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: 24

Answer

Vertical asymptote at $t=-2.$

Work Step by Step

$h(t)=\dfrac{t^2-2t}{t^4-16}=\dfrac{t(t-2)}{(t^2+4)(t^2-4)}=\dfrac{t(t-2)}{(t^2+4)(t-2)(t+2)}$ $=\dfrac{t}{(t^2+4)(t+2)}.$ Vertical asymptotes occur when the denominator alone is $0\to$ $(t+2)(t^2+4)=0\to t=-2.$
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