Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 222

\[1)\ k = 2,\ \ \ x_{0} = 1,\ \ \ y_{0} = - 1:\]

\[y = 2x + b\]

\[- 1 = 2 + b\]

\[b = - 3.\]

\[Ответ:\ \ y = 2x - 3.\]

\[2)\ k = 3,\ \ \ x_{0} = - 2,\ \ \ y_{0} = 1:\]

\[y = 3x + b\]

\[1 = - 6 + b\]

\[b = 7.\]

\[Ответ:\ \ y = 3x + 7.\]

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

\[y = - 2x + b\]

\[- 4 = - 6 + b\]

\[b = 2.\]

\[Ответ:\ \ y = - 2x + 2.\]

\[4)\ k = \frac{1}{3},\ \ \ x_{0} = 1,\ \ \ \ y_{0} = 0:\]

\[y = \frac{1}{3}x + b\]

\[0 = \frac{1}{3} + b\]

\[b = - \frac{1}{3}.\]

\[Ответ:\ \ y = \frac{1}{3}x - \frac{1}{3}.\]

\[5)\ k = \frac{2}{3},\ \ \ x_{0} = \frac{1}{3},\ \ \ y_{0} = \frac{1}{3}:\]

\[y = \frac{2}{3}x + b\]

\[\frac{1}{3} = \frac{2}{9} + b\]

\[b = \frac{1}{9}.\]

\[Ответ:\ \ \ y = \frac{2}{3}x + \frac{1}{9}.\]

\[6)\ k = - \frac{1}{2},\ \ \ x_{0} = 0,\ \ \ y_{0} = 0:\]

\[y = - \frac{1}{2}x + b\]

\[0 = 0 + b\]

\[b = 0.\]

\[Ответ:\ \ y = - \frac{1}{2}\text{x.}\]