Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-2,4)\\(0,5)\\(1,-11)$
Substitute: $4=a(-2)^2+b(-2)+c\\5=a(0)^2+b(0)+c\\-11=a(1)^2+b(1)+c$
We have the system: $4a-2b+c=4\\c=5\\a+b+c=-11$
Substitute $c$ into the equations. We have:
$4a-2b=-1\\a+b= -16$
Multiply the second equation with $2$ and add to the first equation:
$6a=-41\\a=-\frac{11}{2}$
Substitute $a$ into the second equation to find $b$:
$\frac{-11}{2}+b=-16\\
b=\frac{-21}{2}$
Hence, $a=\frac{-11}{2}\\b=\frac{-21}{2}\\c=5$
Substitute back to the initial equation: $y=\frac{-11}{2}x^2-\frac{21}{2}x+5$
