(P): y=ax^2+bx+c(a<>0)
Theo đề, ta có:
\(\left\{{}\begin{matrix}\dfrac{-b}{2a}=3\\a\cdot\left(-5\right)^2+b\cdot\left(-5\right)+c=6\\a\cdot0^2+b\cdot0+c=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\c=-2\\25a-5\cdot\left(-2a\right)-2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-2\\b=-2a\\35a=8\end{matrix}\right.\)
=>c=-2; a=8/35; b=-16/35