\(M\left(0;1\right)\in\left(P\right)\Rightarrow c=1\)
Lại có \(I\left(-1;2\right)\) là đỉnh \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{b}{2a}=-1\\-\dfrac{b^2-4ac}{4a}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b^2+4a=0\\b=2a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-1\\b=-2\end{matrix}\right.\left(\text{Vì }a\ne0\right)\)
\(\Rightarrow y=-x^2-2x+1\)