Thay x=8 và y=0 vào (P), ta được:
\(a\cdot8^2+b\cdot8+c=0\)
=>64a+8b+c=0(1)
Đỉnh I(-6;12) nên ta có:
\(\left\{{}\begin{matrix}-\dfrac{b}{2a}=-6\\-\dfrac{b^2-4ac}{4a}=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=12a\\b^2-4ac=-48a\end{matrix}\right.\)(2)
Từ (1),(2) ta có hệ phương trình:
\(\left\{{}\begin{matrix}b=12a\\b^2-4ac=-48a\\64a+8b+c=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=12a\\144a^2+48a-4ac=0\\64a+96a+c=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=12a\\36a+12-c=0\\160a+c=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=12a\\36a-c=-12\\160a+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=12a\\196a=-12\\c=-160a\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=-\dfrac{3}{49}\\b=-\dfrac{36}{49}\\c=-160\cdot\dfrac{-3}{49}=\dfrac{480}{49}\end{matrix}\right.\)
