1. **Problem Statement:** We are given a differential equation $y' = f(y)$ and a graph of $y = f(y)$. 2. **Goal:** Sketch a representative family of solutions for the differential equation based on the graph of $f(y)$. 3. **Key Idea:** The slope of the solution curve $y(x)$ at any point depends only on the value of $y$ at that point, since $y' = f(y)$. 4. **Step-by-step analysis:** - Identify equilibrium points where $f(y) = 0$. From the graph, these occur approximately at $y \approx -1$, $y \approx 0$, and $y \approx 0.6$. - Determine stability of equilibria: - At $y \approx -1$, $f(y)$ crosses from positive to negative, so this is a stable equilibrium. - At $y \approx 0$, $f(y)$ crosses from negative to positive, so this is an unstable equilibrium. - At $y \approx 0.6$, $f(y)$ crosses from positive to negative, so this is a stable equilibrium. - For $y$ values between equilibria, the sign of $f(y)$ tells us whether solutions increase or decrease: - For $y$ between $-1$ and $0$, $f(y) < 0$, so $y$ decreases. - For $y$ between $0$ and $0.6$, $f(y) > 0$, so $y$ increases. 5. **Sketching solutions:** - Solutions starting near $y = -1$ tend to stay near $-1$ (stable). - Solutions starting just above $-1$ decrease toward $-1$. - Solutions starting between $-1$ and $0$ decrease toward $-1$. - Solutions starting just below $0$ increase away from $0$ (unstable). - Solutions starting between $0$ and $0.6$ increase toward $0.6$. - Solutions starting above $0.6$ decrease toward $0.6$. 6. **Summary:** The family of solutions will show horizontal asymptotes at $y = -1$ and $y = 0.6$, with solutions moving away from the unstable equilibrium at $y=0$. This qualitative analysis allows sketching solution curves without explicit formulas.