Đáp án:
\(\left( C \right):\,\,\,\,y = {x^2} - 4x + 7.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( C \right):\,\,\,y = a{x^2} + bx + c\\
\left( C \right)\,\,\,co\,\,\,truc\,\,\,doi\,\,\,xung:\,\,\,x = 2\\
\Rightarrow - \frac{b}{{2a}} = 2 \Leftrightarrow b = - 4a.\,\,\,\left( 1 \right)\\
\left( C \right)\,\,\,di\,\,\,qua\,\,\,A\left( {1;\,\,4} \right)\\
\Rightarrow 4 = a + b + c\,\,\,\,\,\left( 2 \right)\\
Dinh\,\,\,I\left( { - \frac{b}{{2a}};\,\, - \frac{\Delta }{{4a}}} \right)\,\,\,\,cua\,\,\,\,\left( C \right) \in y = 2x - 1\\
\Rightarrow {y_I} = 2.2 - 1 = 3\\
\Rightarrow I\left( {2;\,\,\,3} \right) \in \left( C \right)\\
\Rightarrow 4a + 2a + c = 3\,\,\,\,\,\,\left( 3 \right)\\
Tu\,\,\,\left( 1 \right),\,\,\,\left( 2 \right),\,\,\,\left( 3 \right)\,\,\,ta\,\,co:\\
\left\{ \begin{array}{l}
b = - 4a\\
a + b + c = 4\\
4a + 2a + c = 3
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
b = - 4\\
c = 7
\end{array} \right.\\
\Rightarrow \left( C \right):\,\,\,\,y = {x^2} - 4x + 7.
\end{array}\)
