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)?

The intercepts can be read from the equation as (-2, 0), (0, 18), (3, 0). These define a triangle with a base of 5 and an altitude of 18, so the area is

... A = (1/2)·5·18 = 45 . . . . square units

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The x-intercepts are those values of x that make one or another of the factors be zero. (x-3) is zero when x=3; (x+2) is zero when x=-2.

The y-intercept is the value when x=0. That is (-3)²·(2) = 9·2 = 18.

i'm not sure i quite understand, could you provide more information?