1. **State the problem:** Find the vertical asymptote $x=c$ of the function $$f(x) = \frac{x^2 - x - 2}{x - 2}.$$\n\n2. **Recall the rule for vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at that point.\n\n3. **Find where the denominator is zero:** Solve $$x - 2 = 0 \implies x = 2.$$\n\n4. **Check the numerator at $x=2$:** $$x^2 - x - 2 = 2^2 - 2 - 2 = 4 - 2 - 2 = 0.$$\nSince the numerator is also zero at $x=2$, this means there might be a removable discontinuity instead of a vertical asymptote.\n\n5. **Factor numerator:** $$x^2 - x - 2 = (x - 2)(x + 1).$$\n\n6. **Simplify the function:** $$f(x) = \frac{(x - 2)(x + 1)}{x - 2}.$$\nCancel the common factor $x - 2$ (except at $x=2$ where the function is undefined): $$f(x) = \cancel{\frac{(x - 2)(x + 1)}{x - 2}} = x + 1, \quad x \neq 2.$$\n\n7. **Conclusion:** Since the factor cancels, there is no vertical asymptote at $x=2$, but a hole (removable discontinuity).\n\n**Answer:** There is no vertical asymptote for this function. The point $x=2$ is a hole, not an asymptote.