Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-2,-4)\\(0,-10)\\(3,-7)$
Substitute: $-4=a(-2)^2+b(-2)+c\\-10=a(0)^2+b(0)+c\\-7=a(3)^2+b(3)+c$
We have the system: $4a-2b+c=-4\\c=-10\\9a+3b+c=-7$
Substitute $c$ into the equations. We have:
$4a-2b=6\\9a+3b= 3$
Multiply the first equation with $3$ and the second equation with $2$:
$12-6b=18\\18a+6b=6$
Add equations:
$30a=24\rightarrow a=\frac{4}{5}$
Substitute $a$ into the first equation:
$4\frac{4}{5}-2b=6\\
-2b=\frac{14}{5}\\
b=\frac{-7}{5}$
Hence, $a=\frac{4}{5}\\b=\frac{-7}{5}\\c=-10$
Substitute back to the initial equation: $y=\frac{4}{5}x^2-\frac{7}{5}x-10$
