1. **State the problem:** Mr. Hernandez's pool had 2000 gallons of water at 8:00 A.M. and 500 gallons at 11:00 A.M. We need to find the rate of change in the amount of water in gallons per hour. 2. **Formula used:** The rate of change (slope) is given by the formula: $$\text{Rate of change} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$ where $y$ represents the amount of water and $x$ represents time. 3. **Identify values:** - Initial amount $y_1 = 2000$ gallons at time $x_1 = 8$ hours (8:00 A.M.) - Final amount $y_2 = 500$ gallons at time $x_2 = 11$ hours (11:00 A.M.) 4. **Calculate the change in water and time:** $$\Delta y = y_2 - y_1 = 500 - 2000 = -1500$$ gallons $$\Delta x = x_2 - x_1 = 11 - 8 = 3$$ hours 5. **Calculate the rate of change:** $$\text{Rate} = \frac{-1500}{3} = -500$$ gallons per hour 6. **Interpretation:** The negative sign indicates the water is decreasing. So, the pool is draining at a rate of 500 gallons per hour. **Final answer:** $$\boxed{-500 \text{ gallons per hour}}$$