Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(4,8)\\(7,-4)\\(8,0)$
Substitute: $8=a(4)^2+b(4)+c\\-4=a(7)^2+b(7)+c\\0=a(8)^2+b(8)+c$
We have the system: $16a+4b+c=8\\49a+7b+c=-4\\64a+8b+c=0$
This gives:
$33a+3b=-12\\
\rightarrow 11a+b=-4$ (1)
Subtract the second equation:
$15a+b=4$ (2)
Subtract equation (1) from equation (2):
$4a=8\\
\rightarrow a=2$
Find $b$:
$30+b=4\\
\rightarrow b=-26$
Find $c$:
$16(2)+8(-26)+c=0\\
\rightarrow c=80$
Hence, $a=2\\b=-26\\c=80$
Substitute back to the initial equation: $y=2x^2-26x+80$
