Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-2,-13)\\(2,3)\\(4,5)$
Substitute: $-13=a(-2)^2+b(-2)+c\\3=a(2)^2+b(2)+c\\5=a(4)^2+b(4)+c$
We have the system: $4a-2b+c=-13\\4a+2b+c=3\\16a+4b+c=5$
Add the second equation to the first equation:
$8a+2c=-10\\
\rightarrow 4a+c=-5$ (1)
Multiply the first equation by $2$ and add to the third equation:
$24a+3c=-21\\
\rightarrow 8a+c=-7$ (2)
Subtract equation (1) from equation (2):
$4a=-2\\a=\frac{-1}{2}$
Substitute $a$:
$4(\frac{-1}{2})+c=-5\\c=-3$
Find b:
$-2+2b-3=3\\
\rightarrow b=4$
Hence, $a=\frac{-1}{2}\\b=4\\c=-3$
Substitute back to the initial equation: $y=\frac{-1}{2}x^2+4x-3$
