Answer :

naǫ
[tex]y=ax^2+bx+c \\ \\ (0,3) \\ 3=a \times 0^2+b \times 0 + c \\ 3=c \\ \\ \hbox{so the equation is:} \\ y=ax^2+bx+3 \\ \\ (-1,-2) \\ -2=a \times (-1)^2+b \times (-1) + 3 \\ -2=a-b+3 \\ -2-3=a-b \\ -5=a-b[/tex]

[tex](-2,-5) \\ -5=a \times (-2)^2+b \times (-2) + 3 \\ -5=4a-2b+3 \\ -5-3=4a-2b \\ -8=4a-2b \\ \\ \hbox{solve the system of equations:} \\ -5=a-b \ \ \ |\times (-2) \\ -8=4a-2b \\ \\ 10=-2a+2b \\ \underline{-8=4a-2b \ \ } \\ 10-8=4a-2a \\ 2=2a \\ \frac{2}{2}=a \\ a=1 \\ \\ -5=a-b \\ -5=1-b \\ -5-1=-b \\ -6=-b \\ b=6 \\ \\ \boxed{y=x^2+6x+3}[/tex]