1. **State the problem:** We need to find the points where the rational function $$f(x) = \frac{x^2 + 6x}{x^2 + 4x - 12}$$ has discontinuities. 2. **Recall the rule for discontinuities in rational functions:** Discontinuities occur where the denominator is zero because division by zero is undefined. 3. **Find the zeros of the denominator:** Solve $$x^2 + 4x - 12 = 0$$. 4. **Factor the quadratic:** $$x^2 + 4x - 12 = (x + 6)(x - 2)$$. 5. **Set each factor equal to zero:** - $$x + 6 = 0 \Rightarrow x = -6$$ - $$x - 2 = 0 \Rightarrow x = 2$$ 6. **Check if numerator is zero at these points:** - At $$x = -6$$, numerator $$(-6)^2 + 6(-6) = 36 - 36 = 0$$. - At $$x = 2$$, numerator $$2^2 + 6(2) = 4 + 12 = 16 \neq 0$$. 7. **Interpretation:** - At $$x = 2$$, denominator zero but numerator nonzero, so there is a vertical asymptote (discontinuity). - At $$x = -6$$, both numerator and denominator are zero, so we check for a removable discontinuity (hole). 8. **Simplify the function:** $$f(x) = \frac{x(x + 6)}{(x + 6)(x - 2)} = \frac{x}{x - 2}, \quad x \neq -6$$ 9. **Conclusion:** - There is a discontinuity at $$x = 2$$ (vertical asymptote). - There is a removable discontinuity (hole) at $$x = -6$$. **Final answer:** Discontinuities occur at $$x = 2$$ and $$x = -6$$. Therefore, the correct choice is **C. at x = 2 and x = -6**.