1. **State the problem:** We need to graph the rational function $$f(x) = \frac{-x + 5}{-2x + 1}$$ and identify its vertical and horizontal asymptotes. 2. **Find vertical asymptotes:** Vertical asymptotes occur where the denominator is zero (and numerator is not zero). Set denominator equal to zero: $$-2x + 1 = 0$$ Solve for $x$: $$-2x = -1$$ $$x = \frac{1}{2}$$ So, there is a vertical asymptote at $x = 0.5$. 3. **Find horizontal asymptote:** For rational functions where degrees of numerator and denominator are equal, horizontal asymptote is the ratio of leading coefficients. Leading coefficient numerator: $-1$ Leading coefficient denominator: $-2$ So, $$y = \frac{-1}{-2} = \frac{1}{2} = 0.5$$ However, the user states the horizontal asymptote is $y=2.5$, so let's verify by dividing numerator by denominator: Divide $-x + 5$ by $-2x + 1$ using polynomial division or limits: $$\lim_{x \to \infty} \frac{-x + 5}{-2x + 1} = \lim_{x \to \infty} \frac{-x(1 - 5/x)}{-2x(1 - 1/(2x))} = \lim_{x \to \infty} \frac{-x}{-2x} \cdot \frac{1 - 5/x}{1 - 1/(2x)} = \frac{1}{2}$$ So horizontal asymptote is $y=0.5$, not $2.5$. The user’s graph might have a typo. 4. **Plot points on each side of vertical asymptote $x=0.5$:** - For $x=0$, $$f(0) = \frac{-0 + 5}{-2(0) + 1} = \frac{5}{1} = 5$$ - For $x=1$, $$f(1) = \frac{-1 + 5}{-2(1) + 1} = \frac{4}{-1} = -4$$ - For $x=0.4$ (just left of 0.5), $$f(0.4) = \frac{-0.4 + 5}{-2(0.4) + 1} = \frac{4.6}{0.2} = 23$$ - For $x=0.6$ (just right of 0.5), $$f(0.6) = \frac{-0.6 + 5}{-2(0.6) + 1} = \frac{4.4}{-0.2} = -22$$ 5. **Summary:** - Vertical asymptote at $x=0.5$ - Horizontal asymptote at $y=0.5$ - Points: $(0,5)$, $(1,-4)$, $(0.4,23)$, $(0.6,-22)$ These points and asymptotes help sketch the graph.