Решебник по алгебре 9 класс Макарычев Задание 247

 

\[\boxed{\text{247}\text{\ (247)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = x^{2} + px + q\]

\[\textbf{а)}\ по\ теореме\ Виета:\ \]

\[\left\{ \begin{matrix} x_{1} \cdot x_{2} = q = 12\ \ \ \ \ \ \ \\ x_{1} + x_{2} = - p = - 7 \\ \end{matrix} \right.\ ;\]

\[\textbf{б)}\ (0;6):\ \ \]

\[\ q = 6;\]

\[(2;0):\ \ \ \ \]

\[0 = 4 + 2p + q;\]

\[\left\{ \begin{matrix} q = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4 + 2p + 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} q = 6\ \ \ \ \ \ \ \ \\ 2p = - 10 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} q = 6\ \ \ \\ p = - 5 \\ \end{matrix} \right.\ ;\]

\[\textbf{в)}\ координаты\ вершины:\ \ \]

\[(6;24).\]

\[\left\{ \begin{matrix} 6 = - \frac{p}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 24 = 36 + 6p + q \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} p = - 12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 24 = 36 - 6 \cdot 12 + q \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} p = - 12 \\ q = 60\ \ \ \\ \end{matrix} \right.\ .\]

\[\boxed{\text{247.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

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

\[x^{2} - 3 = 0\]

\[x^{2} = 3\]

\[x = \pm \sqrt{3}.\]

\[\textbf{а)}\ y > 0\ \ при\ \ \]

\[x \in \left( - \infty;\ - \sqrt{3} \right) \cup \left( \sqrt{3}; + \infty \right);\]

\[\textbf{б)}\ y < 0\ \ \ при\ \]

\[\ x \in \left( - \sqrt{3};\sqrt{3} \right).\]