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 - Page 88: 21

Answer

Vertical asymptote at $x=0$ or $x=3$

Work Step by Step

$f(x)=\dfrac{4x^2+4x-24}{x^4-2x^3-9x^2+18x}=\dfrac{4(x+3)(x-2)}{x(x-2)(x+3)(x-3)}$ $=\dfrac{4}{x(x-3)}.$ Vertical asymptotes occur when the denominator alone is $0\to$ $x(x-3)=0\to x=0$ or $x=3.$
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