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: 22


Vertical asymptote is $x=3$ and there are no holes.

Work Step by Step

If there is a rational function $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ where $p\left( x \right)$ is a numerator and $q\left( x \right)$ is a denominator then $x=a$ is a vertical asymptote of the function $f\left( x \right)$ if $x=a$ is a zero of the denominator $q\left( x \right)$. Equate the denominator to zero. $\begin{align} & x-3=0 \\ & x=3 \end{align}$ Thus, $x=3$ is a vertical asymptote. If there is a rational function $f\left( x \right)=\frac{p\left( x \right)}{q\left( x \right)}$ where $p\left( x \right)$ is the numerator and $q\left( x \right)$ is the denominator then $x=a$ is called a hole if $x-a$ is a common factor of the numerator and denominator. There is no common factor between $x$ and $x-3$. So, there are no holes. Hence, $x=3$ is a vertical asymptote and there are no holes.
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