1. **State the problem:** We are given the function $$f(x) = \frac{3x^2 - 14x - 5}{3x^3 - x - 2}$$ and need to analyze its domain, vertical asymptotes, and horizontal asymptote. 2. **Domain:** The domain consists of all real numbers except where the denominator is zero because division by zero is undefined. 3. **Find the zeros of the denominator:** Solve $$3x^3 - x - 2 = 0$$. 4. **Factor or find roots:** Try possible rational roots using the Rational Root Theorem: factors of 2 over factors of 3, i.e., $$\pm1, \pm2, \pm\frac{1}{3}, \pm\frac{2}{3}$$. 5. **Test x = 1:** $$3(1)^3 - 1 - 2 = 3 - 1 - 2 = 0$$, so $$x=1$$ is a root. 6. **Divide denominator by (x - 1):** Using polynomial division or synthetic division: $$3x^3 - x - 2 \div (x - 1) = 3x^2 + 3x + 2$$ 7. **Factor quadratic:** $$3x^2 + 3x + 2$$ has discriminant $$\Delta = 3^2 - 4 \times 3 \times 2 = 9 - 24 = -15 < 0$$, so no real roots. 8. **Domain conclusion:** The denominator is zero only at $$x=1$$, so domain is $$(-\infty, 1) \cup (1, \infty)$$. 9. **Vertical asymptote:** Vertical asymptotes occur where denominator is zero and numerator is not zero. At $$x=1$$, numerator: $$3(1)^2 - 14(1) - 5 = 3 - 14 - 5 = -16 \neq 0$$ So vertical asymptote at $$x=1$$. 10. **Horizontal asymptote:** Compare degrees of numerator (2) and denominator (3). Since degree denominator > degree numerator, horizontal asymptote is $$y=0$$. **Final answers:** - Domain: $$(-\infty, 1) \cup (1, \infty)$$ - Vertical asymptote: $$x=1$$ - Horizontal asymptote: $$y=0$$