steps} - \end{pccsolution} - \end{pccexample} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{2}); - \addplot[soldot] coordinates{(-3,0)(5,0)}; - \addplot[soldot] coordinates{(0,1.5)}node[axisnode,anchor=north west]{$\left( 0,\frac{3}{2} \right)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:sketchtwoasymptp1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=2*(x+3)*(x-5)/( (x+5)*(x-4));}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{2}); - \addplot[soldot] coordinates{(-3,0)(5,0)}; - \addplot[soldot] coordinates{(0,1.5)}node[axisnode,anchor=north west]{$\left( 0,\frac{3}{2} \right)$}; - \addplot[pccplot] expression[domain=-10:-5.56708]{f}; - \addplot[pccplot] expression[domain=-4.63511:3.81708]{f}; - \addplot[pccplot] expression[domain=4.13511:10]{f}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:sketchtwoasymptp2} - \end{subfigure}% - \caption{$y=\dfrac{2(x+3)(x-5)}{(x+5)(x-4)}$} - \end{figure} - - The rational functions that we have considered so far have had simple - factors in the denominator; each function has behaved like $\frac{1}{x}$ - around each of its vertical asymptotes. \Cref{rat:ex:2asympnozeros,rat:ex:2squaredasymp} - consider functions that have a repeated factor in the denominator. - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{rat:ex:2asympnozeros} - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $f$ - that has formula - \[ - f(x)=\frac{100}{(x+5)(x-4)^2} - \] - \begin{pccsolution} - \begin{steps} - \item $f$ has vertical asymptotes at $-5$ and $4$. $f$ does not have - any holes. The curve of $f$ will have $3$ branches. - \item $f$ does not have any zeros (since $100\ne 0$). The vertical intercept of $f$ - is $\left( 0,\frac{5}{4} \right)$. - \item $f$ behaves like $\frac{1}{x}$ around $-5$ and behaves like $\frac{1}{x^2}$ - around $4$. - \item The degree of the numerator of $f$ is $0$ and the degree of the - denominator of $f$ is $2$. $f$ has a horizontal asymptote with - equation $y=0$. - \item We put the details we have obtained so far on \cref{rat:fig:2asympnozerosp1}. - - The function $f$ is similar to the function $w$ that we considered in \cref{rat:ex:sketchtwoasymp}| - it has two vertical asymptotes and $3$ branches, but in contrast to $w$ it does not have any zeros. - - We sketch $f$ in \cref{rat:fig:2asympnozerosp2}, using the middle branch as our guide - because we have the most information about the function on the interval $(-5,4)$. - - Once we have drawn the middle branch, there is only one way to complete the graph - because of our observations about the behavior of $f$ around its vertical asymptotes (it behaves like $\frac{1}{x}$), - which we have done in \cref{rat:fig:2asympnozerosp2}. - - Note that we are not yet able to find the local minimum of $f$ algebraically on the interval $(-5,4)$, - so we make a reasonable guess as to where it is| we can be confident that it is above the horizontal axis - since $f$ has no zeros. You may think that this is unsatisfactory, but once we have the tools of calculus, we will - be able to find local minimums more precisely. - \end{steps} - \end{pccsolution} - \end{pccexample} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{0}); - \addplot[soldot] coordinates{(0,1.25)}node[axisnode,anchor=south east]{$\left( 0,\frac{5}{4} \right)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:2asympnozerosp1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=100/( (x+5)*(x-4)^2);}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{0}); - \addplot[soldot] coordinates{(0,1.25)}node[axisnode,anchor=south east]{$\left( 0,\frac{5}{4} \right)$}; - \addplot[pccplot] expression[domain=-10:-5.12022]{f}; - \addplot[pccplot] expression[domain=-4.87298:2.87298,samples=50]{f}; - \addplot[pccplot] expression[domain=5:10]{f}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:2asympnozerosp2} - \end{subfigure}% - \caption{$y=\dfrac{100}{(x+5)(x-4)^2}$} - \end{figure} - - %=================================== - % Author: