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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?

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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 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 at 7:51
    
I get it now thank you both. It becomes $-(x^2-2x+1)+9 -> -x^2 + 2x + 8$ –  Orbit Jan 29 at 7:53

1 Answer 1

up vote 3 down vote accepted

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 $c$.

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