Finding the rule of a quadratic graph

Question

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 $

Could I please have some help/advice?

Answer 1

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 $c$.