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

Question

I need help!!!!

Answer 1

The parabola is h (x ) = 3x ^2 - 6x - 9.

Vertex form of a parabola is y  = a  (x - h )^2 + k.

Here x² coefficient is 3, for perfect square make x² coefficient 1 by dividing each side by 3.

h (x ) = 3x ^2 - 6x - 9

h (x )/3 = x ^2 - 2x - 3

To change the expression into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

x  coefficient = - 2 then (half the x coefficient)² is (- 1)^2 = 1.

So, Add 1 to each side.

h (x )/3 + 1 = x ^2 - 2x  + 1 - 3

h (x )/3 + 1 = (x - 1)^2 - 3

h (x )/3 = (x - 1)^2 - 3 - 1

h (x )/3 = (x - 1)^2 - 4

h (x ) = 3(x - 1)^2 - 12.

Therefore, vertex form of the parabola is h (x ) = 3(x - 1)^2 - 12.