College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 4 - Section 4.2 - Exponential Functions - 4.2 Exercises - Page 409: 40

Answer

See the picture below.

Work Step by Step

The parent function is $f(x)=2^x$ (with red) the given function is $g(x)=2^{x-3}-1$ (with blue). The parent function can be graphed by calculating a few coordinates and connecting them with a smooth curve: $f(-2)=2^{-2}=(\frac{1}{4})$ $f(-1)=2^{-1}=(\frac{1}{2}$ $f(0)=2^0=1$ $f(1)=2^1=2$ $f(2)=2^2=4$ For every corresponding x-value the following equation is true: $f(x-3)-1=g(x)$ This means that the graph is translated 3 units to the right and 1 unit down ($g(x)$ involves a horizontal shift of 3 to the right and also a vertical shift of 1 downwards.). First, the horizontal shift. We only consider the g(x) function as $g'(x)=2^{x-3}$ For example if $f(0)=1$ in the original $f(x)$, this will be equal to $g'(3)=f(3-3)=f(0)=1$. Here,$ f(0)=g'(3)$ also, $f(1)=g'(4)$ We can see that here, each point in the parent function was moved to the right by 3 units. Second, we translate this $g'(x)$ function to get the originally given function. Now, the following equation is true: $g'(x)-1=g(x)$ Every $g'(x)$ value will be decreased by 1. For example if $g'(3)=1$, this will be translated as $g'(3)-1=1-1=0$. We can see that here, the $g'(x)$ is greater than $g(x)$ for every corresponding x-value by 1.)
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