## College Algebra (11th Edition)

The parent function is $f(x)=(\frac{1}{3})^x$ (with green) the given function is $g(x)=(\frac{1}{3})^{x+3}-2$ (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: $f(x+3)-2=g(x)$ This means that the graph is translated 3 units left and 2 units down. ( $g(x)$ involves a horizontal shift of 3 to the left and also a vertical shift of 2 downwards.). First, the horizontal shift. We only consider the g(x) function as $g'(x)=\frac{1}{3}^{x+3}$ 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: $g'(x)-2=g(x)$ Every $g'(x)$ value will be decreased by 2. For example if $g'(-3)=1$, this will be translated as $g'(-3)-2=1-2=-1$. We can see that here, the $g'(x)$ is greater than $g(x)$ for every corresponding x-value by 2.)