Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Cumulative Review Exercises - Page 1181: 13

Answer

See below:

Work Step by Step

Consider the function, $ f\left( x \right)=\sqrt{x}$ and $ g\left( x \right)=\sqrt{x-2}+1$ The parent function for $ g\left( x \right)=\sqrt{x-2}+1$ is $ f\left( x \right)=\sqrt{x}$. If h is a positive real number, the graph of $ f\left( x \right)-h $ is the graph of $ y=f\left( x \right)$ shifted downward h units. If c is a positive real number, then the graph of $ f\left( x+c \right)$ is the graph of $ y=f\left( x \right)$ shifted to the left c units. Therefore, the graph of $ g\left( x \right)=\sqrt{x-2}+1$ is the graph of $ f\left( x \right)=\sqrt{x}$ shifted to the right $2$ units and upward $1$ unit. The point $\left( 2,\sqrt{2} \right)$ on the graph of $ f\left( x \right)=\sqrt{x}$ corresponds to $\left( 2,1 \right)$ on the graph of $ g\left( x \right)=\sqrt{x-2}+1$. The point $\left( 6,\sqrt{6} \right)$ on the graph of $ f\left( x \right)=\sqrt{x}$ corresponds to $\left( 6,3 \right)$ on the graph of $ g\left( x \right)=\sqrt{x-2}+1$. The point $\left( 3,\sqrt{3} \right)$ on the graph of $ f\left( x \right)=\sqrt{x}$ corresponds to $\left( 3,2 \right)$ on the graph of $ g\left( x \right)=\sqrt{x-2}+1$.
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