In order to find the values of a, b and c, we need to use the three ordered pairs given in the equation and solve the system with 3 variables and 3 equations:
[tex]\begin{gathered} y=ax^2+bx+c \\ (-2,-1)\colon \\ -1=4a-2b+c \\ (1,11)\colon \\ 11=a+b+c \\ (2,27)\colon \\ 27=4a+2b+c \end{gathered}[/tex]
Adding the first and third equations:
[tex]\begin{gathered} 26=8a+2c \\ 13=4a+c \\ c=13-4a \end{gathered}[/tex]
Using this value of c in the first and second equations:
[tex]\begin{gathered} -1=4a-2b+13-4a \\ -1=-2b+13 \\ 2b=13+1 \\ 2b=14 \\ b=7 \\ \\ 11=a+b+13-4a \\ 11=a+7+13-4a \\ 11=-3a+20 \\ 3a=20+11 \\ 3a=31 \\ a=\frac{31}{3} \end{gathered}[/tex]
Calculating the value of c:
[tex]\begin{gathered} c=13-4a \\ c=13-4\cdot\frac{31}{3} \\ c=13-\frac{124}{3} \\ c=\frac{39-124}{3} \\ c=-\frac{85}{3} \end{gathered}[/tex]
So we have a = 31/3, b = 7 and c = -85/3.
