Precalculus (6th Edition) Blitzer

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

Chapter 2 - Cumulative Review Exercises - Page 435: 16

Answer

The graph is shown below:
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Work Step by Step

To find the x-intercepts of the function, equate the function $f\left( x \right)=\frac{\left( x-1 \right)}{\left( x-2 \right)}$ to zero. $\begin{align} & \frac{\left( x-1 \right)}{\left( x-2 \right)}=0 \\ & x=1 \end{align}$ Thus, $x=1$ is the x-intercept. Equate the denominator to zero for the vertical asymptote: $\left( x-2 \right)=0$ The vertical asymptote is $x=2$. Compare the degrees of the numerator and denominator to find the horizontal asymptote. Since the degree of the numerator and denominator is 1, so the horizontal asymptote is $y=1$. To find the y-intercept of the function, find the value of $f\left( 0 \right)$. $\begin{align} & f\left( 0 \right)=\frac{\left( 0-1 \right)}{\left( 0-2 \right)} \\ & =\frac{1}{2} \end{align}$ Thus, the y-intercept is $\frac{1}{2}$. Substitute x with $-x$ to check the symmetry of the function: $\begin{align} & f\left( -x \right)=\frac{\left( -x-1 \right)}{\left( -x-2 \right)} \\ & =\frac{\left( x+1 \right)}{\left( x+2 \right)} \end{align}$ Since $f\left( x \right)\ne f\left( -x \right)$, the graph is not symmetric with respect to the y-axis and since $f\left( -x \right)\ne -f\left( x \right)$, the graph is not symmetric through the origin.
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