При каких значениях a и b парабола y=ax^2+bx-1 проходит через точки M

(-1; 3) и N (2; 4)?

\[y = ax^{2} + bx - 1;\ \ M\ ( - 1;3);\ \ \]

\[\text{N\ }(2;4)\]

\[\left\{ \begin{matrix}
a - b - 1 = 3\ \ \ \ \ \\
4a + 2b - 1 = 4 \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
a - b = 4\ \ | \cdot 4 \\
4a + 2b = 5\ \ \ \ \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
4a - 4b = 16 \\
4a + 2b = 5\ \ \\
\end{matrix} \right.\ \text{\ \ }( - )\]

\[\left\{ \begin{matrix}
- 6b = 11 \\
a = 4 + b \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
b = - \frac{11}{6} \\
a = \frac{13}{6}\text{\ \ \ \ } \\
\end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix}
b = - 1\frac{6}{5} \\
a = 2\frac{1}{6}\text{\ \ \ } \\
\end{matrix} \right.\ \]

\[Ответ:\ при\ a = 2\frac{1}{6};\ b = - 1\frac{5}{6}.\]