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 $(0,3)$. - $f'(2) = 0$ and $f'(-2) = 0$ mean the slope of the tangent line to the curve is zero at $x = 2$ and $x = -2$. - $f'(x) > 0$ for $-2 < x < 2$ means the function is strictly increasing between $-2$ and $2$. 3. **Analyzing the options:** - Graph a: Has local maxima at $x = -2$ and $x = 2$, so $f'(x)$ changes sign from positive to negative at these points, but the function is decreasing between $-2$ and $2$, which contradicts $f'(x) > 0$ there. - Graph b: The slope is negative in some intervals between $-2$ and $2$, so $f'(x) > 0$ is not satisfied. - Graph c: Has a local maximum at $(0,3)$, so $f'(x)$ changes from positive to negative at $0$, contradicting $f'(x) > 0$ for all $x$ in $(-2,2)$. - Graph d: Passes through $(0,3)$, has $f'(2) = f'(-2) = 0$, and the slope is positive between $-2$ and $2$. Although it has a sharp cusp at $0$, the problem does not forbid this, and the slope condition is met. 4. **Conclusion:** Graph d satisfies all the given conditions. **Final answer:** The correct graph is **d**.