Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(0,-22)\\(-6,2)\\(5,-12)$
Substitute: $-22=a(0)^2+b(0)+c\\-6=a(2)^2+b(2)+c\\-12=a(5)^2+b(5)+c$
We have the system: $c=-22\\4a+2b+c=-6\\25a+5b+c=-12$
Substitute $c=-22$:
$2a+b=8\\5a+b=2$
Multiply the first equation by $-1$ and add to the second one:
$3a=-6\\
\rightarrow a=-2$
Find b: $4(-2)+2b+(-22)=-12\\
\rightarrow b=12$
Hence, $a=-2\\b=12\\c=-22$
Substitute back to the initial equation: $y=-2x^2+12x-22=-2(x-3)^2-4$
Hence, the vertex is $(3,-4)$
