How many square units are in the area of the triangle whose vertices are the $x$ and $y$ intercepts of the curve $y = (x-3)^2 (x+2)$?
I can use the formula $A = \frac{1}{2} bh$ for the area of a triangle, where $b$ is the base and $h$ is the height. To find the base of the triangle, I need to find the $x$-intercepts of the curve, where $y = 0$. I can factor the curve as $y = (x-3)(x-3)(x+2)$, and set it equal to zero. This gives me three solutions: $x = 3, 3, -2$. The base of the triangle is the distance between the two distinct solutions, which is $|3 - (-2)| = 5$ units. To find the height of the triangle, I need to find the $y$-intercept of the curve, where $x = 0$. I can plug in $x = 0$ into the curve and get $y = (0-3)^2 (0+2) = 18$. The height of the triangle is $|18 - 0| = 18$ units. Now I can use the formula for the area and get $A = \frac{1}{2} (5)(18) = 45$ square units. 