jmc

We begin with a drawing of the graph and the polygon in question (it's possible to solve the problem without drawing this picture, but we provide it for clarity): The $y$-intercept of the graph is $(0,f(0))=\left(0,-\frac{5}{3}\right)$. To find the $x$-intercepts, we solve the equation $$\frac{(x-2{)}^2-9}{3}=0,$$which yields $$(x-2{)}^2=9$$and thus $x=2\pm 3$. So, the $x$-intercepts are $(-1,0)$ and $(5,0)$.

The triangle whose vertices are $(-1,0),$ $(5,0),$ and $\left(0,-\frac{5}{3}\right)$ has base $6$ and height $\frac{5}{3}$, so its area is $$\frac{1}{2}\cdot 6\cdot \frac{5}{3}=\boxed{5}.$$