Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-4,-6)\\(0,-2)\\(2,6)$
Substitute: $-6=a(-4)^2+b(-4)+c\\-2=a(0)^2+b(0)+c\\6=a(2)^2+b(2)+c$
We have the system: $16a-4b+c=-6\\c=-2\\4a+2b+c=6$
Substitute $c$ into the equations, we have:
$16a-4b=-4\\4a+2b=8$
Multiply the second equation with $2$ and add to the first equation:
$24a=12\rightarrow a=\frac{1}{2}$
Substitute $a$ into the second equation:
$4(\frac{1}{2})+2b=8\\
2b=6\\\rightarrow b=3$
Hence, $a=\frac{1}{2}\\b=3\\c=-2$
Substitute back to the initial equation: $y=\frac{1}{2}x^2+3x-2$
