《2026年点金训练精讲巧练高中数学选择性必修第二册人教A版》

求曲线 $ y = 2x - x^3 $ 过点 $ (-1, -2) $ 的切线的方程.

答案： 解：$y^{\prime} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$
$ = \lim_{\Delta x \to 0} \frac{2(x + \Delta x) - (x + \Delta x)^{3} - 2x + x^{3}}{\Delta x}$
$ = \lim_{\Delta x \to 0} [2 - 3x^{2} - 3x\Delta x - (\Delta x)^{2}] = 2 - 3x^{2}$.
设切点坐标为$(x_{0}, 2x_{0} - x_{0}^{3})$，则切线方程为$y - 2x_{0} + x_{0}^{3} = (2 - 3x_{0}^{2})(x - x_{0})$.
因为切线过点$(-1, -2)$，所以$-2 - 2x_{0} + x_{0}^{3} = (2 - 3x_{0}^{2}) · (-1 - x_{0})$，
即$2x_{0}^{3} + 3x_{0}^{2} = 0$，解得$x_{0} = 0$或$x_{0} = -\frac{3}{2}$.
所以切点坐标为$(0, 0)$或$(-\frac{3}{2}, \frac{3}{8})$.
当切点坐标为$(0, 0)$时，切线斜率$k = \frac{-2 - 0}{-1 - 0} = 2$，切线方程为$y = 2x$；
当切点坐标为$(-\frac{3}{2}, \frac{3}{8})$时，切线斜率$k = \frac{3}{8} - (-2) = \frac{19}{4} ÷ (-\frac{3}{2} - (-1)) = -\frac{19}{4} × \frac{1}{-\frac{1}{2}} = \frac{19}{2} × \frac{1}{2}（此处原解析计算有误，应为\frac{\frac{3}{8}-(-2)}{-\frac{3}{2}-(-1)}=\frac{\frac{19}{8}}{-\frac{1}{2}}=-\frac{19}{4}）$，切线方程为$y + 2 = -\frac{19}{4}(x + 1)$，即$19x + 4y + 27 = 0$.
综上可知，曲线$y = 2x - x^{3}$过点$(-1, -2)$的切线的方程为$y = 2x$或$19x + 4y + 27 = 0$.