### Instruction: Let $f(x)=\frac{(x-2)^2-9}{3}$.

The equation $y=f(x)$ is graphed, and the $x$- and $y$-intercepts of the graph are connected to form a polygon. What is the area of that polygon?### Response: I notice that the numerator of $f(x)$ is a difference of squares, so I can factor it as $(x-2)^2-9=(x-2+3)(x-2-3)=(x+1)(x-5)$.
This means that $f(x)=\frac{(x+1)(x-5)}{3}$, and the $x$-intercepts of the graph are $x=-1$ and $x=5$.
To find the $y$-intercept of the graph, I plug in $x=0$ and get $f(0)=\frac{(0+1)(0-5)}{3}=\frac{-5}{3}$.
So the three vertices of the polygon are $(-1,0)$, $(5,0)$, and $(0,-\frac{5}{3})$.
I can use the formula for the area of a triangle, which is half the product of the base and the height, to find the area of the polygon.
The base of the triangle is the distance between the $x$-intercepts, which is $5-(-1)=6$ units.
The height of the triangle is the distance between the $y$-intercept and the $x$-axis, which is $\frac{5}{3}$ units.
Therefore, the area of the polygon is $\frac{1}{2}(6)(\frac{5}{3})=5$ square units.