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: 55



Work Step by Step

Zero at $3$ implies numerator is equal to $0$ when $x=3\to(x-3)=0.$ Vertical asymptote at $-2$ and $6$ implies a denominator of equation $(x+2)(x-6).$ Putting everything together gives us $\dfrac{(x-3)}{(x+2)(x-6)}.$
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