1. **State the problem:** Find the equations of the vertical asymptotes of the function $$f(x) = \frac{x^2 + 4x}{x^2 - 2x - 24}$$. 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:** $$x^2 - 2x - 24 = (x - 6)(x + 4)$$ 4. **Set denominator equal to zero to find potential vertical asymptotes:** $$x - 6 = 0 \Rightarrow x = 6$$ $$x + 4 = 0 \Rightarrow x = -4$$ 5. **Check numerator at these points:** $$x^2 + 4x = x(x + 4)$$ At $$x = 6$$, numerator is $$6(6 + 4) = 6 \times 10 = 60 \neq 0$$. At $$x = -4$$, numerator is $$-4(-4 + 4) = -4 \times 0 = 0$$. 6. **Interpretation:** Since numerator is zero at $$x = -4$$, this point is a hole, not a vertical asymptote. 7. **Final answer:** The only vertical asymptote is at $$x = 6$$. **Answer: C. x = 6**