1. **State the problem:** We want to determine all vertical asymptotes of the function $$f(x) = \frac{x - 3}{9 - x^2}$$. 2. **Identify where the function is undefined:** Vertical asymptotes occur where the denominator is zero (and numerator is not zero there). 3. **Set denominator equal to zero:** $$9 - x^2 = 0$$ 4. **Solve for x:** $$x^2 = 9$$ $$x = \pm 3$$ 5. **Check numerator at these points:** - At $$x = 3$$, numerator $$3 - 3 = 0$$ (0 in numerator and denominator, check factorization). - At $$x = -3$$, numerator $$-3 - 3 = -6 \neq 0$$. 6. **Factor denominator:** $$9 - x^2 = (3 - x)(3 + x)$$ 7. **Simplify function:** $$f(x) = \frac{x - 3}{(3 - x)(3 + x)} = \frac{x - 3}{-(x - 3)(x + 3)} = \frac{x - 3}{-(x - 3)(x + 3)} = -\frac{1}{x + 3}$$ for $$x \neq 3$$. 8. **Interpretation:** - At $$x = 3$$, factor cancels out, so no vertical asymptote but a hole. - At $$x = -3$$, denominator zero without cancellation, so vertical asymptote present. **Final answer:** The function has a vertical asymptote at $$x = -3$$ only.