Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 138

\[\boxed{\mathbf{138}.}\]

\[1)\ y = x^{2} + 3\]

\[a > 0 - ветви\ вверх;\]

\[x_{0} = 0;\ \ y_{0} = 3.\]

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

\[a > 0 - ветви\ вверх.\]

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

\[3)\ y = x^{2} - 6x + 9 = (x - 3)^{2}\]

\[a > 0 - ветви\ вверх.\]

\[x_{0} = 3;\ \ y_{0} = 0.\]

\[4)\ y = x^{2} + x + \frac{1}{4} = \left( x + \frac{1}{2} \right)^{2}\]

\[a > 0 - ветви\ вверх.\]

\[x_{0} = - \frac{1}{2} = - 0,5;\ \ y_{0} = 0.\]

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

\[a > 0 - ветви\ вверх.\]

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

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

\[a > 0 - ветви\ вверх.\]

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

\[7)\ y = (x - 3)(x + 1) = x^{2} -\]

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

\[a > 0 - ветви\ вверх.\]

\[x_{0} = \frac{2}{2} = 1;y_{0} = 1 - 2 - 3 =\]

\[= - 4.\]

\[8)\ y = (x - 1)(x + 5) = x^{2} -\]

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

\[a > 0 - ветви\ вверх.\]

\[x_{0} = - \frac{4}{2} = - 2;y_{0} = 4 - 8 -\]

\[- 5 = - 9.\]

\[9)\ y = x^{2} + 3x - 4\]

\[a > 0 - ветви\ вверх.\]

\[x_{0} = - \frac{3}{2} = - 1,5;\]

\[y_{0} = \frac{9}{4} - 3 \cdot \frac{3}{2} - 4 = \frac{9}{4} - \frac{18}{4} -\]

\[- 4 = - \frac{9}{4} - 4 = - 2,25 - 4 =\]

\[= - 6,25.\]

\[10)\ y = x^{2} - 3x - 4\]

\[a > 0 - ветви\ вверх.\]

\[x_{0} = \frac{3}{2} = 1,5;\]

\[y_{0} = \frac{9}{4} - 3 \cdot \frac{3}{2} - 4 = \frac{9}{4} + \frac{9}{2} - 4 =\]

\[= \frac{9}{4} - \frac{18}{4} - 4 = - \frac{9}{4} - 4 =\]

\[= - 6,25.\]