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 399: 77

Answer

Graph the function as:
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Work Step by Step

Step 1: Substitute $x=-x$. $\begin{align} & f\left( x \right)=\frac{{{x}^{2}}+x-12}{{{x}^{2}}-4} \\ & f\left( -x \right)=\frac{{{\left( -x \right)}^{2}}+\left( -x \right)-12}{{{\left( -x \right)}^{2}}-4} \\ & =\frac{{{x}^{2}}-x-12}{{{x}^{2}}-4} \end{align}$ Therefore, the function $f\left( -x \right)$ is not equal to $-f\left( x \right)$ and the $f\left( x \right)$. So, the graph of the function is not symmetrical about the $y$ axis and the origin. Step 2: To calculate the x intercepts equate $f\left( x \right)=0$. $\begin{align} & \frac{{{x}^{2}}+x-12}{{{x}^{2}}-4}=0 \\ & x=3,-4 \end{align}$ , Step 3: To calculate the y intercepts evaluate $f\left( 0 \right)$. $f\left( 0 \right)=3$ Step 4: Since the degree of the numerator is equal to the denominator, the horizontal asymptote is: $y=1$ Step 5: For the vertical asymptote, equate the denominator to 0. $x=\pm 2$
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