1. **State the problem:** We need to sketch the graph of the function $$y=\frac{x^2 + 3x + 2}{x^2 - 1}$$ and understand its behavior. 2. **Identify the function and domain:** The function is a rational function where the numerator is $$x^2 + 3x + 2$$ and the denominator is $$x^2 - 1$$. 3. **Factor numerator and denominator:** $$x^2 + 3x + 2 = (x+1)(x+2)$$ $$x^2 - 1 = (x-1)(x+1)$$ 4. **Simplify the function:** Since $$x+1$$ is common in numerator and denominator, for $$x \neq -1$$, $$y = \frac{(x+1)(x+2)}{(x-1)(x+1)} = \frac{x+2}{x-1}$$ 5. **Domain restrictions:** The original denominator cannot be zero, so $$x \neq 1$$ and $$x \neq -1$$. 6. **Vertical asymptotes:** At points where denominator is zero and not canceled, vertical asymptotes occur. Here, at $$x=1$$ vertical asymptote. At $$x=-1$$, the factor cancels, so there is a hole (removable discontinuity). 7. **Horizontal asymptote:** Since degrees of numerator and denominator are equal (both degree 1 after simplification), horizontal asymptote is ratio of leading coefficients: $$y = \frac{1}{1} = 1$$ 8. **Find holes:** At $$x=-1$$, hole exists. Find $$y$$ value at hole by limit or simplified function: $$y = \frac{-1+2}{-1-1} = \frac{1}{-2} = -\frac{1}{2}$$ 9. **Intercepts:** - **x-intercept:** Set numerator zero: $$x+2=0 \Rightarrow x=-2$$ - **y-intercept:** Set $$x=0$$: $$y=\frac{0+2}{0-1} = \frac{2}{-1} = -2$$ 10. **Summary:** - Vertical asymptote at $$x=1$$ - Hole at $$(-1, -\frac{1}{2})$$ - Horizontal asymptote at $$y=1$$ - x-intercept at $$x=-2$$ - y-intercept at $$y=-2$$ This information helps sketch the graph accurately. **Final answer:** The function simplifies to $$y=\frac{x+2}{x-1}$$ with a hole at $$x=-1$$ and vertical asymptote at $$x=1$$.