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

Answer 1

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)