1. **Problem Statement:** We are given a differential equation of the form $$\frac{dP}{dt} = f(P)$$ with an initial value $$P(t_0) = P_0$$ and a slope field graph. The graph shows that the slope is zero at $$y=1$$, positive below $$y=1$$, and negative above $$y=1$$. We need to determine for which values of $$P_0$$ the solution $$P(t)$$ is decreasing. 2. **Understanding the slope field:** The slope field represents the derivative $$\frac{dP}{dt}$$ at various points. If $$\frac{dP}{dt} < 0$$, the function $$P(t)$$ is decreasing at that point. If $$\frac{dP}{dt} > 0$$, $$P(t)$$ is increasing. 3. **Analyzing the graph:** From the description: - At $$y=1$$, slope is zero: $$f(1) = 0$$. - For $$P > 1$$, slope is negative: $$f(P) < 0$$. - For $$P < 1$$, slope is positive: $$f(P) > 0$$. 4. **Conclusion:** The solution $$P(t)$$ is decreasing when $$\frac{dP}{dt} = f(P) < 0$$, which happens for $$P_0 > 1$$. **Final answer:** $$\boxed{\text{The solution } P(t) \text{ is decreasing for initial values } P_0 > 1.}$$