Решебник по алгебре 9 класс Мерзляк Задание 384

\[\boxed{\text{384\ (384).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ y = \frac{8x + 2x^{2} - x^{3}}{x};\ \ \ x \neq 0\]

\[y = - x^{2} + 2x + 8;\ \ \]

\[a = - 1 < 0 - ветви\ вниз.\]

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

\[y_{0} = - 1 + 2 + 8 = 9;\]

\[вершина\ параболы - \text{\ \ }(1;9).\]

\[y = 0:\ \ \]

\[- x^{2} + 2x + 8 = 0\]

\[x_{1} + x_{2} = 2,\ \ x_{1} = 4\]

\[x_{1} \cdot x_{2} = - 8,\ \ x_{2} = - 2\]

\[( - 2;0),\ \ (4;0).\]

\[x = 0:\ \ \]

\[y = 8,\ \ (0;8).\]

\[y( - 1) = - 1 - 2 + 8 = 5.\]

\[2)\ y = \frac{x^{3} - 8}{x - 2} - 3\]

\[y = \frac{(x - 2)\left( x^{2} + 2x + 4 \right)}{(x - 2)} - 3 =\]

\[= x^{2} + 2x + 4 - 3;\ \ \ \ x \neq 2\]

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

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

\[a = 1 > 0 - ветви\ вверх;\]

\[( - 1;0) - вершина\ параболы.\]

\[3)\ y = \frac{x^{4} - 16}{x^{2} - 4}\]

\[y = \frac{\left( x^{2} - 4 \right)\left( x^{2} + 4 \right)}{x^{2} - 4};\ \ x \neq \pm 2\]

\[y = x^{2} + 4\]

\[a = 1 > 0 - ветви\ вверх.\]

\[(4;0) - вершина\ параболы.\]

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

\[y = \frac{\left( x^{2} - 1 \right)\left( x^{2} + 5 \right)}{\left( x^{2} - 1 \right)};\ \ x \neq \pm 1\]

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

\[a = 1 > 0 - ветви\ вверх;\]

\[(0;5) - вершина\ параболы.\]