1. **Problem Statement:** Find the asymptotes, intercepts, and graph of the function $$f(x) = \frac{x^3 + x^2 - 4x - 4}{x^3 - x^2 - 8x + 12}.$$ 2. **Asymptotes:** - Vertical asymptotes occur where the denominator is zero and the numerator is not zero. - Horizontal or slant asymptotes depend on the degrees of numerator and denominator. 3. **Find vertical asymptotes:** Factor denominator: $$x^3 - x^2 - 8x + 12 = (x-3)(x+2)(x-2).$$ Set denominator to zero: $$x=3, x=-2, x=2.$$ Check numerator at these points: - At $x=3$: numerator $= 27 + 9 - 12 - 4 = 20 \neq 0$ vertical asymptote. - At $x=-2$: numerator $= -8 + 4 + 8 - 4 = 0$ numerator zero, so no vertical asymptote here. - At $x=2$: numerator $= 8 + 4 - 8 - 4 = 0$ numerator zero, so no vertical asymptote here. 4. **Simplify function by factoring numerator:** Factor numerator: $$x^3 + x^2 - 4x - 4 = (x+2)(x-2)(x+1).$$ 5. **Simplify $f(x)$:** $$f(x) = \frac{(x+2)(x-2)(x+1)}{(x-3)(x+2)(x-2)} = \frac{x+1}{x-3}, \quad x \neq -2, 2.$$ 6. **Asymptotes from simplified form:** - Vertical asymptote at $x=3$. - Holes (removable discontinuities) at $x=-2$ and $x=2$. 7. **Horizontal/slant asymptote:** Since degrees of numerator and denominator are equal (both 1 after simplification), horizontal asymptote is ratio of leading coefficients: $$y = \frac{1}{1} = 1.$$ 8. **Intercepts:** - **x-intercept:** Set numerator zero: $$x+1=0 \Rightarrow x=-1.$$ - **y-intercept:** Set $x=0$: $$f(0) = \frac{0+1}{0-3} = -\frac{1}{3}.$$ 9. **Summary:** - Vertical asymptote: $x=3$ - Holes at $x=-2$ and $x=2$ - Horizontal asymptote: $y=1$ - x-intercept: $(-1,0)$ - y-intercept: $(0,-\frac{1}{3})$ 10. **Graph:** The graph behaves like $y=\frac{x+1}{x-3}$ except it has holes at $x=-2$ and $x=2$. It approaches $y=1$ as $x \to \pm \infty$ and has a vertical asymptote at $x=3$. **Final answer:** $$\boxed{\text{Vertical asymptote: } x=3, \quad \text{Holes at } x=-2,2, \quad \text{Horizontal asymptote: } y=1, \quad \text{x-intercept } (-1,0), \quad \text{y-intercept } (0,-\frac{1}{3})}.$$