Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 137

\[\boxed{\mathbf{137}.}\]

\[Координаты\ вершины\ \]

\[параболы\ \left( x_{0};y_{0} \right).\]

\[x_{0} = - \frac{b}{2a}.\]

\[1)\ y = (x - 1)^{2} + 5 = x^{2} -\]

\[- 2x + 1 + 5 = x^{2} - 2x + 6\]

\[x_{0} = \frac{2}{2} = 1;\]

\[y_{0} = 5.\]

\[Ответ:(1;5).\]

\[2)\ y = - (x + 2)^{2} - 3\]

\[x_{0} = - 2;\]

\[y_{0} = - 3.\]

\[Ответ:( - 2;\ - 3).\]

\[3)\ y = - (x + 3)²\]

\[x_{0} = - 3;\]

\[y_{0} = 0.\]

\[Ответ:( - 3;0).\]

\[4)\ y = x^{2} - 7\]

\[x_{0} = 0;\ \ y_{0} = - 7.\]

\[Ответ:(0;\ - 7).\]

\[5)\ y = 2x^{2} - 4x + 1\]

\[x_{0} = \frac{4}{4} = 1;\]

\[y_{0} = 2 - 4 + 1 = - 1.\]

\[Ответ:(1;\ - 1).\]

\[6)\ y = 3x^{2} + 6x - 7\]

\[x_{0} = - \frac{6}{6} = - 1;\]

\[y_{0} = 3 - 6 - 7 = - 10.\]

\[Ответ:( - 1;\ - 10).\]

\[7)\ y = - 4x^{2} + 16x - 2\]

\[x_{0} = \frac{16}{8} = 2;\]

\[y_{0} = - 16 + 32 - 2 = 14.\]

\[Ответ:(2;\ 14).\]

\[8)\ y = - 5x^{2} - 20x - 13\]

\[x_{0} = - \frac{20}{10} = - 2;\]

\[y_{0} = - 20 + 40 - 13 = 7.\]

\[Ответ:( - 2;7).\]