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 and the numerator is not zero at those points. 3. **Factor the denominator:** $$x^2 - x - 2 = (x - 2)(x + 1)$$ 4. **Set denominator equal to zero to find potential vertical asymptotes:** $$ (x - 2)(x + 1) = 0 $$ This gives $$ x = 2 \quad \text{or} \quad x = -1 $$ 5. **Check numerator at these points:** Numerator is $$x^2 + 2x + 1 = (x + 1)^2$$. At $$x = -1$$, numerator is $$(-1 + 1)^2 = 0$$. At $$x = 2$$, numerator is $$(2 + 1)^2 = 9 \neq 0$$. 6. **Interpretation:** Since numerator is zero at $$x = -1$$ and denominator is also zero there, this may be a removable discontinuity, not a vertical asymptote. At $$x = 2$$, denominator is zero but numerator is not zero, so there is a vertical asymptote at $$x = 2$$. 7. **Final answer:** The vertical asymptote is at $$x = 2$$. **Answer choice:** c. $$x=2$$