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Asked by kingsarmiento

find the point on the curve y=x^2-6x+10 where tangent is horizontal
4 years ago

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Answered by oobleck
y' = 2x-6
2x-6=0 at x=3

or, remembering your Algebra I, that is the vertex of the parabola, which is always at
x = -b/2a = 6/2 = 3 in this case
Now just find y(3)
4 years ago

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