2.
Theo đề bài ta có:
$
\left\{\begin{matrix}
\dfrac{-b}{2a} = -1\\
\dfrac{4ac - b^2}{4a} = \dfrac{3}{2}
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
b = 2a\\
\dfrac{8a - b^2}{4a} = \dfrac{3}{2}
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
b = 2a\\
\dfrac{8a - 4a^2}{4a} = \dfrac{3}{2}
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
b = 2a\\
2 - a = \dfrac{3}{2}
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
b = 2a\\
a = \dfrac{1}{2}
\end{matrix}\right. \\ \Leftrightarrow \left\{\begin{matrix}
a = \dfrac{1}{2}\\
b = 1
\end{matrix}\right. \\ y = \frac{1}{2}x^2 + x + 2 $