1. **Problem:** Find the vertical asymptote(s) for the function $$f(x) = \frac{x+2}{x^2 - x - 6}$$. 2. **Recall:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at those points. 3. **Step 1:** Factor the denominator: $$x^2 - x - 6 = (x - 3)(x + 2)$$ 4. **Step 2:** Set the denominator equal to zero to find potential vertical asymptotes: $$ (x - 3)(x + 2) = 0 $$ $$x - 3 = 0 \Rightarrow x = 3$$ $$x + 2 = 0 \Rightarrow x = -2$$ 5. **Step 3:** Check if the numerator is zero at these points: - At $$x = 3$$, numerator $$3 + 2 = 5 \neq 0$$ - At $$x = -2$$, numerator $$-2 + 2 = 0$$ 6. **Step 4:** Since numerator is zero at $$x = -2$$, this may be a hole, not a vertical asymptote. 7. **Step 5:** Simplify the function by canceling common factors: $$f(x) = \frac{x+2}{(x-3)(x+2)} = \frac{\cancel{x+2}}{(x-3)\cancel{(x+2)}} = \frac{1}{x-3}$$ 8. **Step 6:** The vertical asymptote is at $$x = 3$$. **Final answer:** The vertical asymptote is at $$x=3$$.