## College Algebra (11th Edition)

The parent function is $f(x)=(\frac{1}{3})^x$ (with green) the given function is $g(x)=-(\frac{1}{3})^{-x}$ (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: $-f(-x)=g(x)$ This means that the graph is reflected across both the x-axis and the y-axis. First, the reflection across the y-axis. We only consider the g(x) function as $g'(x)=\frac{1}{3}^{-x}$ For every corresponding x-value the following equation is true: $f(x)=g'(-x)$ This means that the graph is reflected across the y-axis. Because when f(x)=g'(x), the g'(x) function acts like the f(x). For example if $f(-1)=3$ in the original $f(x)$, this will be equal to $g'(1)=f(-1)=3$. Here,$f(-1)=g'(1)$ also, $f(-2)=g'(2)$ We can see that here, each point in the parent function was reflected across the y-axis. Second, the reflection across the x-axis. Now, the following equation is true: $-g'(x)=g(x)$ Every $g'(x)$ value will be multiplied by (-1). For example if $g'(1)=3$, this will be translated as $g(1)=-g'(1)=-3$. Also, $g(2)=-g'(2)=-9$ We can see that here, every $g'(x)$ is multiplied by (-1), therefore g'(x) is reflected across the x-axis.)