1. The problem asks to identify which function corresponds to the given slope field shown in the top-right graph. 2. A slope field represents the direction of the tangent (slope) to the solution curves of a differential equation at various points. 3. The given options are functions: (A) $y=\sin x$, (B) $y=\cos x$, (C) $y=x^2$, (D) $y=\frac{1}{6}x^3$, (E) $y=\frac{1}{4}x^4$. 4. To match the slope field to a function, we consider the derivative $y'$ of each function, since the slope field shows $y'$ at points $(x,y)$. 5. Compute derivatives: - For (A) $y=\sin x$, $y'=\cos x$. - For (B) $y=\cos x$, $y'=-\sin x$. - For (C) $y=x^2$, $y'=2x$. - For (D) $y=\frac{1}{6}x^3$, $y'=\frac{1}{2}x^2$. - For (E) $y=\frac{1}{4}x^4$, $y'=x^3$. 6. The slope field vectors depend on $x$ and $y$. Since the slope field is shown on the Cartesian plane, the slope at each point $(x,y)$ corresponds to $y' = f'(x)$ evaluated at $x$. 7. The slope field for $y=x^2$ has slopes $y'=2x$, which depend only on $x$ and increase linearly. 8. The slope field for $y=\frac{1}{6}x^3$ has slopes $y'=\frac{1}{2}x^2$, which are always non-negative and quadratic in $x$. 9. The slope field for $y=\frac{1}{4}x^4$ has slopes $y'=x^3$, which change sign depending on $x$. 10. The slope field for $y=\sin x$ or $y=\cos x$ involves trigonometric derivatives, which oscillate. 11. Given the vector field in the top-right graph shows slopes increasing with $x$ and positive for positive $x$, the best match is (D) $y=\frac{1}{6}x^3$ with derivative $y'=\frac{1}{2}x^2$. 12. Therefore, the function corresponding to the slope field is $y=\frac{1}{6}x^3$. **Final answer:** (D) $y=\frac{1}{6}x^3$