The turning point on the curve

y=x2−4x
is at?

2 answers

dy/dx = 2x - 4
= 0 at the vertex of this parabola
x = 2, then y = -4

Your turning point is (2,-4)
You have y = x^2 - 4x
as with all parabolas y = ax^2+bx+c, the vertex is at x = -b/2a