Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(5,2)\\(0,2)\\(8,-6)$
Substitute: $5=a(2)^2+b(2)+c\\2=a(0)^2+b(0)+c\\-6=a(8)^2+b(8)+c$
We have the system: $4a+2b+c=5\\c=2\\64a+8b+c=-6$
Subtract the second equation from the first equation:
$25a+5b=0\\
\rightarrow 5a+b=0$ (1)
Subtract the first equation from the second equation:
$64a+8b=-8\\
\rightarrow 8a+b=-1$ (2)
Subtract equation (1) from equation (2):
$3a=-1\\
\rightarrow a=-\frac{1}{3}$
Find $b$:
$\frac{-25}{3})+5b+2=-2\\
\rightarrow b=\frac{5}{3}$
Hence, $a=\frac{-1}{3}\\b=\frac{5}{3}\\c=2$
Substitute back to the initial equation: $y=\frac{-1}{3}x^2+\frac{5}{3}x+2$
