Question

f (x) = ax^2 + bx + c đi qua điểm A(1,2) và có đỉnh I(-1,1)

Answer 1

Theo đề, ta co hệ:

\(\left\{{}\begin{matrix}a+b+c=2\\-\dfrac{b}{2a}=-1\\-\dfrac{b^2-4ac}{4a}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+c=2\\b=2a\\b^2-4ac=-4a\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=2\\b=2a\\4a^2-4ac=-4a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b+c=2\\b=2a\\a-c=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=2\\b=2a\\c=a+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+2a+a+1=2\\b=2a\\c=a+1\end{matrix}\right.\)

=>a=1/4; b=1/2; c=5/4