Hughes - % Date: May 2012 - %=================================== - \begin{pccexample}\label{rat:ex:2squaredasymp} - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $g$ - that has formula - \[ - g(x)=\frac{50(2-x)}{(x+3)^2(x-5)^2} - \] - \begin{pccsolution} - \begin{steps} - \item $g$ has vertical asymptotes at $-3$ and $5$. $g$ does - not have any holes. The curve of $g$ will have $3$ branches. - \item $g$ has a simple zero at $2$. The vertical intercept of $g$ is - $\left( 0,\frac{4}{9} \right)$. - \item $g$ behaves like $\frac{1}{x^2}$ around both of its - vertical asymptotes. - \item The degree of the numerator of $g$ is $1$ and the degree of the denominator - of $g$ is $4$. Using \vref{rat:def:longrun}, we calculate that - the horizontal asymptote of $g$ has equation $y=0$. - \item The details that we have found so far have been drawn in - \cref{rat:fig:2squaredasymp1}. The function $g$ is similar to the functions - we considered in \cref{rat:ex:sketchtwoasymp,rat:ex:2asympnozeros} because - it has $2$ vertical asymptotes and $3$ branches. - - We sketch $g$ using the middle branch as our guide because we have the most information - about $g$ on the interval $(-3,5)$. Note that there is no other way to draw this branch - without introducing other zeros which $g$ does not have. - - Once we have drawn the middle branch, there is only one way to complete the graph - because of our observations about the behavior of $g$ around its vertical asymptotes| it - behaves like $\frac{1}{x^2}$. - - \end{steps} - \end{pccsolution} - \end{pccexample} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-3},{x}); - \addplot[asymptote,domain=-10:10]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{0}); - \addplot[soldot] coordinates{(2,0)(0,4/9)}node[axisnode,anchor=south west]{$\left( 0,\frac{4}{9} \right)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:2squaredasymp1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=50*(2-x)/( (x+3)^2*(x-5)^2);}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - ytick={-5,5}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-3},{x}); - \addplot[asymptote,domain=-10:10]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{0}); - \addplot[soldot] coordinates{(2,0)(0,4/9)}node[axisnode,anchor=south west]{$\left( 0,\frac{4}{9} \right)$}; - \addplot[pccplot] expression[domain=-10:-3.61504]{f}; - \addplot[pccplot] expression[domain=-2.3657:4.52773]{f}; - \addplot[pccplot] expression[domain=5.49205:10]{f}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:2squaredasymp2} - \end{subfigure}% - \caption{$y=\dfrac{50(2-x)}{(x+3)^2(x-5)^2}$} - \end{figure} - - Each of the rational functions that we have considered so far has had either - a \emph{simple} zero, or no zeros at all. Remember from our work on polynomial - functions, and particularly \vref{poly:def:multzero}, that a \emph{repeated} zero - corresponds to the curve of the function behaving differently at the zero - when compared to how the curve behaves at a simple zero. \Cref{rat:ex:doublezero} details a - function that has a non-simple zero. - - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{pccexample}\label{rat:ex:doublezero} - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of the function $g$ - that has formula - \[ - h(x)=\frac{(x-3)^2}{(x+4)(x-6)} - \] - \begin{pccsolution} - \begin{steps} - \item $h$ has vertical asymptotes at $-4$ and $6$. $h$ does - not have any holes. The curve of $h$ will have $3$ branches. - \item $h$ has a zero at $3$ that has \emph{multiplicity $2$}. - The vertical intercept of $h$ is - $\left( 0,-\frac{3}{8} \right)$. - \item $h$ behaves like $\frac{1}{x}$ around both of