Постройте график функции y=-корень из x. Найдите координаты точки пересечения графика этой функции с прямой x+4y+3=0.

Question

Вопрос:

Ответ:

Accepted Answer

\[y = - \sqrt{x}\]

\[x\]	\[9\]	\[4\]	\[1\]	\[0\]
\[y\]	\[- 3\]	\[- 2\]	\[- 1\]	\[0\]

\[\left\{ \begin{matrix} x + 4y + 3 = 0 \\ y = - \sqrt{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 4y = - x - 3 \\ y = - \sqrt{x}\text{\ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = \frac{- x - 3}{4} \\ y = - \sqrt{x}\text{\ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\frac{- (x + 3)}{4} = - \sqrt{x}\ \ \ \ \ \ \ \ | \cdot 4\]

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

\[x^{2} + 6x + 9 = 16x\]

\[x^{2} - 10x + 9 = 0\]

\[D = 100 - 36 = 64\]

\[x_{1,2} = \frac{10 \pm 8}{2} = 9;\ \ 1.\]

\[x_{1} = 9;\ \ \ \ \ \ \ \ \ x_{2} = 1\]

\[y(9) = - \sqrt{9} = - 3\]

\[y(1) = - \sqrt{1} = - 1\]

\[Ответ:(9; - 3);\ \ (1; - 1)\text{.\ }\]