Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-6,-1)\\(-3,-4)\\(3,8)$
Substitute: $-1=a(-6)^2+b(-6)+c\\-4=a(-3)^2+b(-3)+c\\8=a(3)^2+b(3)+c$
We have the system: $36a-6b+c=-1\\9a-3b+c=-4\\9a+3b+c=8$
Add the second equation to the third equation:
$18a+2c=4\\
\rightarrow 9a+c=2$ (1)
Multiply the third equation by $2$ and add to the first equation:
$54a+3c=15\\
\rightarrow 18a+c=5$ (2)
Subtract equation (1) from equation (2):
$3a=1\\a=\frac{1}{3}$
Substitute $a$:
$9(\frac{1}{3})+c=2\\c=-1$
Find b:
$9(\frac{1}{3})+b+(-1)=8 \\
\rightarrow b=2$
Hence, $a=\frac{1}{3}\\b=2\\c=-1$
Substitute back to the initial equation: $y=\frac{1}{3}x^2+2x-1$
