Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
up vote 2 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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.