1. **State the problem:** Find the vertical asymptotes of the function $$f(x)=\frac{x^2 + 2x + 1}{x^2 - x - 2}$$. 2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero but the numerator is not zero at those points. 3. **Factor numerator and denominator:** $$x^2 + 2x + 1 = (x+1)^2$$ $$x^2 - x - 2 = (x-2)(x+1)$$ 4. **Simplify the function:** $$f(x) = \frac{(x+1)^2}{(x-2)(x+1)}$$ Cancel the common factor $(x+1)$: $$f(x) = \frac{\cancel{(x+1)}(x+1)}{(x-2)\cancel{(x+1)}} = \frac{x+1}{x-2}$$ 5. **Determine vertical asymptotes:** - The factor $(x+1)$ was canceled, so $x=-1$ is a removable discontinuity, not a vertical asymptote. - The denominator zero at $x=2$ remains, so $x=2$ is a vertical asymptote. **Final answer:** The vertical asymptote is at $$x=2$$.