1. **State the problem:** We need to find and graph all vertical and horizontal asymptotes of the rational function $$f(x) = \frac{-7x + 1}{2x^2 - 6x - 8}$$. 2. **Find vertical asymptotes:** Vertical asymptotes occur where the denominator is zero and the numerator is not zero. Set denominator equal to zero: $$2x^2 - 6x - 8 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=-6$, $c=-8$. Calculate discriminant: $$b^2 - 4ac = (-6)^2 - 4 \times 2 \times (-8) = 36 + 64 = 100$$ Calculate roots: $$x = \frac{-(-6) \pm \sqrt{100}}{2 \times 2} = \frac{6 \pm 10}{4}$$ So, $$x_1 = \frac{6 + 10}{4} = \frac{16}{4} = 4$$ $$x_2 = \frac{6 - 10}{4} = \frac{-4}{4} = -1$$ 4. **Check numerator at these points:** $$-7(4) + 1 = -28 + 1 = -27 \neq 0$$ $$-7(-1) + 1 = 7 + 1 = 8 \neq 0$$ Since numerator is not zero at $x=4$ and $x=-1$, vertical asymptotes are at: $$x=4 \quad \text{and} \quad x=-1$$ 5. **Find horizontal asymptotes:** Compare degrees of numerator and denominator. Degree numerator = 1, degree denominator = 2. Since degree denominator > degree numerator, horizontal asymptote is: $$y=0$$ 6. **Summary:** - Vertical asymptotes at $x=-1$ and $x=4$. - Horizontal asymptote at $y=0$. These lines can be graphed as dashed lines to indicate asymptotes.