1. The problem is to analyze the function $$y = \frac{1}{x} + 2$$. 2. This function is a rational function with a vertical asymptote where the denominator is zero, i.e., at $$x=0$$. 3. The horizontal asymptote is found by considering the limit as $$x \to \pm \infty$$: $$\lim_{x \to \pm \infty} \left(\frac{1}{x} + 2\right) = 2$$ 4. The function is undefined at $$x=0$$, so there is a vertical asymptote there. 5. To find the intercepts: - For the y-intercept, set $$x=0$$, but the function is undefined there, so no y-intercept. - For the x-intercept, set $$y=0$$: $$0 = \frac{1}{x} + 2$$ $$\Rightarrow \frac{1}{x} = -2$$ $$\Rightarrow x = \frac{1}{-2} = -\frac{1}{2}$$ 6. So the x-intercept is at $$\left(-\frac{1}{2}, 0\right)$$. 7. The function is decreasing on $$(-\infty, 0)$$ and increasing on $$(0, \infty)$$ because the derivative is: $$y' = -\frac{1}{x^2}$$ which is always negative except undefined at $$x=0$$. 8. Summary: - Vertical asymptote at $$x=0$$ - Horizontal asymptote at $$y=2$$ - x-intercept at $$\left(-\frac{1}{2}, 0\right)$$ - No y-intercept Final answer: The function $$y = \frac{1}{x} + 2$$ has a vertical asymptote at $$x=0$$, a horizontal asymptote at $$y=2$$, and an x-intercept at $$\left(-\frac{1}{2}, 0\right)$$.