Find all points on the graph of f(x)=-x^3+3x^2-2 where there is a horizontal tangent line.

Answer 1

f'=0=2x^2+6x

= 2x(x+3) so at x=0, x=-3, the tangent is horizontal.

Answer 2

since it is a cubic, there will presumably be two

f' = 0 = 3 x^2 + 6 x = 3x ( x+2)
x = 0 and x = -2
you find f(x) there

Answer 3

Would the points be (0,-2) and -2, 18) when I plug them into f(x)?