Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-6,-2)\\(0,-2)\\(-3,4)$
Substitute: $-3=a(-6)^2+b(-6)+c\\-2=a(-4)^2+b(-4)+c\\4=a(-3)^2+b(-3)+c$
We have the system: $36a-6b+c=-2\\16a-4b+c=-2\\9a-3b+c=4$
The last equation can be written as: $c=3b-9a+4$
Substitute this into the second equation:
$16a-4b+(3b-9a+4)=-2\\
\rightarrow b=10a$
Substitute $b$ into the the third equation:
$9a-3(10a)+(24a-2)=4\\
\rightarrow a=2$
Hence, $a=2\\b=20\\c=46$
Substitute back to the initial equation: $y=2x^2+20x+46$