its - vertical asymptotes. - \item The degree of the numerator of $h$ is $2$ and the degree of the denominator - of $h$ is $2$. Using \vref{rat:def:longrun}, we calculate that - the horizontal asymptote of $h$ has equation $y=1$. - \item The details that we have found so far have been drawn in - \cref{rat:fig:doublezerop1}. The function $h$ is different - from the functions that we have considered in previous examples because - of the multiplicity of the zero at $3$. - - We sketch $h$ using the middle branch as our guide because we have the most information - about $h$ on the interval $(-4,6)$. Note that there is no other way to draw this branch - without introducing other zeros which $h$ does not have| also note how - the curve bounces off the horizontal axis at $3$. - - Once we have drawn the middle branch, there is only one way to complete the graph - because of our observations about the behavior of $h$ around its vertical asymptotes| it - behaves like $\frac{1}{x}$. - - \end{steps} - \end{pccsolution} - \end{pccexample} - - \begin{figure}[!htbp] - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-5,ymax=5, - xtick={-8,-6,...,8}, - ytick={-3,3}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-4},{x}); - \addplot[asymptote,domain=-10:10]({6},{x}); - \addplot[asymptote,domain=-10:10]({x},{1}); - \addplot[soldot] coordinates{(3,0)(0,-3/8)}node[axisnode,anchor=north west]{$\left( 0,-\frac{3}{8} \right)$}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:doublezerop1} - \end{subfigure}% - \hfill - \begin{subfigure}{.45\textwidth} - \begin{tikzpicture}[/pgf/declare function={f=(x-3)^2/((x+4)*(x-6));}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-5,ymax=5, - xtick={-8,-6,...,8}, - ytick={-3,3}, - width=\textwidth, - ] - \addplot[asymptote,domain=-10:10]({-4},{x}); - \addplot[asymptote,domain=-10:10]({6},{x}); - \addplot[asymptote,domain=-10:10]({x},{1}); - \addplot[soldot] coordinates{(3,0)(0,-3/8)}node[axisnode,anchor=north west]{$\left( 0,-\frac{3}{8} \right)$}; - \addplot[pccplot] expression[domain=-10:-5.20088]{f}; - \addplot[pccplot] expression[domain=-3.16975:5.83642,samples=50]{f}; - \addplot[pccplot] expression[domain=6.20088:10]{f}; - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:doublezerop2} - \end{subfigure}% - \caption{$y=\dfrac{(x-3)^2}{(x+4)(x-6)}$} - \end{figure} - \begin{exercises} - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{problem}[\Cref{rat:step:last}]\label{rat:prob:deduce} - \pccname{Katie} is working on graphing rational functions. She - has been concentrating on functions that have the form - \begin{equation}\label{rat:eq:deducecurve} - f(x)=\frac{a(x-b)}{x-c} - \end{equation} - Katie notes that functions with this type of formula have a zero - at $b$, and a vertical asymptote at $c$. Furthermore, these functions - behave like $\frac{1}{x}$ around their vertical asymptote, and the - curve of each function will have $2$ branches. - - Katie has been working with $3$ functions that have the form given - in \cref{rat:eq:deducecurve}, and has followed \crefrange{rat:step:first}{rat:step:penultimate}; - her results are shown in \cref{rat:fig:deducecurve}. There is just one - more thing to do to complete the graphs| follow \cref{rat:step:last}. - Help Katie finish each graph by deducing the curve of each function. - \begin{shortsolution} - \Vref{rat:fig:deducecurve1} - - \begin{tikzpicture}[/pgf/declare function={f=3*(x+4)/(x+5);}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\solutionfigurewidth, - ] - \addplot[soldot] coordinates{(-4,0)(0,12/5)}; - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({x},{3}); - \addplot[pccplot] expression[domain=-10:-5.42857]{f}; - \addplot[pccplot] expression[domain=-4.76923:10,samples=50]{f}; - \end{axis} - \end{tikzpicture} - - \Vref{rat:fig:deducecurve2} - - \begin{tikzpicture}[/pgf/declare