Find the points at which f(x)=x^2(x+1)^3 has a horizontal tangent.
f"(x) = x^2 (3)(x+1)^2 + 2x(x+1)^3
= x(x+1)^2 [3x + 2(x+1)]
= 0 for a horizontal tangent , (slope = 0)
x(x+1)(5x+2) = 0
x = 0, x = -1, x = -2/5
sub those values into the original to get the corresponding y values for the points