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

Ответ:

(-1; 3) и N (2; 4)?

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

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

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

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

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

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

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

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

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