function={f=-3*(x-2)/(x-4);}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\solutionfigurewidth, - ] - \addplot[soldot] coordinates{(2,0)(0,-3/2)}; - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{-3}); - \addplot[pccplot] expression[domain=-10:3.53846,samples=50]{f}; - \addplot[pccplot] expression[domain=4.85714:10]{f}; - \end{axis} - \end{tikzpicture} - - \Vref{rat:fig:deducecurve4} - - \begin{tikzpicture}[/pgf/declare function={f=2*(x-6)/(x-4);}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\solutionfigurewidth, - ] - \addplot[soldot] coordinates{(6,0)(0,3)}; - \addplot[asymptote,domain=-10:10]({x},{2}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[pccplot] expression[domain=-10:3.5,samples=50]{f}; - \addplot[pccplot] expression[domain=4.3333:10]{f}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{problem} - - \begin{figure}[!htb] - \begin{widepage} - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-4,0)(0,12/5)}; - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({x},{3}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:deducecurve1} - \end{subfigure}% - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(2,0)(0,-3/2)}; - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{-3}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:deducecurve2} - \end{subfigure}% - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(6,0)(0,3)}; - \addplot[asymptote,domain=-10:10]({x},{2}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:deducecurve4} - \end{subfigure} - \caption{Graphs for \cref{rat:prob:deduce}} - \label{rat:fig:deducecurve} - \end{widepage} - \end{figure} - - %=================================== - % Author: Hughes - % Date: June 2012 - %=================================== - \begin{problem}[\Cref{rat:step:last} for more complicated rational functions]\label{rat:prob:deducehard} - \pccname{David} is also working on graphing rational functions, and - has been concentrating on functions that have the form - \[ - r(x)=\frac{a(x-b)(x-c)}{(x-d)(x-e)} - \] - David notices that functions with this type of formula have simple zeros - at $b$ and $c$, and vertical asymptotes at $d$ and $e$. Furthermore, - these functions behave like $\frac{1}{x}$ around both vertical asymptotes, - and the curve of the function will have $3$ branches. - - David has followed \crefrange{rat:step:first}{rat:step:penultimate} for - $3$ separate functions, and drawn the results in \cref{rat:fig:deducehard}. - Help David finish each graph by deducing the curve of each function. - \begin{shortsolution} - \Vref{rat:fig:deducehard1} - - \begin{tikzpicture}[/pgf/declare function={f=(x-6)*(x+3)/( (x-4)*(x+1));}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\solutionfigurewidth, - ] - \addplot[soldot] coordinates{(-3,0)(6,0)(0,9/2)}; - \addplot[asymptote,domain=-10:10]({-1},{x}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{2}); - \addplot[pccplot] expression[domain=-10:-1.24276]{f}; - \addplot[pccplot] expression[domain=-0.6666:3.66667]{f}; - \addplot[pccplot] expression[domain=4.24276:10]{f}; - \end{axis} - \end{tikzpicture} - - \Vref{rat:fig:deducehard2} - - \begin{tikzpicture}[/pgf/declare function={f=3*(x-2)*(x+3)/( (x-6)*(x+5));}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\solutionfigurewidth, - ] - \addplot[soldot] coordinates{(-3,0)(2,0)(0,3/5)}; - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({6},{x}); - \addplot[asymptote,domain=-10:10]({x},{3}); - \addplot[pccplot] expression[domain=-10:-5.4861]{f}; - \addplot[pccplot] expression[domain=-4.68395:5.22241]{f}; - \addplot[pccplot] expression[domain=7.34324:10]{f}; - \end{axis} - \end{tikzpicture} - - \Vref{rat:fig:deducehard3} - - \begin{tikzpicture}[/pgf/declare