1. **Problem Statement:** Find the instantaneous rate of change at each given x-value from the graph. 2. **Formula:** The instantaneous rate of change at a point is the slope of the tangent line at that point. It can be approximated by the average rate of change over a very small interval around the point: $$\text{slope} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$$ 3. **Step-by-step solution:** - For **x = 2**: The problem does not provide explicit points around x=2, so we use the graph or given data to estimate the slope of the tangent line at x=2. - For **x = -1**: Use points (-1, 0) and (0, 16) from the graph. $$\text{slope} = \frac{16 - 0}{0 - (-1)} = \frac{16}{1} = 16$$ - For **x = 3**: The problem gives the slope as 4 divided by -1, which is: $$\text{slope} = \frac{4}{-1} = -4$$ 4. **Explanation:** - At x = -1, the slope is positive 16, indicating a steep upward tangent. - At x = 3, the slope is -4, indicating a downward slope. - At x = 2, the slope should be estimated from the graph or data; if not given, it can be approximated by the average rate of change near x=2. **Final answers:** - Instantaneous rate of change at x = 2: (estimate from graph or data) - Instantaneous rate of change at x = -1: 16 - Instantaneous rate of change at x = 3: -4