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: 38

Answer

See the picture below.
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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 left and 1 unit up ($g(x)$ involves a horizontal shift of 3 to the left and also a vertical shift of 1 upwards.). 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, $f(1)=g'(-2)$ We can see that here, each point in the parent function was moved to the left 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 increased by 1. For example if $g'(-3)=1$, this will be translated as $g'(-3)+1=1+1=2$. We can see that here, the $g(x)$ is greater than $g'(x)$ for every corresponding x-value by 1.)
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