function={f=2*(x-7)*(x+3)/( (x+6)*(x-5));}] - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\solutionfigurewidth, - ] - \addplot[soldot] coordinates{(-3,0)(7,0)(0,1.4)}; - \addplot[asymptote,domain=-10:10]({-6},{x}); - \addplot[asymptote,domain=-10:10]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{2}); - \addplot[pccplot] expression[domain=-10:-6.91427]{f}; - \addplot[pccplot] expression[domain=-5.42252:4.66427]{f}; - \addplot[pccplot] expression[domain=5.25586:10]{f}; - \end{axis} - \end{tikzpicture} - - \end{shortsolution} - \end{problem} - - \begin{figure}[!htb] - \begin{widepage} - \setlength{\figurewidth}{0.3\textwidth} - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-3,0)(6,0)(0,9/2)}; - \addplot[asymptote,domain=-10:10]({-1},{x}); - \addplot[asymptote,domain=-10:10]({4},{x}); - \addplot[asymptote,domain=-10:10]({x},{2}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:deducehard1} - \end{subfigure}% - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-3,0)(2,0)(0,3/5)}; - \addplot[asymptote,domain=-10:10]({-5},{x}); - \addplot[asymptote,domain=-10:10]({6},{x}); - \addplot[asymptote,domain=-10:10]({x},{3}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:deducehard2} - \end{subfigure}% - \hfill - \begin{subfigure}{\figurewidth} - \begin{tikzpicture} - \begin{axis}[ - xmin=-10,xmax=10, - ymin=-10,ymax=10, - xtick={-8,-6,...,8}, - width=\textwidth, - ] - \addplot[soldot] coordinates{(-3,0)(7,0)(0,1.4)}; - \addplot[asymptote,domain=-10:10]({-6},{x}); - \addplot[asymptote,domain=-10:10]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{2}); - \end{axis} - \end{tikzpicture} - \caption{} - \label{rat:fig:deducehard3} - \end{subfigure}% - \hfill - \caption{Graphs for \cref{rat:prob:deducehard}} - \label{rat:fig:deducehard} - \end{widepage} - \end{figure} - %=================================== - % Author: Adams (Hughes) - % Date: March 2012 - %=================================== - \begin{problem}[\Crefrange{rat:step:first}{rat:step:last}] - Use \crefrange{rat:step:first}{rat:step:last} to sketch a graph of - each of the following functions - \fixthis{need 2 more subproblems here} - \begin{multicols}{4} - \begin{subproblem} - $y=\dfrac{4}{x+2}$ - \begin{shortsolution} - Vertical intercept: $(0,2)$; vertical asymptote: $x=-2$, horizontal asymptote: $y=0$. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-5,xmax=5, - ymin=-5,ymax=5, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot] expression[domain=-5:-2.8]{4/(x+2)}; - \addplot[pccplot] expression[domain=-1.2:5]{4/(x+2)}; - \addplot[soldot]coordinates{(0,2)}; - \addplot[asymptote,domain=-5:5]({-2},{x}); - \addplot[asymptote,domain=-5:5]({x},{0}); - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=\dfrac{2x-1}{x^2-9}$ - \begin{shortsolution} - Vertical intercept:$\left( 0,\frac{1}{9} \right)$; - horizontal intercept: $\left( \frac{1}{2},0 \right)$; - vertical asymptotes: $x=-3$, $x=3$, horizontal asymptote: $y=0$. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-5,xmax=5, - ymin=-5,ymax=5, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot] expression[domain=-5:-3.23974]{(2*x-1)/(x^2-9)}; - \addplot[pccplot,samples=50] expression[domain=-2.77321:2.83974]{(2*x-1)/(x^2-9)}; - \addplot[pccplot] expression[domain=3.17321:5]{(2*x-1)/(x^2-9)}; - \addplot[soldot]coordinates{(0,1/9)(1/2,0)}; - \addplot[asymptote,domain=-5:5]({-3},{x}); - \addplot[asymptote,domain=-5:5]({3},{x}); - \addplot[asymptote,domain=-5:5]({x},{0}); - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=\dfrac{x+3}{x-5}$ - \begin{shortsolution} - Vertical intercept $\left( 0,-\frac{3}{5} \right)$; horizontal - intercept: $(-3,0)$; vertical asymptote: $x=5$; horizontal asymptote: $y=1$. - - \begin{tikzpicture} - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-5,ymax=5, - xtick={-8,-6,...,8}, - minor