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: 71

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 $x=-8.$
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