1. **Problem Statement:** Find the x- and y-intercepts, state the asymptotes, and sketch the graph of the rational function $$f(x) = \frac{(x - 3)(x + 1)}{(x - 1)(x + 2)}.$$ 2. **Finding x-intercepts:** The x-intercepts occur where the numerator is zero (and denominator is not zero). Set numerator equal to zero: $$ (x - 3)(x + 1) = 0 $$ So, $$ x - 3 = 0 \Rightarrow x = 3 $$ $$ x + 1 = 0 \Rightarrow x = -1 $$ Check denominator at these points to ensure no division by zero: $$ (3 - 1)(3 + 2) = 2 \times 5 = 10 \neq 0 $$ $$ (-1 - 1)(-1 + 2) = (-2) \times 1 = -2 \neq 0 $$ Thus, x-intercepts are at $x=3$ and $x=-1$. 3. **Finding y-intercept:** The y-intercept occurs at $x=0$: $$ f(0) = \frac{(0 - 3)(0 + 1)}{(0 - 1)(0 + 2)} = \frac{(-3)(1)}{(-1)(2)} = \frac{-3}{-2} = \frac{3}{2} $$ So, the y-intercept is at $\left(0, \frac{3}{2}\right)$. 4. **Finding vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and numerator is not zero. Set denominator equal to zero: $$ (x - 1)(x + 2) = 0 $$ So, $$ x - 1 = 0 \Rightarrow x = 1 $$ $$ x + 2 = 0 \Rightarrow x = -2 $$ Check numerator at these points: $$ (1 - 3)(1 + 1) = (-2)(2) = -4 \neq 0 $$ $$ (-2 - 3)(-2 + 1) = (-5)(-1) = 5 \neq 0 $$ Thus, vertical asymptotes at $x=1$ and $x=-2$. 5. **Finding horizontal asymptote:** Since numerator and denominator are both degree 2 polynomials, horizontal asymptote is ratio of leading coefficients. Expand numerator: $$ (x - 3)(x + 1) = x^2 + x - 3x - 3 = x^2 - 2x - 3 $$ Expand denominator: $$ (x - 1)(x + 2) = x^2 + 2x - x - 2 = x^2 + x - 2 $$ Leading coefficients are both 1, so horizontal asymptote is: $$ y = \frac{1}{1} = 1 $$ 6. **Summary:** - x-intercepts: $x=3$, $x=-1$ - y-intercept: $\left(0, \frac{3}{2}\right)$ - Vertical asymptotes: $x=1$, $x=-2$ - Horizontal asymptote: $y=1$ 7. **Sketching the graph:** - Plot intercepts and asymptotes. - The function is undefined at vertical asymptotes. - The graph approaches $y=1$ as $x \to \pm \infty$. Final answer: $$\boxed{\text{x-intercepts: } x=3, -1; \quad y\text{-intercept: } \left(0, \frac{3}{2}\right); \quad \text{Vertical asymptotes: } x=1, -2; \quad \text{Horizontal asymptote: } y=1}$$