Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - 2.2 - Functions - 2.2 Exercises - Page 182: 25


(a) $q(0)=-\frac{1}{9}$ (b) It is not possible. (c) $q(y+3)=\frac{1}{y^2+6y}$

Work Step by Step

$q(x)=\frac{1}{x^2-9}$ (a) $q(0)=\frac{1}{0^2-9}=-\frac{1}{9}$ (b) $q(3)=\frac{1}{3^2-9}=\frac{1}{0}$ (it is not possible to divide by $0$) (c) $q(y+3)=\frac{1}{(y+3)^2-9}=\frac{1}{y^2+6y+9-9}=\frac{1}{y^2+6y}$
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