#### Answer

See the picture below

#### Work Step by Step

The parent function is $f(x)=(\frac{1}{3})^x$ (with green) the given function is $g(x)=(\frac{1}{3})^{x+2}-1$ (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: $f(x+2)-1=g(x)$
This means that the graph is translated 2 units left and 1 units down.
($g(x)$ involves a horizontal shift of 2 to the left and also a vertical shift of 1 downwards.).
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, $f(1)=g'(-1)$
We can see that here, each point in the parent function was moved to the left by 2 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 decreased by 1.
For example if $g'(-2)=1$, this will be translated as $g'(-2)-1=1-1=0$.
We can see that here, the $g'(x)$ is greater than $g(x)$ for every corresponding x-value by 1.)