1. **Problem:** Determine if the function $$y=\frac{7}{6-x}-9$$ is decreasing over its entire domain. 2. **Step 1:** Find the derivative $$y'$$ to analyze increasing/decreasing behavior. $$y=7(6-x)^{-1}-9$$ Using the chain rule: $$y' = 7 \cdot (-1)(6-x)^{-2} \cdot (-1) = \frac{7}{(6-x)^2}$$ 3. **Step 2:** Analyze the sign of $$y'$$. Since $$7 > 0$$ and $$(6-x)^2 > 0$$ for all $$x \neq 6$$, we have $$y' > 0$$ everywhere in the domain. 4. **Step 3:** Conclusion for function 1. Because $$y' > 0$$, the function is strictly increasing on its entire domain (all real numbers except $$x=6$$). --- 5. **Problem:** Determine if the function $$y=-\frac{3}{9-x}+8$$ is decreasing over its entire domain. 6. **Step 1:** Find the derivative. Rewrite: $$y = -3(9-x)^{-1} + 8$$ Derivative: $$y' = -3 \cdot (-1)(9-x)^{-2} \cdot (-1) = -\frac{3}{(9-x)^2}$$ 7. **Step 2:** Analyze the sign of $$y'$$. Since $$3 > 0$$ and $$(9-x)^2 > 0$$ for all $$x \neq 9$$, the derivative is: $$y' = -\frac{3}{(9-x)^2} < 0$$ 8. **Step 3:** Conclusion for function 2. Because $$y' < 0$$ everywhere in the domain, the function is strictly decreasing on its entire domain (all real numbers except $$x=9$$). --- 9. **Problem:** Determine if the function $$y=\frac{1}{2-3x}+4$$ is decreasing over its entire domain. 10. **Step 1:** Find the derivative. Rewrite: $$y = (2-3x)^{-1} + 4$$ Derivative: $$y' = -1 \cdot (2-3x)^{-2} \cdot (-3) = \frac{3}{(2-3x)^2}$$ 11. **Step 2:** Analyze the sign of $$y'$$. Since $$3 > 0$$ and $$(2-3x)^2 > 0$$ for all $$x \neq \frac{2}{3}$$, we have $$y' > 0$$. 12. **Step 3:** Conclusion for function 3. Because $$y' > 0$$ everywhere in the domain, the function is strictly increasing on its entire domain (all real numbers except $$x=\frac{2}{3}$$). --- **Final answer:** - Function 1 is increasing. - Function 2 is decreasing. - Function 3 is increasing. Only function 2 is decreasing over its entire domain.