h(x)=3x^2-6x-9 in vertex form

Question

I need help!!!!

Answer 1

Given h(x) = 3x^2 -6x - 9, the vertex of this parabola can be found where the derivative of h(x) is equal to zero.

h'(x) = 6x - 6 = 0. Solving for x gives x = 1.

Plug this back into h(x) to get h(1) = 3 - 6 - 9 = -12. So y = -12

Therefore the vertex is (1, -12)