1. **Problem statement:** We are given a slope field for a differential equation and asked to find which particular solution satisfies the initial condition $y(0) = 0$. 2. **Step 1: Understand the initial condition.** We need to check which candidate function satisfies $y(0) = 0$. 3. **Step 2: Evaluate each candidate at $x=0$.** - (A) $y = x^2 - x \implies y(0) = 0^2 - 0 = 0$ - (B) $y = -\sin x \implies y(0) = -\sin 0 = 0$ - (C) $y = e^{-x} - 1 \implies y(0) = e^{0} - 1 = 1 - 1 = 0$ - (D) $y = e^{-x^2} - 1 \implies y(0) = e^{0} - 1 = 1 - 1 = 0$ All four satisfy $y(0) = 0$. 4. **Step 3: Analyze the slope field and behavior of solutions.** The slope field shows slopes slanting downward from left to right, denser near the origin, indicating negative slopes near $x=0$. 5. **Step 4: Check the derivative of each candidate at $x=0$ to match slope field behavior.** - (A) $y' = 2x - 1 \implies y'(0) = -1$ (negative slope) - (B) $y' = -\cos x \implies y'(0) = -1$ (negative slope) - (C) $y' = -e^{-x} \implies y'(0) = -1$ (negative slope) - (D) $y' = -2x e^{-x^2} \implies y'(0) = 0$ (zero slope) 6. **Step 5: Since the slope field near $x=0$ shows negative slopes, option (D) is unlikely because its slope at 0 is zero. Options (A), (B), and (C) have slope -1 at 0, matching the slope field. 7. **Step 6: Consider the shape of the slope field and the functions.** The slope field slopes downward from left to right, suggesting the solution decreases near zero. - (A) $y = x^2 - x$ is a parabola opening upward with vertex at $x=\frac{1}{2}$. - (B) $y = -\sin x$ oscillates. - (C) $y = e^{-x} - 1$ is a decreasing exponential shifted down by 1. 8. **Step 7: The slope field suggests a smooth decreasing curve near zero, consistent with (C).** **Final answer:** (C) $y = e^{-x} - 1$.