Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 3 - Linear and Quadratic Functions - Chapter Review - Cumulative Review - Page 165: 11

Answer

a) no; b) $-1; (-2,-1)$; c) $-8; (-8,2)$

Work Step by Step

We are given the function: $f(x)=\dfrac{x}{x+4}$ a) Check if the point $\left(1,\dfrac{1}{4}\right)$ is on the graph of $f$: $\dfrac{1}{1+4}\stackrel{?}{=}\dfrac{1}{4}$ $\dfrac{1}{5}\not=\dfrac{1}{4}$ Therefore the point $\left(1,\dfrac{1}{4}\right)$ is not on the graph. b) Compute $f(-2)$: $f(-2)=\dfrac{-2}{-2+4}=-1$ The point on the graph is $(-2,-1)$. c) Solve the equation $f(x)=2$: $\dfrac{x}{x+4}=2$ $x=2(x+4)$ $x=2x+8$ $x=-8$ The point on the graph is $(-8,2)$.
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