1. **State the problem:** Find the vertical asymptote of the function $$f(x) = \frac{x^2 - x - 2}{x - 2}$$. 2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at that point. 3. **Find where the denominator is zero:** Set $$x - 2 = 0$$ which gives $$x = 2$$. 4. **Check the numerator at $$x=2$$:** $$x^2 - x - 2 = 2^2 - 2 - 2 = 4 - 2 - 2 = 0$$. 5. Since the numerator is also zero at $$x=2$$, we need to simplify the function to see if the factor cancels. 6. **Factor the numerator:** $$x^2 - x - 2 = (x - 2)(x + 1)$$. 7. **Simplify the function:** $$f(x) = \frac{(x - 2)(x + 1)}{x - 2}$$. 8. **Cancel the common factor:** $$f(x) = \frac{\cancel{(x - 2)}(x + 1)}{\cancel{(x - 2)}} = x + 1$$ for $$x \neq 2$$. 9. Since the factor cancels, there is a **hole** at $$x=2$$, not a vertical asymptote. 10. **Conclusion:** There is no vertical asymptote for this function. **Final answer:** The function has no vertical asymptote because the factor causing the zero in the denominator cancels with the numerator, resulting in a removable discontinuity (hole) at $$x=2$$.