1. **State the problem:** Find the values of $x$ where the function $$y = \frac{x^2 - 9}{x^2 - 5x + 6}$$ is discontinuous. 2. **Recall the rule:** A rational function is discontinuous where its denominator is zero because division by zero is undefined. 3. **Factor numerator and denominator:** $$x^2 - 9 = (x - 3)(x + 3)$$ $$x^2 - 5x + 6 = (x - 2)(x - 3)$$ 4. **Rewrite the function:** $$y = \frac{(x - 3)(x + 3)}{(x - 2)(x - 3)}$$ 5. **Simplify by canceling common factors:** $$y = \frac{\cancel{(x - 3)}(x + 3)}{(x - 2)\cancel{(x - 3)}}$$ 6. **Analyze discontinuities:** - The factor $(x - 3)$ cancels, so at $x=3$ there is a removable discontinuity (a hole). - The factor $(x - 2)$ remains in the denominator, so at $x=2$ the function has a vertical asymptote (non-removable discontinuity). 7. **Conclusion:** The function is discontinuous at $x=2$ and $x=3$. **Final answer:** The values of $x$ where the function is discontinuous are $\boxed{2 \text{ and } 3}$. This corresponds to option a. 2 & 3.