1. **State the problem:** We are analyzing the function $$y=\frac{x+1}{x-1}$$ and its graph. 2. **Find vertical asymptotes:** Vertical asymptotes occur where the denominator equals zero because the function is undefined there. Set denominator to zero: $$x-1=0 \implies x=1$$ So, there is a vertical asymptote at $$x=1$$. 3. **Find horizontal asymptotes:** To find the horizontal asymptote, evaluate the limit of $$y$$ as $$x \to \pm \infty$$: $$\lim_{x \to \infty} \frac{x+1}{x-1} = \lim_{x \to \infty} \frac{x\left(1+\frac{1}{x}\right)}{x\left(1-\frac{1}{x}\right)} = \lim_{x \to \infty} \frac{1+\frac{1}{x}}{1-\frac{1}{x}} = \frac{1+0}{1-0}=1$$ Similarly as $$x \to -\infty$$, the limit is also $$1$$. Thus, the horizontal asymptote is $$y=1$$. 4. **Simplify and interpret:** The function has a hyperbolic shape with vertical asymptote at $$x=1$$ and horizontal asymptote at $$y=1$$. 5. **Summary:** This matches the description given and confirms the graph's properties. **Final answer:** The vertical asymptote is at $$x=1$$ and the horizontal asymptote is at $$y=1$$.