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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 $y=-(x-2)^{2}-36$

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1 Answer 1

up vote 1 down vote accepted

Optimal value $= 18$ means that the value of the parabola at the vertex is $18$. Thus, $k=18$.

The other two conditions mean that

$$a (h+4)^2+18 = 0$$ $$a(h-8)^2+18=0$$

Use substitution to solve, i.e. $a=-18/(h+4)^2$, so that

$$18 = 18 \frac{(h-8)^2}{(h+4)^2} \implies (h-8)^2 = (h+4)^2 \implies h=2$$

which then means that

$$a = -18/6^2 = -1/2$$

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