Решебник по алгебре и начала математического анализа 10 класс Алимов Задание 1479

\[\boxed{\mathbf{1479}\mathbf{.}}\]

\[y = ax^{2} + bx + c\]

\[y(0) = - 3:\]

\[- 3 = a \bullet 0^{2} + b \bullet 0 + c\]

\[c = - 3.\]

\[y(3) = 0:\]

\[0 = a \bullet 3^{2} + b \bullet 3 - 3\]

\[9a + 3b - 3 = 0\]

\[3a + b - 1 = 0\]

\[b = 1 - 3a.\]

\[y( - 2) = 15:\]

\[15 = a \bullet ( - 2)^{2} + (1 - 3a) \bullet ( - 2) - 3\]

\[4a - 2 + 6a - 3 = 15\]

\[10a = 20\]

\[a = 2;\]

\[b = 1 - 3 \bullet 2 = 1 - 6 = - 5.\]

\[Подставим:\]

\[y = 2x^{2} - 5x - 3;\]

\[Вершина\ параболы:\]

\[x = - \frac{b}{2a} = - \frac{- 5}{2 \bullet 2} = \frac{5}{4} = 1\frac{1}{4};\]

\[y\left( \frac{5}{4} \right) = 2 \bullet \frac{25}{16} - 5 \bullet \frac{5}{4} - 3 =\]

\[= \frac{25}{8} - \frac{50}{8} - 3 =\]

\[= - 3 - 3\frac{1}{8} = - 6\frac{1}{8}.\]