Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-3,-2)\\(3,10)\\(6,-2)$
Substitute: $-2=a(-3)^2+b(-3)+c\\10=a(3)^2+b(3)+c\\-2=a(6)^2+b(6)+c$
We have the system: $9a-9b+c=-2\\9a+3b+c=10\\36a+6b+c=-2$
Subtract the second equation from the third equation:
$27a+3b=-12\\
\rightarrow 9a+b=-4$ (1)
Subtract the first equation from the second equation:
$6b=12\\
\rightarrow b=2$ (2)
Substitute $b$ in equation (1):
$9a+2=-4\\
\rightarrow a=-\frac{2}{3}$
Find $c$:
$9(\frac{-2}{3})+3(2)+c=-2\\
\rightarrow c=10$
Hence, $a=\frac{-2}{3}\\b=2\\c=10$
Substitute back to the initial equation: $y=\frac{-2}{3}x^2+2x+10$
