Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1476

\[\boxed{\mathbf{1476}\mathbf{.}}\]

\[y = ax^{2} + bx - 4;\]

\[y(1) = 0;\ y(4) = 0:\]

\[\left\{ \begin{matrix} 0 = a \bullet 1^{2} + b \bullet 1 - 4 \\ 0 = a \bullet 4^{2} + b \bullet 4 - 4 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} a + b - 4 = 0\ \ \ \ \ \ \ \\ 16a + 4b - 4 = 0 \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} b = 4 - a\ \ \ \ \ \ \ \\ 16a + 4b = 4 \\ \end{matrix} \right.\ \]

\[16a + 4(4 - a) = 4\]

\[16a + 16 - 4a = 4\]

\[12a = - 12\]

\[a = - 1;\]

\[b = 4 - ( - 1) = 4 + 1 = 5.\]

\[Наибольшее\ значение:\]

\[x_{0} = - \frac{b}{2a} = \frac{- 5}{2 \bullet ( - 1)} = \frac{5}{2};\]

\[y\left( x_{0} \right) = - 1 \bullet \left( \frac{5}{2} \right)^{2} + 5 \bullet \frac{5}{2} - 4 =\]

\[= - \frac{25}{4} + \frac{50}{4} - \frac{16}{4} = \frac{9}{4} = 2,25.\]

\[Ответ:\ \ y_{\max} = 2,25.\]