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Find the points on the graph y = -x^3 + 6x^2 at which the tangent line is horizontal.

I found -x^2 + 4 = 0
are the points [4,32]

1 answer

y'=0=-3x^2+12x or
x(-x+4)=0
so the x points are 0,4
so the graph points are
(0,0) and (4,32)

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