ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 223

\[1)\ a = \frac{\pi}{4}\ ,\ \ \ x_{0} = - 3,\ \ \ y_{0} = 2:\]

\[y = tg\frac{\pi}{4} \bullet x + b = x + b\]

\[2 = - 3 + b\]

\[b = 5.\]

\[Ответ:\ \ y = x + 5.\]

\[2)\ a = \frac{3\pi}{4};x_{0} = - 1;\ y_{0} = - 1:\]

\[y = tg\frac{3\pi}{4} \bullet x + b = - x + b\]

\[- 1 = 1 + b\ \]

\[b = - 2.\]

\[Ответ:\ \ y = - x - 2.\]

\[3)\ a = \frac{\pi}{6},\ \ \ x_{0} = 6,\ \ \ y_{0} = - 5:\]

\[y = tg\frac{\pi}{6} \bullet x + b = \frac{\sqrt{3}}{3}x + b\]

\[- 5 = 2\sqrt{3} + b\]

\[b = - 5 - 2\sqrt{3}.\]

\[Ответ:\ \ y = \frac{\sqrt{3}}{3}x - 5 - 2\sqrt{3}.\]

\[4)\ a = \frac{2\pi}{3},\ \ \ x_{0} = 4,\ \ \ y_{0} = - 3:\]

\[y = tg\frac{2\pi}{3} \bullet x + b = - \sqrt{3}x + b\]

\[- 3 = - 4\sqrt{3} + b\]

\[b = 4\sqrt{3} - 3.\]

\[Ответ:\ \ y = - \sqrt{3}x + 4\sqrt{3} - 3.\]