## College Algebra (11th Edition)

Published by Pearson

# Chapter 4 - Section 4.2 - Exponential Functions - 4.2 Exercises: 49

#### Answer

See the picture below.

#### Work Step by Step

The parent function is $f(x)=(\frac{1}{3})^x$ (with red) the given function is $g(x)=(\frac{1}{3})^{x-2}+2$ (with blue). The parent function can be graphed by calculating a few coordinates and connecting them with a smooth curve: $f(-2)=(\frac{1}{3})^{-2}=9$ $f(-1)=(\frac{1}{3})^{-1}=3$ $f(0)=(\frac{1}{3})^0=1$ $f(1)=(\frac{1}{3})^1=\frac{1}{3}$ $f(2)=(\frac{1}{3})^2=\frac{1}{9}$ For every corresponding x-value the following equation is true: $f(x-2)+2=g(x)$ This means that the graph is translated 2 units right and 2 units up ($g(x)$ involves a horizontal shift of 2 to the right and also a vertical shift of 2 upwards.). First, the horizontal shift. We only consider the g(x) function as $g'(x)=\frac{1}{3}^{x-2}$ For example if $f(0)=1$ in the original $f(x)$, this will be equal to $g'(2)=f(2-2)=f(0)=1$. $Here, f(0)=g'(2)$ also, $f(1)=g'(3)$ We can see that here, each point in the parent function was moved to the right by 2 units. Second, we translate this $g'(x)$ function to get the originally given function. Now, the following equation is true: $g'(x)+2=g(x)$ Every $g'(x)$ value will be increased by 2. For example if $g'(2)=1$, this will be translated as $g'(2)+2=1+2=3$. We can see that here, the $g(x)$ is greater than $g'(x)$ for every corresponding x-value by 2.)

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