1. **State the problem:** We need to graph the rational function $$f(x) = \frac{-2}{-x - 2}$$ and identify its vertical and horizontal asymptotes. 2. **Rewrite the function:** Simplify the denominator: $$-x - 2 = -(x + 2)$$ So, $$f(x) = \frac{-2}{-(x + 2)} = \frac{-2}{-1 \cdot (x + 2)} = \frac{-2}{-1} \cdot \frac{1}{x + 2} = \frac{2}{x + 2}$$ 3. **Vertical asymptote:** The function is undefined where the denominator is zero: $$x + 2 = 0 \implies x = -2$$ So, the vertical asymptote is at $$x = -2$$. 4. **Horizontal asymptote:** For large $$|x|$$, the function behaves like: $$f(x) \approx \frac{2}{x}$$ As $$x \to \pm \infty$$, $$f(x) \to 0$$. So, the horizontal asymptote is at $$y = 0$$. 5. **Plot points:** Choose two points on each side of the vertical asymptote $$x = -2$$. - For $$x < -2$$: - $$x = -3$$: $$f(-3) = \frac{2}{-3 + 2} = \frac{2}{-1} = -2$$ - $$x = -4$$: $$f(-4) = \frac{2}{-4 + 2} = \frac{2}{-2} = -1$$ - For $$x > -2$$: - $$x = -1$$: $$f(-1) = \frac{2}{-1 + 2} = \frac{2}{1} = 2$$ - $$x = 0$$: $$f(0) = \frac{2}{0 + 2} = \frac{2}{2} = 1$$ 6. **Graph shape:** The graph is a rational hyperbola with vertical asymptote at $$x = -2$$ and horizontal asymptote at $$y = 0$$. - For $$x < -2$$, the function values are negative and approach zero from below. - For $$x > -2$$, the function values are positive and approach zero from above. **Final answer:** $$f(x) = \frac{2}{x + 2}$$ Vertical asymptote: $$x = -2$$ Horizontal asymptote: $$y = 0$$ Points plotted: $$(-4, -1), (-3, -2), (-1, 2), (0, 1)$$