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Which of the following is a point at which the tangent to y=(x-2)^2(x+3) is horizontal?

a) (-2,16)
b) (-1, 18)
c) (1,4)
d) (2,0)

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?

  • Calculus -

    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?

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