Answer
Vertical asymptote at $x=6$ and $x=-6.$
Work Step by Step
A function has vertical asymptotes at values that make the denominator only $0$.
$h(x)=\dfrac{6x}{36-x^2}=\dfrac{6x}{(6-x)(6+x)}\to(6-x)(6+x)=0\to x=6$ or $x=-6.$
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: 70
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