1. **Problem:** Determine the domain, intercepts, asymptotes, and sketch the graph of $$f(x) = \frac{x^2 + 3x + 2}{x^2 - 1}$$. 2. **Formula and rules:** - Domain excludes values making denominator zero. - Vertical asymptotes occur where denominator is zero and numerator is nonzero. - Horizontal asymptote for rational functions with equal degree numerator and denominator is ratio of leading coefficients. - x-intercepts are roots of numerator. - y-intercept is value at $$x=0$$. 3. **Find domain:** Denominator: $$x^2 - 1 = (x-1)(x+1)$$. Set denominator $$\neq 0$$: $$x \neq \pm 1$$. Domain: $$\{x \in \mathbb{R} : x \neq -1, 1\}$$. 4. **Find intercepts:** - Numerator: $$x^2 + 3x + 2 = (x+1)(x+2)$$. - x-intercepts: $$x = -1, -2$$. - y-intercept: $$f(0) = \frac{0 + 0 + 2}{0 - 1} = \frac{2}{-1} = -2$$. 5. **Find asymptotes:** - Vertical asymptotes at $$x = -1, 1$$ (denominator zeros). - Horizontal asymptote: degrees equal (2), leading coefficients both 1, so $$y = 1$$. 6. **Summary:** - Domain: $$x \neq \pm 1$$. - x-intercepts: $$-1, -2$$. - y-intercept: $$-2$$. - Vertical asymptotes: $$x = -1, 1$$. - Horizontal asymptote: $$y = 1$$. 7. **Graph behavior:** - Near vertical asymptotes, function tends to $$\pm \infty$$. - For large $$|x|$$, $$f(x) \to 1$$. Final answer: $$\boxed{\text{Domain: } x \neq \pm 1, \quad x\text{-intercepts: } -1, -2, \quad y\text{-intercept: } -2, \quad \text{Vertical asymptotes: } x=\pm 1, \quad \text{Horizontal asymptote: } y=1}$$