Chắc là \(a\ne0\)
Pt hoành độ giao điểm: \(ax^2+bx+c=0\Rightarrow\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)
Do tọa độ đỉnh là (1;8) \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{b}{2a}=1\\\dfrac{4ac-b^2}{4a}=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\4ac-\left(-2a\right)^2=32a\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-2a\\c=a+8\end{matrix}\right.\)
Mà \(MN=4\Leftrightarrow\left|x_1-x_2\right|=4\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=16\)
\(\Leftrightarrow\left(\dfrac{-2a}{a}\right)^2-4\dfrac{a+8}{a}=16\)
\(\Leftrightarrow a=-2\Rightarrow b=4\Rightarrow c=6\)
