Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Spot three points: $(-2,19)\\(2,-5)\\(4,-11)$
Substitute: $19=a(-2)^2+b(-2)+c\\-5=a(2)^2+b(2)+c\\-11=a(4)^2+b(4)+c$
We have the system: $4a-2b+c=19\\4a+2b+c=-5\\16a+4b+c=-11$
Multiply the first equation by $-1$
$-9a+3b-c=-19$
Add it to the second equation:
$4b=-24\\
\rightarrow b=-6$
Substitute $b=-6$ into the second and third equation:
$4a+c=7\\16a+c=13$
Multiply both sides of the second equation by $-1$ and add it to the third one:
$12a=6\\
\rightarrow a=\frac{1}{2}$
Find $c$:
$4(\frac{1}{2})+2(-6)+c=-5\\
\rightarrow c=5$
Hence, $a=\frac{1}{2}\\b=-6\\c=5$
Substitute back to the initial equation: $y=\frac{1}{2}x^2-6x+5$