ytick={-3,-1,...,3}, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot] expression[domain=-10:3.666]{(x+3)/(x-5)}; - \addplot[pccplot] expression[domain=7:10]{(x+3)/(x-5)}; - \addplot[asymptote,domain=-5:5]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{1}); - \addplot[soldot]coordinates{(0,-3/5)(-3,0)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=\dfrac{2x+3}{3x-1}$ - \begin{shortsolution} - Vertical intercept: $(0,-3)$; horizontal intercept: $\left( -\frac{3}{2},0 \right)$; - vertical asymptote: $x=\frac{1}{3}$, horizontal asymptote: $y=\frac{2}{3}$. - - \begin{tikzpicture}[/pgf/declare function={f=(2*x+3)/(3*x-1);}] - \begin{axis}[ - framed, - xmin=-5,xmax=5, - ymin=-5,ymax=5, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot] expression[domain=-5:0.1176]{f}; - \addplot[pccplot] expression[domain=0.6153:5]{f}; - \addplot[asymptote,domain=-5:5]({1/3},{x}); - \addplot[asymptote,domain=-5:5]({x},{2/3}); - \addplot[soldot]coordinates{(0,-3)(-3/2,0)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=\dfrac{4-x^2}{x^2-9}$ - \begin{shortsolution} - Vertical intercept: $\left( 0,-\frac{4}{9} \right)$; - horizontal intercepts: $(2,0)$, $(-2,0)$; - vertical asymptotes: $x=-3$, $x=3$; horizontal asymptote: $y=-1$. - - \begin{tikzpicture}[/pgf/declare function={f=(4-x^2)/(x^2-9);}] - \begin{axis}[ - framed, - xmin=-5,xmax=5, - ymin=-5,ymax=5, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot] expression[domain=-5:-3.20156]{f}; - \addplot[pccplot,samples=50] expression[domain=-2.85774:2.85774]{f}; - \addplot[pccplot] expression[domain=3.20156:5]{f}; - \addplot[asymptote,domain=-5:5]({-3},{x}); - \addplot[asymptote,domain=-5:5]({3},{x}); - \addplot[asymptote,domain=-5:5]({x},{-1}); - \addplot[soldot] coordinates{(-2,0)(2,0)(0,-4/9)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=\dfrac{(4x+5)(3x-4)}{(2x+5)(x-5)}$ - \begin{shortsolution} - Vertical intercept: $\left( 0,\frac{4}{5} \right)$; - horizontal intercepts: $\left( -\frac{5}{4},0 \right)$, $\left( \frac{4}{3},0 \right)$; - vertical asymptotes: $x=-\frac{5}{2}$, $x=5$; horizontal asymptote: $y=6$. - - \begin{tikzpicture}[/pgf/declare function={f=(4*x+5)*(3*x-4)/((2*x+5)*(x-5));}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-20,ymax=20, - xtick={-8,-6,...,8}, - ytick={-10,0,...,10}, - minor ytick={-15,-5,...,15}, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot] expression[domain=-10:-2.73416]{f}; - \addplot[pccplot] expression[domain=-2.33689:4.2792]{f}; - \addplot[pccplot] expression[domain=6.26988:10]{f}; - \addplot[asymptote,domain=-20:20]({-5/2},{x}); - \addplot[asymptote,domain=-20:20]({5},{x}); - \addplot[asymptote,domain=-10:10]({x},{6}); - \addplot[soldot]coordinates{(0,4/5)(-5/4,0)(4/3,0)}; - \end{axis} - \end{tikzpicture} - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Inverse functions] - Each of the following rational functions are invertible - \[ - F(x)=\frac{2x+1}{x-3}, \qquad G(x)= \frac{1-4x}{x+3} - \] - \begin{subproblem} - State the domain of each function. - \begin{shortsolution} - \begin{itemize} - \item The domain of $F$ is $(-\infty,3)\cup(3,\infty)$. - \item The domain of $G$ is $(-\infty,-3)\cup(-3,\infty)$. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Find the inverse of each function, and state its domain. - \begin{shortsolution} - \begin{itemize} - \item $F^{-1}(x)=\frac{3x+1}{x-2}$; the domain of $F^{-1}$ is $(-\infty,2)\cup(2,\infty)$. - \item $G^{-1}(x)=\frac{3x+1}{x+4}$; the domain of $G^{-1}$ is $(-\infty,-4)\cup(-4,\infty)$. