Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-6,29)\\(-4,12)\\(2,-3)$
Substitute: $29=a(-6)^2+b(-6)+c\\12=a(-4)^2+b(-4)+c\\-3=a(2)^2+b(2)+c$
We have the system: $36a-6b+c=29\\16a-4b+c=12\\4a+2b+c=-3$
Subtract the third equation from the second equation:
$12a-6b=15\\
\rightarrow 4a-2b=5$ (1)
Subtract the third equation from the first equation:
$32a-8b=32\\
\rightarrow 4a-b=4$ (2)
Subtract equation (1) from equation (2):
$b=-1$
Substitute $b$:
$4a-(-1)=4\\a=\frac{3}{4}$
Find $c$:
$4(\frac{3}{4})+2(-1)+c=-3\\
\rightarrow c=-4$
Hence, $a=\frac{3}{4}\\b=-1\\c=-4$
Substitute back to the initial equation: $y=\frac{3}{4}x^2-x-4$
