1. **Problem Statement:** Given the rational function $$g(x) = \frac{2x^3 - 2x^2 + 8x - 8}{x^3 - x^2 - 9x + 9},$$ find the holes, horizontal asymptotes, vertical asymptotes, x-intercepts, y-intercepts, domain, range, and sketch the graph. 2. **Step 1: Factor numerator and denominator to find holes and asymptotes.** Factor numerator: $$2x^3 - 2x^2 + 8x - 8 = 2(x^3 - x^2 + 4x - 4).$$ Group terms: $$x^3 - x^2 + 4x - 4 = (x^3 - x^2) + (4x - 4) = x^2(x - 1) + 4(x - 1) = (x - 1)(x^2 + 4).$$ So numerator factors as: $$2(x - 1)(x^2 + 4).$$ Factor denominator: $$x^3 - x^2 - 9x + 9 = (x^3 - x^2) - (9x - 9) = x^2(x - 1) - 9(x - 1) = (x - 1)(x^2 - 9).$$ Further factor: $$x^2 - 9 = (x - 3)(x + 3).$$ So denominator factors as: $$(x - 1)(x - 3)(x + 3).$$ 3. **Step 2: Identify holes.** Holes occur where factors cancel in numerator and denominator. Common factor: $(x - 1)$. So hole at $x = 1$. 4. **Step 3: Simplify function after canceling hole factor.** $$g(x) = \frac{2\cancel{(x - 1)}(x^2 + 4)}{\cancel{(x - 1)}(x - 3)(x + 3)} = \frac{2(x^2 + 4)}{(x - 3)(x + 3)}.$$ 5. **Step 4: Horizontal asymptotes.** Degree numerator = 2, degree denominator = 2. For equal degrees, horizontal asymptote is ratio of leading coefficients. Leading coefficient numerator: $2$ (from $2x^2$ term). Leading coefficient denominator: $1$ (from $x^2$ term). So horizontal asymptote: $$y = \frac{2}{1} = 2.$$ 6. **Step 5: Vertical asymptotes.** Vertical asymptotes occur where denominator is zero and not canceled. Denominator factors: $(x - 3)(x + 3)$. So vertical asymptotes at: $$x = 3, \quad x = -3.$$ 7. **Step 6: X-intercepts.** Set numerator equal to zero: $$2(x^2 + 4) = 0 \implies x^2 + 4 = 0 \implies x^2 = -4.$$ No real solutions, so no x-intercepts. 8. **Step 7: Y-intercept.** Set $x=0$: $$g(0) = \frac{2(0^2 + 4)}{(0 - 3)(0 + 3)} = \frac{2(4)}{-3 \times 3} = \frac{8}{-9} = -\frac{8}{9}.$$ So y-intercept at $(0, -\frac{8}{9})$. 9. **Step 8: Domain.** Domain excludes values making denominator zero: $$x \neq 1, 3, -3.$$ Note $x=1$ is a hole, so excluded. 10. **Step 9: Range.** Range is all real numbers except the horizontal asymptote value $y=2$ because the function approaches but never reaches it. 11. **Step 10: Summary:** - Holes: $x=1$ - Horizontal asymptote: $y=2$ - Vertical asymptotes: $x=3, x=-3$ - X-intercepts: none - Y-intercept: $(0, -\frac{8}{9})$ - Domain: $\{x \in \mathbb{R} \mid x \neq -3, 1, 3\}$ - Range: $\{y \in \mathbb{R} \mid y \neq 2\}$