Answer
See below
Work Step by Step
The standard form of the equation is: $y=ax^2+bx+c$
Given three points: $(-1,2)\\(4,-23)\\(2,-7)$
Substitute: $2=a(-1)^2+b(-1)+c\\-23=a(4)^2+b(4)+c\\-7=a(2)^2+b(2)+c$
We have the system: $a-b+c=2\\16a+4b+c=-23\\4a+2b+c=-7$
Subtract the first equation from the second equation:
$15a+5b=-25\\
\rightarrow 3a+b=-5$ (1)
Subtract the third equation from the second equation:
$12a+2b=-6\\
\rightarrow 6a+b=-3$ (2)
Substitute equation (1) from equation (2):
$3a=-3\\
\rightarrow a=-1$
Substitute $a$ to equation (1):
$3(-1)+b=-5\\
\rightarrow b=-2$
Find $c$:
$-1-(-2)+c=2\\
\rightarrow c=1$
Hence, $a=-1\\b=-2\\c=1$
Substitute back to the initial equation: $y=-x^2-x+1$
