1. **Problem statement:** Given the curve with equation $$y = x + \frac{1}{x}$$, we need to find: a. The equations of the asymptotes. b. The minimum and maximum values of $$y$$. c. Sketch the graph of the curve. 2. **Asymptotes:** - Vertical asymptote occurs where the function is undefined, i.e., where the denominator is zero. - Here, $$x = 0$$ is a vertical asymptote because $$\frac{1}{x}$$ is undefined at $$x=0$$. - For large $$|x|$$, $$y = x + \frac{1}{x} \approx x$$, so the oblique asymptote is $$y = x$$. 3. **Finding minimum and maximum values:** - To find extrema, differentiate $$y$$ with respect to $$x$$: $$y' = 1 - \frac{1}{x^2}$$ - Set derivative equal to zero to find critical points: $$1 - \frac{1}{x^2} = 0$$ $$\Rightarrow 1 = \frac{1}{x^2}$$ $$\Rightarrow x^2 = 1$$ $$\Rightarrow x = \pm 1$$ 4. **Evaluate $$y$$ at critical points:** - At $$x=1$$: $$y = 1 + \frac{1}{1} = 2$$ - At $$x=-1$$: $$y = -1 + \frac{1}{-1} = -1 -1 = -2$$ 5. **Determine nature of critical points:** - Second derivative: $$y'' = \frac{2}{x^3}$$ - At $$x=1$$: $$y'' = 2 > 0$$, so minimum at $$x=1$$. - At $$x=-1$$: $$y'' = -2 < 0$$, so maximum at $$x=-1$$. 6. **Summary:** - Vertical asymptote: $$x=0$$ - Oblique asymptote: $$y=x$$ - Minimum value: $$y_{min} = 2$$ at $$x=1$$ - Maximum value: $$y_{max} = -2$$ at $$x=-1$$ 7. **Graph sketch description:** - The curve has two branches: one in the first quadrant approaching $$x=0^+$$ and $$y=x$$, with a minimum at (1,2). - The other branch is in the third quadrant approaching $$x=0^-$$ and $$y=x$$, with a maximum at (-1,-2).