`a)` `y=ax^2+bx+c (P)`
`text{(P) đi qua A(-1;2);B(2;-7);C(5;1)}`
$⇒\begin{cases}a-b+c=2\\4a+2b+c=-7\\25a+5b+c=1\end{cases}$
$⇔\begin{cases}a-b+c=2\\3a+3b=-7-2=-9\\21a+3b=1+7=8\end{cases}$
$⇔\begin{cases}c=2-a+b\\3a+3b=-7-2=-9\\18a=8+9=17\end{cases}$
$⇔\begin{cases}c=2-\dfrac{17}{18}-\dfrac{71}{18}=\dfrac{-26}{9}\\b=-3-\dfrac{17}{18}=\dfrac{-71}{18}\\a=\dfrac{17}{18}\end{cases}$
Vậy: `a={17}/18;b=-{71}/{18};c=-{26}/{9}`
`y={17}/{18} x^2 - {71}/{18} x -{26}/9`
`text{b) (P) đi qua M(3;-1)}`
`=>9a+3b+c=-1<=>c=-1-9a-3b (1)`
`text{(P) có đỉnh I(-2;4)}`
`=> -b/{2a}=-2<=>b=4a (2)`
`text{Thay x=-2;y=4 vào (P) ta có:}`
`4a-2b+c=4<=>c=4-4a+2b (3)`
`text{Từ (1);(2);(3)}`
`=>-1-9a-3b=4-4a+2b`
$⇔5a+5b=-5⇔a+b=-1$
`<=>a+4a=-1<=>5a=-1<=>a={-1}/5`
`=>b=4a={-4}/5`
`=>c=4+4/ 5+2.{-4}/5={16}/5`
Vậy `a=-1/ 5 ;b= -4/ 5 ; c={16}/5`
`y=-1 /5 x^2-4 /5 x + {16}/5`
