1. **Problem:** Find the vertical and horizontal asymptotes of the function $$f(x) = \frac{x^2 - 9}{2x^2 - 5x - 3}$$. 2. **Vertical asymptote:** occurs where the denominator is zero and numerator is not zero. Solve denominator zero: $$2x^2 - 5x - 3 = 0$$ Use quadratic formula: $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$$ So, $$x = \frac{5 + 7}{4} = 3$$ $$x = \frac{5 - 7}{4} = -\frac{1}{2}$$ Check numerator at these points: $$x=3: 3^2 - 9 = 9 - 9 = 0$$ numerator zero, so no vertical asymptote at 3. $$x=-\frac{1}{2}: \left(-\frac{1}{2}\right)^2 - 9 = \frac{1}{4} - 9 = -\frac{35}{4} \neq 0$$ numerator not zero, so vertical asymptote at $$x = -\frac{1}{2}$$. 3. **Horizontal asymptote:** compare degrees of numerator and denominator. Both numerator and denominator are degree 2. Horizontal asymptote is ratio of leading coefficients: $$y = \frac{1}{2}$$ 4. **Answer:** vertical asymptote at $$x = -\frac{1}{2}$$ and horizontal asymptote at $$y = \frac{1}{2}$$. 5. **Check options:** (B) a = -1/2, b = 1/2. --- **Final answer:** (B) a = -\frac{1}{2}, b = \frac{1}{2}