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МатематикаГДЗ по алгебре 9 класс Мерзляк Задание 345
Задание 345
\[\boxed{\text{345\ (345).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[1)\ y = x^{2} - 4x - 5\]
\[a = 1 > 0 - \ \ ветви\ \]
\[направлены\ вверх.\]
\[x_{0} = \frac{4}{2} = 2;\]
\[y_{0} = 4 - 8 - 5 = - 9.\]
\[Ox:\ \]
\[x^{2} - 4x - 5 = 0\]
\[x_{1} + x_{2} = 4,\ \ \]
\[x_{1} = 5,\ \ (5;0)\]
\[x_{1}x_{2} = - 5,\ \ \]
\[x_{2} = - 1,\ \ ( - 1;0)\]
\[Oy:\ \ \]
\[y(0) = - 5;\ \ (0;\ - 5).\]
\[y(1) = 1 - 4 - 5 = - 8\]
\[2)\ y = - x^{2} + 2x + 3\]
\[a = - 1 < 0 - \ \ ветви\ вниз.\]
\[x_{0} = \frac{- 2}{- 2} = 1;\]
\[y_{0} = - 1 + 2 + 3 = 4.\]
\[Ox:\ \ \]
\[- x^{2} + 2x + 3 = 0\]
\[x_{1} + x_{2} = 2,\ \ x_{1} = 3,\ \ \]
\[(3;0)\]
\[x_{1}x_{2} = - 3,\ \ x_{2} = - 1,\ \ \]
\[( - 1;0).\]
\[Oy:\ \ \ \]
\[y = 3;\ \ \ (0;\ 3).\]
\[3)\ y = 6x - x^{2} = - x^{2} + 6x\ \]
\[a = - 1 < 0 - \ ветви\ вниз.\]
\[x_{0} = \frac{- 6}{- 2} = 3;\]
\[y_{0} = 18 - 9 = 9.\]
\[Ox:\ \ \]
\[- x^{2} + 6x = 0\]
\[x(6 - x) = 0\ \ \]
\[x = 0,\ \ x = 6,\ \ \]
\[(0;0),\ \ (6;0).\]
\[Oy:\ \ \ \ \]
\[y = 0;\ \ (0;\ 0).\]
\[y(1) = 6 - 1 = 5;\]
\[y(2) = 12 - 4 = 8.\]
\[4)\ y = 2x^{2} - 8x + 8\]
\[a = 2 > 0 - \ ветви\ вверх.\]
\[x_{0} = \frac{8}{4} = 2;\]
\[y_{0} = 8 - 16 + 8 = 0.\]
\[Ox:\ \ \]
\[2x^{2} - 8x + 8 = 0\ \ |\ :2\]
\[x^{2} - 4x + 4 = 0\]
\[(x - 2)^{2} = 0,\ \ x = 0,\ \ \]
\[(2;0)\]
\[Oy:\ \ \ \]
\[y = 8\ \ (0;8).\]
\[y(1) = 8 - 8 + 2 = 2.\]
\[5)\ y = x^{2} - 2x + 4\]
\[a = 1 > 0\ \ ветви\ вверх.\]
\[x_{0} = \frac{2}{2} = 1;\]
\[y_{0} = 1 - 2 + 4 = 3.\]
\[Ox:\ \ \]
\[x^{2} - 2x + 4 = 0\]
\[D = 4 - 16 < 0 - \ \ точек\ нет.\]
\[Oy:\ \ \ \]
\[y = 4\ \ (0;4).\]
\[y( - 1) = 1 + 2 + 4 = 7.\]
\[6)\ y = - \frac{1}{2}x^{2} + 3x - 4\]
\[a = - \frac{1}{2} < 0 - \ ветви\ вниз.\]
\[x_{0} = \frac{- 3}{- 2 \cdot \frac{1}{2}} = 3;\]
\[y_{0} = - \frac{1}{2} \cdot 9 + 9 - 4 = 0,5.\]
\[Ox:\ \ \]
\[- \frac{1}{2}x^{2} + 3x - 4 = 0\ \ \ | \cdot ( - 2)\]
\[x^{2} - 6x + 8 = 0\]
\[x_{1} + x_{2} = 6,\ \ x_{1} = 4,\ \ \]
\[(4;0)\]
\[x_{1}x_{2} = 8,\ \ x_{2} = 2,\ \ (2;0).\]
\[\text{Oy}:\ \ \ \]
\[y = - 4;\ \ (0;\ - 4).\]
\[7)\ y = x^{2} - 6x + 5\]
\[a = 1 > 0 - \ \ ветви\ вверх\]
\[x_{0} = \frac{6}{2} = 3\]
\[y_{0} = 9 - 18 + 5 = - 4\]
\[Ox:\ \ x^{2} - 6x + 5 = 0\]
\[x_{1} + x_{2} = 6,\ \ x_{1} = 5,\ \ \]
\[(5;0)\]
\[x_{1}x_{2} = 5,\ \ x_{2} = 1,\ \ (1;0).\]
\[\text{Oy}:\ \ \]
\[y = 5;\ \ (0;5).\]
\[y(2) = 4 - 12 + 5 = - 3.\]
\[8)\ y = 2x^{2} - 5x + 2\]
\[a = 2 > 0 - ветви\ вверх.\]
\[x_{0} = \frac{5}{4} = 1\frac{1}{4};\]
\[y_{0} = \frac{2 \cdot 25}{16} - \frac{25}{4} + 2 =\]
\[= - \frac{9}{8} = - 1\frac{1}{8}.\]
\[Ox:\ \ \ \]
\[2x^{2} - 5x + 2 = 0\]
\[D = 25 - 16 = 9\]
\[x = \frac{5 \pm 3}{4};\ \ \]
\[x_{1} = 2;\ \ x_{2} = \frac{1}{2}.\]
\[(2;0);\ \ \left( \frac{1}{2};0 \right).\]
\[\text{Oy}:\ \ \ \]
\[y = 2;\ \ (0;2).\]
