Calculus 10th Edition

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: 69

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 $x=-3.$
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