Asked by kingsarmiento
find the point on the curve y=x^2-6x+10 where tangent is horizontal
Answers
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)
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)
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