1. **State the problem:** We need to analyze the rational function $$R(x) = \frac{x+6}{x(x+12)}$$ to determine where its graph is above or below the x-axis and identify the correct graph. 2. **Find the domain:** The denominator is zero at $$x=0$$ and $$x=-12$$, so these are vertical asymptotes. The domain is $$(-\infty,-12) \cup (-12,0) \cup (0,\infty)$$. 3. **Find zeros of the numerator:** Set numerator equal to zero: $$x+6=0 \implies x=-6$$. This is the x-intercept. 4. **Determine sign of $$R(x)$$ on intervals:** - For $$x < -12$$: Choose $$x=-13$$. Numerator: $$-13+6=-7$$ (negative). Denominator: $$-13(-13+12) = -13(-1) = 13$$ (positive). So $$R(x)$$ is negative (negative/positive = negative). - For $$-12 < x < 0$$: Choose $$x=-1$$. Numerator: $$-1+6=5$$ (positive). Denominator: $$-1(-1+12) = -1(11) = -11$$ (negative). So $$R(x)$$ is negative (positive/negative = negative). - For $$x > 0$$: Choose $$x=1$$. Numerator: $$1+6=7$$ (positive). Denominator: $$1(1+12) = 1(13) = 13$$ (positive). So $$R(x)$$ is positive (positive/positive = positive). 5. **Summary of sign:** - Below x-axis on $$(-\infty,-12) \cup (-12,0)$$. - Above x-axis on $$(0,\infty)$$. 6. **Choose the correct graph:** - The graph has vertical asymptotes at $$x=0$$ and $$x=-12$$. - The function is below the x-axis on intervals $$(-\infty,-12)$$ and $$(-12,0)$$. - The function is above the x-axis on $$(0,\infty)$$. This matches **Graph A** description: asymptotes at $$x=0$$ and $$x=-12$$, curve below x-axis on the left, approaching zero from below on the right. **Final answers:** - B. Below x-axis on $$(-\infty,-12) \cup (-12,0)$$. - C. Above x-axis on $$(0,\infty)$$. - Correct graph: A.