1. **State the problem:** Analyze the rational function $$g(x) = \frac{2x^3 - 2x^2 + 8x - 8}{x^3 - x^2 - 9x + 9}$$ to find holes, asymptotes, intercepts, domain, and range. 2. **Factor numerator and denominator:** Given steps show: $$g(x) = \frac{2(x-1)(x^2+4)}{(x-1)(x-3)(x+3)}$$ 3. **Simplify the function by canceling common factors:** Cancel $(x-1)$: $$g(x) = \frac{2\cancel{(x-1)}(x^2+4)}{\cancel{(x-1)}(x-3)(x+3)}$$ $$= \frac{2(x^2+4)}{(x-3)(x+3)}$$ 4. **a) Holes:** Hole occurs where factor canceled: $x=1$ Evaluate $g(1)$: $$g(1) = \frac{2(1^2+4)}{(1-3)(1+3)} = \frac{2(5)}{-2 \times 4} = \frac{10}{-8} = -1.25$$ **Hole at** $(1, -1.25)$ 5. **b) Horizontal asymptotes:** Degree numerator = degree denominator = 2 Horizontal asymptote is ratio of leading coefficients: $$y = \frac{2}{1} = 2$$ 6. **c) Vertical asymptotes:** Set denominator zero: $$(x-3)(x+3) = 0 \Rightarrow x=3, x=-3$$ Vertical asymptotes at $x=3$ and $x=-3$ 7. **d) X-intercepts:** Set numerator zero: $$2(x^2+4) = 0 \Rightarrow x^2+4=0$$ No real roots, so **no x-intercepts** 8. **e) Y-intercepts:** Evaluate $g(0)$: $$g(0) = \frac{2(0^2+4)}{(0-3)(0+3)} = \frac{2(4)}{-3 \times 3} = \frac{8}{-9} = -\frac{8}{9} \approx -0.89$$ Y-intercept at $(0, -\frac{8}{9})$ 9. **f) Domain:** Exclude values making denominator zero or hole: $$x \neq -3, 1, 3$$ Domain: $\{x \in \mathbb{R} \mid x \neq -3, 1, 3\}$ 10. **g) Range:** From analysis and asymptotes: $$y \neq 2, -1, 25$$ Range: $\{y \in \mathbb{R} \mid y \neq 2, -1, 25\}$ **Final answers:** - Hole: $(1, -1.25)$ - Horizontal asymptote: $y=2$ - Vertical asymptotes: $x=-3, 3$ - X-intercepts: none - Y-intercept: $(0, -\frac{8}{9})$ - Domain: $x \neq -3, 1, 3$ - Range: $y \neq 2, -1, 25$