# Finding the rule of a quadratic graph

I am trying to find the rule. Insofar:

$y = a(x-b)^2 + c$

Turning point is $(1,9)$

So $b = 1$ and $c = 9$

$y = a(-4-1)^2 + 9$

$-16 = a(-4-1)^2 + 9$

$-25 = a (-4-1)^2$

$-25 = a (-5)^2$

$-25 = 25a$

$a = -1$

So;

$y = -1(x-1)^2 + 9$

$-x^2 + 8$

However this is wrong the answer is $y = 8 + 2x - x^2$

Carefully multiply out the binomial square in $\ -1 \cdot (x-1)^2 \ + \ 9 \$ , collect the terms, and you will see that you have a different form of the correct answer... It does not equal $\ -x^2 \ + \ 8 \ .$ – RecklessReckoner Jan 29 '14 at 7:48
Yes, it's all right until the last step where you simplify $y=-1(x-1)^2+9$ to $-x^2+8$ -- looks like you dropped the term $2ab$ in $(a+b)^2=a^2+2ab+b^2.$ – coffeemath Jan 29 '14 at 7:51
I get it now thank you both. It becomes $-(x^2-2x+1)+9 -> -x^2 + 2x + 8$ – Orbit Jan 29 '14 at 7:53
Assume the rule is $f(x)=ax^2+bx+c$ for some constant $a,b$ and $c$. Since its graph is passing through points $$(1,9),~~(-4,-16),~~(4,0)$$ so we have $$f(1)=9,~~f(-4)=-16,~~f(4)=0$$ so we get $$a(1)^2+b(1)+c=9\to a+b+c=9\\a(-4)^2+b(-4)+c=-16\to 16a-4b+c=-16\\ a(4)^2+b(4)+c=0\to 16a+4b+c=0$$ Now solve the latter three equations simultaneously to find $a,b$ and $c$.