1. **State the problem:** Given the function $$f(x) = \frac{17 - 10x}{2x - 1}$$ for $$x \in \mathbb{R}$$ and $$x \neq \frac{1}{2}$$, we need to: a) Show that $$f(x)$$ can be rewritten as $$f(x) = \frac{12}{2x - 1} - 5$$. b) Find the equation of the vertical asymptote. c) Find the equation of the horizontal asymptote. d) Sketch the graph of $$y = f(x)$$. 2. **Rewrite the function:** Start with the original function: $$f(x) = \frac{17 - 10x}{2x - 1}$$ Rewrite the numerator to express it in terms of the denominator: $$17 - 10x = A(2x - 1) + B$$ where $$A$$ and $$B$$ are constants to find. 3. **Find constants $$A$$ and $$B$$:** Expand the right side: $$A(2x - 1) + B = 2Ax - A + B$$ Equate coefficients with the left side: For $$x$$ terms: $$-10 = 2A \implies A = \frac{-10}{2} = -5$$ For constants: $$17 = -A + B = -(-5) + B = 5 + B \implies B = 17 - 5 = 12$$ 4. **Rewrite $$f(x)$$ using $$A$$ and $$B$$:** $$f(x) = \frac{17 - 10x}{2x - 1} = \frac{-5(2x - 1) + 12}{2x - 1} = \frac{12}{2x - 1} - 5$$ 5. **Vertical asymptote:** The vertical asymptote occurs where the denominator is zero: $$2x - 1 = 0 \implies x = \frac{1}{2}$$ 6. **Horizontal asymptote:** As $$x \to \pm \infty$$, the term $$\frac{12}{2x - 1} \to 0$$, so: $$y = -5$$ is the horizontal asymptote. 7. **Sketch the graph:** - The graph is a rational hyperbola. - Vertical asymptote at $$x = \frac{1}{2}$$. - Horizontal asymptote at $$y = -5$$. - The graph is centered around the point $$\left(\frac{1}{2}, -5\right)$$. - The function approaches the asymptotes but never touches them. **Final answers:** $$f(x) = \frac{12}{2x - 1} - 5$$ Vertical asymptote: $$x = \frac{1}{2}$$ Horizontal asymptote: $$y = -5$$