1. **Problem Statement:** Given the 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 simplify and identify 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: $$2(x - 1)(x^2 + 4).$$ 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: $$(x - 1)(x - 3)(x + 3).$$ 3. **Step 2: Simplify the function by canceling common factors.** $$g(x) = \frac{2\cancel{(x - 1)}(x^2 + 4)}{\cancel{(x - 1)}(x - 3)(x + 3)} = \frac{2(x^2 + 4)}{(x - 3)(x + 3)}.$$ 4. **Step 3: Identify holes.** Holes occur where factors cancel out, here at $$x = 1$$. 5. **Step 4: Horizontal asymptotes.** Degree numerator = 2, degree denominator = 2. For rational functions where degrees are equal, horizontal asymptote is ratio of leading coefficients. Leading coefficient numerator after simplification: $$2 \times 1 = 2$$ (from $$2x^2$$ term). Leading coefficient denominator: $$1$$ (from $$x^2$$ term in $$(x-3)(x+3) = x^2 - 9$$). So horizontal asymptote: $$y = \frac{2}{1} = 2.$$ 6. **Step 5: Vertical asymptotes.** Vertical asymptotes occur where denominator is zero and factor is not canceled. Denominator factors after simplification: $$(x - 3)(x + 3)$$. So vertical asymptotes at $$x = 3$$ and $$x = -3$$. 7. **Step 6: X-intercepts.** Set numerator equal to zero: $$2(x^2 + 4) = 0 \implies x^2 + 4 = 0 \implies x^2 = -4,$$ which has no real solutions. So, no x-intercepts. 8. **Step 7: Y-intercept.** Evaluate $$g(0)$$: $$g(0) = \frac{2(0^2 + 4)}{(0 - 3)(0 + 3)} = \frac{2(4)}{-3 \times 3} = \frac{8}{-9} = -\frac{8}{9}.$$ 9. **Step 8: Domain.** Domain excludes values making denominator zero before simplification: $$x \neq 1, 3, -3.$$ 10. **Step 9: Range.** Range is all real numbers except the horizontal asymptote value if the function never reaches it. Since horizontal asymptote is $$y=2$$ and the function approaches but never equals 2, range is: $$(-\infty, 2) \cup (2, \infty).$$ 11. **Step 10: Summary:** - Holes: $$x=1$$ - Horizontal asymptote: $$y=2$$ - Vertical asymptotes: $$x=3, x=-3$$ - X-intercepts: none - Y-intercept: $$y = -\frac{8}{9}$$ - Domain: $$\{x | x \neq -3, 1, 3\}$$ - Range: $$(-\infty, 2) \cup (2, \infty)$$ 12. **Step 11: Sketch graph** Use vertical asymptotes at $$x=\pm 3$$, hole at $$x=1$$, horizontal asymptote at $$y=2$$, and y-intercept at $$-8/9$$.