1. **State the problem:** We are given the rational function $$f(x) = \frac{x^2(x - 2)}{(x + 1)(x - 2)}$$ and asked to analyze its properties including vertical asymptotes, horizontal/slant asymptotes, intercepts, and holes. 2. **Simplify the function:** Notice that the factor $(x - 2)$ appears in both numerator and denominator. We can cancel it out except where it causes a hole. $$f(x) = \frac{x^2\cancel{(x - 2)}}{(x + 1)\cancel{(x - 2)}} = \frac{x^2}{x + 1}, \quad x \neq 2$$ 3. **Vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the factor is not canceled. Set denominator to zero: $$x + 1 = 0 \implies x = -1$$ So, there is a vertical asymptote at $x = -1$. 4. **Hole:** The canceled factor $(x - 2)$ causes a hole at $x = 2$. Find the $y$-value of the hole by substituting $x=2$ into the simplified function: $$f(2) = \frac{2^2}{2 + 1} = \frac{4}{3}$$ So, there is a hole at $(2, \frac{4}{3})$. 5. **Horizontal or slant asymptote:** Since the degree of numerator (2) is one more than denominator (1), there is a slant asymptote. Perform polynomial division: $$\frac{x^2}{x + 1} = x - 1 + \frac{1}{x + 1}$$ So, the slant asymptote is: $$y = x - 1$$ 6. **x-intercept:** Set numerator equal to zero (using simplified function): $$x^2 = 0 \implies x = 0$$ So, the x-intercept is at $(0,0)$. 7. **y-intercept:** Evaluate $f(0)$: $$f(0) = \frac{0^2}{0 + 1} = 0$$ So, the y-intercept is also at $(0,0)$. **Final summary:** - Vertical asymptote at $x = -1$ - Hole at $(2, \frac{4}{3})$ - Slant asymptote $y = x - 1$ - x-intercept and y-intercept at $(0,0)$