posted by Jessi .
Which of the following is a point at which the tangent to y=(x-2)^2(x+3) is horizontal?
b) (-1, 18)
I found the derivative of y=(x-2)^2(x+3)
which is 3x^2-2x-8
if I solved for 0, it wouldn't give me the answer that are listed above. What did I do wrong?
dy/dx=3x^2-2x-8 is good.
I get x=-3/4 or x=2 for dy/dx=0.
So can you find the answer from above?