1. **State the problem:** Find the intercepts, asymptotes, turning points, and intervals of increase/decrease for the rational function $$f(x) = \frac{5x+3}{2x-1}$$. 2. **Find the x-intercept:** Set the numerator equal to zero: $$5x + 3 = 0 \implies x = -\frac{3}{5}$$ So the x-intercept is $$\left(-\frac{3}{5}, 0\right)$$. 3. **Find the y-intercept:** Set $$x=0$$: $$f(0) = \frac{5(0)+3}{2(0)-1} = \frac{3}{-1} = -3$$ So the y-intercept is $$(0, -3)$$. 4. **Find vertical asymptotes:** Set denominator equal to zero: $$2x - 1 = 0 \implies x = \frac{1}{2}$$ Vertical asymptote at $$x = \frac{1}{2}$$. 5. **Find horizontal or oblique asymptotes:** Degree of numerator and denominator are both 1, so horizontal asymptote is ratio of leading coefficients: $$y = \frac{5}{2}$$ 6. **Find turning points:** Compute derivative using quotient rule: $$f'(x) = \frac{(5)(2x-1) - (5x+3)(2)}{(2x-1)^2} = \frac{10x - 5 - 10x - 6}{(2x-1)^2} = \frac{-11}{(2x-1)^2}$$ Since $$f'(x) = \frac{-11}{(2x-1)^2} < 0$$ for all $$x \neq \frac{1}{2}$$, the function is strictly decreasing everywhere except at the vertical asymptote. 7. **Intervals of increase/decrease:** - Decreasing on $$(-\infty, \frac{1}{2})$$ and $$\left(\frac{1}{2}, \infty\right)$$. **Final answers:** - x-intercept: $$\left(-\frac{3}{5}, 0\right)$$ - y-intercept: $$(0, -3)$$ - Vertical asymptote: $$x = \frac{1}{2}$$ - Horizontal asymptote: $$y = \frac{5}{2}$$ - No turning points (function strictly decreasing) - Decreasing on $$(-\infty, \frac{1}{2}) \cup \left(\frac{1}{2}, \infty\right)$$