pt có dạng
y=ax^2 +bx +c
đỉnh I(2;-1) => \(\left\{{}\begin{matrix}4a+2b+c=-1\\\left(-\dfrac{b}{2a}\right)=2\end{matrix}\right.\)
qua M(4;3) \(\Rightarrow16a+4b+c=3\)
Hệ
\(\left\{{}\begin{matrix}4a+2b+c=-1\\b=-4a\\16a+4b+c=3\end{matrix}\right.\)
giải hê
\(\left\{{}\begin{matrix}4a-8a+c=-1\\b=-4a\\16a-16a+c=3\end{matrix}\right.\)
\(\left\{{}\begin{matrix}c=3\\a=1\\b=-4\end{matrix}\right.\) => y=x^2-4x+3