Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 1058

\[y = x^{2};\ \ \ \ \ \ (0;\ - 9):\]

\[1)\ y^{'}(x) = \left( x^{2} \right)^{'} = 2x;\]

\[y^{'}\left( x_{0} \right) = 2x_{0};\]

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

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

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

\[= 2x_{0}x - x_{0}^{2}.\]

\[2)\ Проходящие\ через\ точку:\]

\[- 9 = 2x_{0} \bullet 0 - x_{0}^{2}\]

\[x_{0}^{2} = 9\]

\[x_{0} = \pm 3.\]

\[k_{1} = 2 \bullet ( - 3) = - 6;\]

\[k_{2} = 2 \bullet 3 = 6.\]

\[3)\ tg\ a = \left| \frac{k_{1} - k_{2}}{1 + k_{1} \bullet k_{2}} \right| =\]

\[= \left| \frac{- 6 - 6}{1 - 6 \bullet 6} \right| = \left| \frac{- 12}{1 - 36} \right| = \frac{12}{35}.\]

\[Ответ:\ \ \frac{12}{35}.\]