1. **State the problem:** Find the vertical and horizontal asymptotes of the function $$f(x) = \frac{x^3 - x^2 - 12x}{-3x^2 + 6x}$$. 2. **Vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. Set denominator equal to zero: $$-3x^2 + 6x = 0$$ Factor out $-3x$: $$-3x(x - 2) = 0$$ So, $x=0$ or $x=2$. Check numerator at these points: - At $x=0$: numerator is $0^3 - 0^2 - 12\cdot0 = 0$. - At $x=2$: numerator is $2^3 - 2^2 - 12\cdot2 = 8 - 4 - 24 = -20 \neq 0$. Since numerator is zero at $x=0$, this is a removable discontinuity, not a vertical asymptote. At $x=2$, numerator is not zero, so vertical asymptote at $x=2$. 3. **Horizontal asymptote:** Compare degrees of numerator and denominator. - Degree numerator: 3 - Degree denominator: 2 Since degree numerator > degree denominator, there is no horizontal asymptote. Instead, there is an oblique (slant) asymptote found by polynomial division: Divide numerator by denominator: $$\frac{x^3 - x^2 - 12x}{-3x^2 + 6x} = \frac{x^3 - x^2 - 12x}{-3x(x - 2)}$$ Perform polynomial division: Divide $x^3$ by $-3x^2$ to get $-\frac{1}{3}x$. Multiply denominator by $-\frac{1}{3}x$: $$-\frac{1}{3}x \cdot (-3x^2 + 6x) = x^3 - 2x^2$$ Subtract: $$ (x^3 - x^2 - 12x) - (x^3 - 2x^2) = (0) + ( -x^2 + 2x^2) - 12x = x^2 - 12x$$ Divide $x^2$ by $-3x^2$ to get $-\frac{1}{3}$. Multiply denominator by $-\frac{1}{3}$: $$-\frac{1}{3} \cdot (-3x^2 + 6x) = x^2 - 2x$$ Subtract: $$ (x^2 - 12x) - (x^2 - 2x) = 0 - 10x = -10x$$ So the quotient is: $$-\frac{1}{3}x - \frac{1}{3}$$ and remainder is: $$-10x$$ Therefore, the oblique asymptote is: $$y = -\frac{1}{3}x - \frac{1}{3}$$ 4. **Summary:** - Vertical asymptote at $x=2$. - No horizontal asymptote. - Oblique asymptote at $y = -\frac{1}{3}x - \frac{1}{3}$. Final answer: Vertical asymptote: $x=2$; Oblique asymptote: $y = -\frac{1}{3}x - \frac{1}{3}$; No horizontal asymptote.