somebody please help so I can pass, pleaseRewrite the following equation in the form y = a(x - h)2 + k. Then, determine the x-coordinate of the minimum.y=2x^2 - 32x + 56The rewritten equation is y = ____ (x - _____ )2 + ____ .The x-coordinate of the minimum is _____

First, we are going to find the vertex of our quadratic. Remember that to find the vertex [tex](h,k)[/tex] of a quadratic equation of the form [tex]y=a x^{2} +bx+c[/tex], we use the vertex formula [tex]h= \frac{-b}{2a} [/tex], and then, we evaluate our equation at [tex]h[/tex] to find [tex]k[/tex].

We now from our quadratic that [tex]a=2[/tex] and [tex]b=-32[/tex], so lets use our formula:
[tex]h= \frac{-b}{2a} [/tex]
[tex]h= \frac{-(-32)}{2(2)} [/tex]
[tex]h= \frac{32}{4} [/tex]
[tex]h=8[/tex]
Now we can evaluate our quadratic at 8 to find [tex]k[/tex]:
[tex]k=2(8)^2-32(8)+56[/tex]
[tex]k=2(64)-256+56[/tex]
[tex]k=128-200[/tex]
[tex]k=-72[/tex]
So the vertex of our function is (8,-72)

Next, we are going to use the vertex to rewrite our quadratic equation:
[tex]y=a(x-h)^2+k[/tex]
[tex]y=2(x-8)^2+(-72)[/tex]
[tex]y=2(x-8)^2-72[/tex]
The x-coordinate of the minimum will be the x-coordinate of the vertex; in other words: 8.

We can conclude that:
The rewritten equation is [tex]y=2(x-8)^2-72[/tex]
The x-coordinate of the minimum is 8