Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.6 - Rational Functions and Their Graphs - Exercise Set - Page 402: 134


The graph is shown below:

Work Step by Step

We know that a rational function $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ can consist of the vertical and horizontal asymptotes. For vertical asymptotes, we keep the denominator function $q\left( x \right)$ equal to 0, and thereby find its roots; that is, a vertical asymptote is the root of the rational function. So, for the vertical asymptote to be given by $x=3$ , we can write the denominator function of the rational function as, $q\left( x \right)=x-3$ We have a horizontal asymptote as: $\begin{align} & y=\frac{0}{1} \\ & =0 \end{align}$ And for the horizontal asymptote to be the x-axis or $y=0$, the degree of the numerator function $p(x)$ has to be less than the degree of the denominator function, $q\left( x \right)$. Let $f\left( x \right)=\frac{p\left( x \right)}{x-3}$. Since the degree of $p\left( x \right)$ is supposed to be one less than that of $q\left( x \right)$ for the function to have a horizontal asymptote, thus the degree of $p\left( x \right)$ is 0, i.e., it is a constant. For calculating $p\left( x \right)$ , we calculate the y intercept. Since the y intercept is given to be -1, $\begin{align} & f\left( 0 \right)=-1 \\ & \frac{p\left( x \right)}{0-3}=-1 \\ & p\left( x \right)=3 \end{align}$ Therefore, the required function is given by, $f\left( x \right)=\frac{3}{x-3}$ Now, for plotting, Press $y=$ to write the equation $y=\frac{3}{x-3}$. Press WINDOW to set the window $\left( -10,10,1 \right)$ and $\left( -10,10,1 \right)$. Press GRAPH to plot the graph of the function. Thus, the graph of the function $f\left( x \right)=\frac{3}{x-3}$ has a vertical asymptote at $x=3$ , horizontal asymptote $y=0$ , $y$ -intercept at $-1$ ,and no $x$ -intercept.
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