求曲线y=x+sqrt(x^2)-x+1的渐近线.

当 $ x \to +\infty $ 时， \[ y = x + \sqrt{x^2 - x + 1} \approx x + x = 2x, \] 但更精确地， \[ y = x + \sqrt{x^2 - x + 1} = x + x\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} \approx x + x\left(1 - \frac{1}{2x}\right) = 2x - \frac{1}{2}, \] 故斜渐近线为 $ y = 2x - \frac{1}{2} $。 当 $ x \to -\infty $ 时， \[ y = x + \sqrt{x^2 - x + 1} \approx x + (-x) = 0, \] 但更精确地， \[ y = x + (-x)\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} \approx x - x\left(1 - \frac{1}{2x}\right) = \frac{1}{2}, \] 故水平渐近线为 $ y = \frac{1}{2} $。 **答案：** \[ \boxed{ \begin{array}{ll} \text{当 } x \to +\infty \text{ 时，斜渐近线 } y = 2x - \frac{1}{2}, \\ \text{当 } x \to -\infty \text{ 时，水平渐近线 } y = \frac{1}{2}. \end{array} } \]