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 - 1.5 Exercises: 22

Answer

Vertical asymptote at $x=1$ and at $x=-1.$

Work Step by Step

$h(x)=\dfrac{x^2-9}{x^3+3x^2-x-3}=\dfrac{(x+3)(x-3)}{(x+3)(x-1)(x+1)}$ $=\dfrac{x-3}{(x-1)(x+1)}.$ Vertical asymptotes occur when the denominator alone is $0\to$ $(x-1)(x+1)=0\to x=1$ or $x=-1.$
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