## College Algebra (11th Edition)

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