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: 33


The vertical asymptote will be $x=3$; there will be a hole at $x=-7$.

Work Step by Step

$\begin{align} & h\left( x \right)=\frac{x+7}{{{x}^{2}}+4x-21} \\ & =\frac{p\left( x \right)}{q\left( x \right)} \end{align}$ So, $p\left( x \right)=x+7$ and $q\left( x \right)={{x}^{2}}+4x-21$. For the vertical asymptotes, put $q\left( x \right)=0$ $\begin{align} & q\left( x \right)={{x}^{2}}+4x-21=0 \\ & {{x}^{2}}+4x-21=0 \\ & {{x}^{2}}+7x-3x-21=0 \\ & x\left( x+7 \right)-3\left( x+7 \right)=0 \\ & \left( x+7 \right)\left( x-3 \right)=0 \end{align}$ Therefore, the vertical asymptotes are: $x=-7,\text{ or }x=3$. But, after rearranging, $x+7$ from the denominator will be eliminated. So, the vertical asymptote will be $x=3$. Now let us find the holes. $\begin{align} & h\left( x \right)=\frac{x+7}{{{x}^{2}}+4x-21} \\ & =\frac{x+7}{\left( x+7 \right)\left( x-3 \right)} \\ & =\frac{1}{x-3} \end{align}$ Here, $x+7$ will be eliminated; when x approaches -7, the polynomial approaches to $-\frac{1}{10}$ but at the point $x=-7$ the polynomial will be discontinuous. So, there will be a hole at $x=-7$ Therefore, the vertical asymptote will be $x=3$; there will be a hole at $x=-7$.
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