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - Hence state the range of the original functions. - \begin{shortsolution} - \begin{itemize} - \item The range of $F$ is the domain of $F^{-1}$, which is $(-\infty,2)\cup(2,\infty)$. - \item The range of $G$ is the domain of $G^{-1}$, which is $(-\infty,-4)\cup(-4,\infty)$. - \end{itemize} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - State the range of each inverse function. - \begin{shortsolution} - \begin{itemize} - \item The range of $F^{-1}$ is the domain of $F$, which is $(-\infty,3)\cup(3,\infty)$. - \item The range of $G^{-1}$ is the domain of $G$, which is $(-\infty,-3)\cup(-3,\infty)$. - \end{itemize}<++> - \end{shortsolution} - \end{subproblem} - \end{problem} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Composition] - Let $r$ and $s$ be the rational functions that have formulas - \[ - r(x)=\frac{3}{x^2},\qquad s(x)=\frac{4-x}{x+5} - \] - Evaluate each of the following. - \begin{multicols}{3} - \begin{subproblem} - $(r\circ s)(0)$ - \begin{shortsolution} - $\frac{75}{16}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(s\circ r)(0)$ - \begin{shortsolution} - $(s\circ r)(0)$ is undefined. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(r\circ s)(2)$ - \begin{shortsolution} - $\frac{147}{4}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(s\circ r)(3)$ - \begin{shortsolution} - $192$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(s\circ r)(4)$ - \begin{shortsolution} - $(s\circ r)(4)$ is undefined. - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $(s\circ r)(x)$ - \begin{shortsolution} - $\dfrac{4x^2-3}{1+5x^2}$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \end{problem} - %=================================== - % Author: Hughes - % Date: March 2012 - %=================================== - \begin{problem}[Piecewise rational functions] - The function $R$ has formula - \[ - R(x)= - \begin{dcases} - \frac{2}{x+3}, & x<-5 \\ - \frac{x-4}{x-10}, & x\geq -5 - \end{dcases} - \] - Evaluate each of the following. - \begin{multicols}{4} - \begin{subproblem} - $R(-6)$ - \begin{shortsolution} - $-\frac{2}{3}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $R(-5)$ - \begin{shortsolution} - $\frac{3}{5}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $R(-3)$ - \begin{shortsolution} - $\frac{7}{13}$ - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $R(5)$ - \begin{shortsolution} - $-\frac{1}{5}$ - \end{shortsolution} - \end{subproblem} - \end{multicols} - \begin{subproblem} - What is the domain of $R$? - \begin{shortsolution} - $(-\infty,10)\cup(10,\infty)$ - \end{shortsolution} - \end{subproblem} - \end{problem} - \end{exercises} - -\section{Graphing rational functions (oblique asymptotes)}\label{rat:sec:oblique} - \begin{subproblem} - $y=\dfrac{x^2+1}{x-4}$ - \begin{shortsolution} - \begin{enumerate} - \item $\left( 0,-\frac{1}{4} \right)$ - \item Vertical asymptote: $x=4$. - \item A graph of the function is shown below - - \begin{tikzpicture}[/pgf/declare function={f=(x^2+1)/(x-4);}] - \begin{axis}[ - framed, - xmin=-20,xmax=20, - ymin=-30,ymax=30, - xtick={-10,10}, - minor xtick={-15,-5,...,15}, - minor ytick={-10,10}, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot,samples=50] expression[domain=-20:3.54724]{f}; - \addplot[pccplot,samples=50] expression[domain=4.80196:20]{f}; - \addplot[asymptote,domain=-30:30]({4},{x}); - \end{axis} - \end{tikzpicture} - \end{enumerate} - \end{shortsolution} - \end{subproblem} - \begin{subproblem} - $y=\dfrac{x^3(x+3)}{x-5}$ - \begin{shortsolution} - \begin{enumerate} - \item $(0,0)$, $(-3,0)$ - \item Vertical asymptote: $x=5$, horizontal asymptote: none. - \item A graph of the function is shown below - - \begin{tikzpicture}[/pgf/declare function={f=x^3*(x+3)/(x-5);}] - \begin{axis}[ - framed, - xmin=-10,xmax=10, - ymin=-500,ymax=2500, - xtick={-8,-6,...,8}, - ytick={500,1000,1500,2000}, - grid=both, - width=\solutionfigurewidth, - ] - \addplot[pccplot,samples=50] expression[domain=-10:4]{f}; - \addplot[pccplot] expression[domain=5.6068:9.777]{f}; - \addplot[asymptote,domain=-500:2500]({5},{x}); - \end{axis} - \end{tikzpicture} - \end{enumerate} - \end{shortsolution} - \end{subproblem} |