1. **Problem Statement:** We need to identify which graph represents a continuous function $f$ such that $f(0) = 3$, $f'(2) = f'(-2) = 0$, and $f''(x) > 0$ for $-2 < x < 2$. 2. **Understanding the conditions:** - $f(0) = 3$ means the function passes through the point $(0,3)$. - $f'(2) = f'(-2) = 0$ means the slope of the tangent line at $x=2$ and $x=-2$ is zero, indicating possible local maxima or minima or inflection points. - $f''(x) > 0$ for $-2 < x < 2$ means the function is concave upward (shaped like a cup) between $-2$ and $2$. 3. **Analyzing the graphs:** - (a) is a downward opening curve (concave down), so $f''(x) < 0$ here, which contradicts $f''(x) > 0$. - (b) is not symmetric and does not have zero slopes at $x= \\pm 2$. - (c) is an upward opening curve (concave up), passes through $(0,3)$, and has zero slopes at $x= \\pm 2$. - (d) is not smooth and has a sharp peak, so it is not continuous with the required smoothness. 4. **Conclusion:** Graph (c) satisfies all conditions: $f(0) = 3$, $f'(2) = f'(-2) = 0$, and $f''(x) > 0$ for $-2 < x < 2$. **Final answer:** The correct graph is (c).