Answer
Vertical asymptotes at $x=3$ and $x=-3.$
Work Step by Step
A function has vertical asymptotes at values that make the denominator only $0$.
$f(x)=\dfrac{x^3}{x^2-9}=\dfrac{x^3}{(x-3)(x+3)}\to(x-3)(x+3)=0\to x=3$ or $x=-3.$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 1 - Limits and Their Properties - Review Exercises - Page 92: 69
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