1. **State the problem:** We need to find the vertical and horizontal asymptotes of the function $$q(x) = \frac{x - 2}{\sqrt{2x - 1}}$$ and then graph it. 2. **Vertical asymptotes:** Vertical asymptotes occur where the denominator is zero or undefined, and the numerator is not zero at the same point. The denominator is $$\sqrt{2x - 1}$$ which is defined only when $$2x - 1 > 0$$, so: $$2x - 1 > 0 \implies x > \frac{1}{2}$$ At $$x = \frac{1}{2}$$, the denominator is zero, so check if the numerator is zero: $$x - 2 = \frac{1}{2} - 2 = -\frac{3}{2} \neq 0$$ Thus, there is a vertical asymptote at $$x = \frac{1}{2}$$. 3. **Horizontal asymptotes:** Horizontal asymptotes describe the behavior as $$x \to \infty$$ or $$x \to -\infty$$. Since the domain is $$x > \frac{1}{2}$$, we only consider $$x \to \infty$$. Evaluate the limit: $$\lim_{x \to \infty} \frac{x - 2}{\sqrt{2x - 1}}$$ Rewrite the denominator: $$\sqrt{2x - 1} = \sqrt{2x}\sqrt{1 - \frac{1}{2x}} = \sqrt{2x} \cdot \sqrt{1 - \frac{1}{2x}}$$ As $$x \to \infty$$, $$\sqrt{1 - \frac{1}{2x}} \to 1$$, so: $$\lim_{x \to \infty} \frac{x - 2}{\sqrt{2x - 1}} \approx \lim_{x \to \infty} \frac{x}{\sqrt{2x}} = \lim_{x \to \infty} \frac{x}{\sqrt{2} \sqrt{x}} = \lim_{x \to \infty} \frac{\cancel{x}}{\sqrt{2} \sqrt{\cancel{x}}} = \lim_{x \to \infty} \frac{\sqrt{x}}{\sqrt{2}} = \infty$$ So, no horizontal asymptote as $$x \to \infty$$. 4. **Summary:** - Vertical asymptote at $$x = \frac{1}{2}$$. - No horizontal asymptote. 5. **Graph:** The function is defined for $$x > \frac{1}{2}$$. - As $$x \to \frac{1}{2}^+$$, denominator $$\to 0^+$$, numerator $$\to -\frac{3}{2}$$, so $$q(x) \to -\infty$$. - As $$x \to \infty$$, $$q(x) \to \infty$$. The graph starts near negative infinity just to the right of $$x=\frac{1}{2}$$ and increases without bound as $$x$$ grows.