jmc

First, we need to find where this curve intersects the $x$ and $y$ axes. If $y=0$, then $(x-3{)}^2(x+2)=0$, which has solutions of $x=3$ and $x=-2$. If $x=0$, then $y=(-3{)}^2(2)=18$. So, the curve has two $x$-intercepts and one $y$-intercept. The length of the base along the $x$-axis is $3-(-2)=5$. The height from this base is equal to the $y$-intercept, 18. The area of the triangle is $\frac{1}{2}\cdot 5\cdot 18=\boxed{45}$.