Answer
Vertical asymptotes at $x=8$ and $x=-8.$
Work Step by Step
A function has vertical asymptotes at values that make only the denominator $0$.
$g(x)=\dfrac{2x+1}{x^2-64}=\dfrac{2x+1}{(x-8)(x+8)}\to(x-8)(x+8)=0\to x=8$ or $x=-8.$
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: 71
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