1. **State the problem:** Find the domain, vertical asymptote(s), and horizontal asymptote(s) of the function $$f(x) = \frac{4 - 2x}{7x - 1}$$. 2. **Domain:** The domain of a rational function is all real numbers except where the denominator is zero. Set the denominator equal to zero: $$7x - 1 = 0$$ Solve for $$x$$: $$7x = 1$$ $$x = \frac{1}{7}$$ So, the domain is all real numbers except $$x = \frac{1}{7}$$. 3. **Vertical asymptote(s):** Vertical asymptotes occur where the denominator is zero and the numerator is not zero at that point. At $$x = \frac{1}{7}$$, denominator is zero. Check numerator at $$x = \frac{1}{7}$$: $$4 - 2\left(\frac{1}{7}\right) = 4 - \frac{2}{7} = \frac{28}{7} - \frac{2}{7} = \frac{26}{7} \neq 0$$ Since numerator is not zero, there is a vertical asymptote at $$x = \frac{1}{7}$$. 4. **Horizontal asymptote(s):** For rational functions, compare degrees of numerator and denominator. Both numerator and denominator are degree 1. Horizontal asymptote is $$y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$$. Leading coefficient numerator: $$-2$$ (from $$-2x$$) Leading coefficient denominator: $$7$$ So, $$y = \frac{-2}{7}$$ 5. **Summary:** - Domain: all real numbers except $$x = \frac{1}{7}$$ - Vertical asymptote: $$x = \frac{1}{7}$$ - Horizontal asymptote: $$y = -\frac{2}{7}$$ Note: The user mentioned horizontal asymptote as 0, but by calculation it is $$-\frac{2}{7}$$. 6. **Graph shape in sentence form:** There is a vertical asymptote at $$x = \frac{1}{7}$$ and a horizontal asymptote at $$y = -\frac{2}{7}$$.