1. **State the problem:** We are given the function $$f(x) = \frac{x^3}{x^2 - x - 2}$$ and asked to analyze its properties including domain, asymptotes, and behavior. 2. **Identify the denominator and find its zeros:** The denominator is $$x^2 - x - 2$$. To find vertical asymptotes, solve $$x^2 - x - 2 = 0$$. 3. **Factor the denominator:** $$x^2 - x - 2 = (x - 2)(x + 1)$$ 4. **Find vertical asymptotes:** Set each factor equal to zero: $$x - 2 = 0 \Rightarrow x = 2$$ $$x + 1 = 0 \Rightarrow x = -1$$ So, vertical asymptotes are at $$x = 2$$ and $$x = -1$$. 5. **Find horizontal or oblique asymptotes:** Since the degree of the numerator (3) is greater than the degree of the denominator (2), there is no horizontal asymptote. Instead, perform polynomial division to find the oblique asymptote. 6. **Divide numerator by denominator:** Divide $$x^3$$ by $$x^2 - x - 2$$: Set up division: $$\frac{x^3}{x^2 - x - 2} = x + \frac{x^2 + 2x}{x^2 - x - 2}$$ Perform the division step by step: - First term: $$x^3 \div x^2 = x$$ - Multiply divisor by $$x$$: $$x(x^2 - x - 2) = x^3 - x^2 - 2x$$ - Subtract: $$x^3 - (x^3 - x^2 - 2x) = x^2 + 2x$$ Since the remainder degree is less than divisor degree, the division stops here. 7. **Oblique asymptote:** $$y = x$$ 8. **Domain:** All real numbers except where denominator is zero: $$x \neq 2, -1$$ 9. **Summary:** - Vertical asymptotes at $$x = 2$$ and $$x = -1$$ - Oblique asymptote at $$y = x$$ - Domain: $$\mathbb{R} \setminus \{2, -1\}$$ 10. **Graph behavior:** The function approaches vertical asymptotes near $$x=2$$ and $$x=-1$$ and follows the line $$y=x$$ for large $$|x|$$. **Final answer:** $$f(x) = \frac{x^3}{x^2 - x - 2}$$ has vertical asymptotes at $$x=2$$ and $$x=-1$$ and an oblique asymptote $$y=x$$.