Finding the Equation of Parabola 
Write the equation in the form $y=a(x-h)^{2}+k$ with zeros -4 and 8, and an optimal value of 18.

I'm not sure what "optimal value" means first of all- I think it means that the maximum value has a y-value of 18. What I've done so far:
$y=a(x+4)(x-8)$. Then to calculate the x-value of the vertex: $\frac{-4+8}{2}=2$ and then you substitute $x=2$ into the original equation to get the y-value of the vertex, which is:
$y=(2+4)(2-8)\implies y=-36$. Then $y=a(x-2)^{2}-36$, and then since we know a point on the line, I subbed in $(-4,0)$, which means that $0=a(-4-2)^{2}-36\implies 0=36a-36\implies a=1.$ So what I'm getting is $y=-(x-2)^{2}-36$

Use factored form: $y=a(x+s)(x+t)$

$y=a(x+4)(x-8)$

'Optimal Value'= $y$ value of $(x,y)$

$18$ = $y$ value of $(x,y)$

$x$ value of $(x,y)= (s+t)/2
                = (-4+8)/2
                = 4/2
                = 2$
Therefore $(x,y) = (2,18)$
Substitute $(2,18)$ into $y=a(x+4)(x-8)$

$18=a(2+4)(2-8)$
$18=a(6)(-6)$
$18=a(-36)$
$-36a=18$
$a=18/-36$
$a=1/-2$

Therefore the equation of the parabola is $y= 1/-2(x+4)(x-8)$
