Решебник по алгебре 11 класс Никольский Параграф 1. Функции и их графики Задание 45

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Год:2020-2021-2022
Тип:учебник

Задание 45

\[\boxed{\mathbf{45.}}\]

\[y = a\left( x - x_{1} \right)^{2} + y_{0}\]

\[\left\lbrack x_{0}; + \infty \right)\ при\ x_{1} < x_{2}:\]

\[y\left( x_{1} \right) = a\left( x_{1} - x_{0} \right)^{2} + y_{0};\]

\[y\left( x_{2} \right) = a\left( x_{2} - x_{0} \right)^{2} + y_{0};\]

\[y\left( x_{1} \right) - y\left( x_{2} \right) = a\left( x_{1} - x_{0} \right)^{2} +\]

\[+ y_{0} - a\left( x_{2} - x_{0} \right)^{2} - y_{0} =\]

\[= a\left( \left( x_{1} - x_{0} \right)^{2} - \left( x_{2} - x_{0} \right)^{2} \right);\]

\[x_{1} - x_{0} < x_{2} - x_{0}\]

\[\ \left( так\ как\ x_{1} < x_{2} \right);\]

\[\left( x_{1} - x_{0} \right)^{2} < \left( x_{2} - x_{0} \right)^{2}\ \]

\[\left( x_{1} - x_{0} \right)^{2} - \left( x_{2} - x_{0} \right)^{2} < 0.\]

\[\left( - \infty;x_{0} \right\rbrack\text{\ \ }при\ x_{1} < x_{2}:\]

\[y\left( x_{1} \right) - y\left( x_{2} \right) =\]

\[= a\left( \left( x_{1} - x_{0} \right)^{2} - \left( x_{2} - x_{0} \right)^{2} \right);\]

\[\left( x_{1} - x_{0} \right)^{2} > \left( x_{2} - x_{0} \right)^{2};\]

\[\left( x_{1} - x_{0} \right)^{2} - \left( x_{2} - x_{0} \right)^{2} > 0.\]

\[\textbf{а)}\ возрастает\ на\ \left\lbrack x_{0}; + \infty \right):\]

\[y\left( x_{1} \right) < y\left( x_{2} \right)\]

\[y\left( x_{1} \right) - y\left( x_{2} \right) < 0\]

\[при\ a > 0.\]

\[\textbf{б)}\ убывает\ на\ \left\lbrack x_{0}; + \infty \right):\]

\[y\left( x_{1} \right) > y\left( x_{2} \right)\]

\[y\left( x_{1} \right) - y\left( x_{2} \right) > 0\]

\[при\ a < 0.\]

\[\textbf{в)}\ возрастает\ на\ \left( - \infty;x_{0} \right\rbrack:\]

\[y\left( x_{1} \right) - y\left( x_{2} \right) < 0\]

\[при\ a < 0.\]

\[\textbf{г)}\ убывает\ на\ \left( - \infty;x_{0} \right\rbrack:\]

\[y\left( x_{1} \right) - y\left( x_{2} \right) > 0\]

\[при\ a > 0.\]

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