A particle which is constrained to move along the $x$-axis, is subjected to a force in the same direction which varies with the distance $x$ of the particle from the origin as $F(x) = -kx + ax^2$. Here $k$ and $a$ are positive constants. For $x \geq 0$, the functional from of the potential energy $U(x)$ of the particle is

Since $F(x)=-\frac{dU}{dx}$, we have $$\frac{dU}{dx}=kx-ax^2$$ Integrating, $$U(x)=\frac{kx^2}{2}-\frac{ax^3}{3}+C$$ So for $x \geq 0$, the curve starts with zero slope at $x=0$ and $$\frac{d^2U}{dx^2}=k-2ax$$ Hence at $x=0$, $\frac{d^2U}{dx^2}=k>0$, so the graph is initially concave upward and has a minimum at $x=0$. Also, $$\frac{dU}{dx}=x(k-ax)$$ Thus the stationary points are at $x=0$ and $x=\frac{k}{a}$. Further, $$\frac{d^2U}{dx^2}\bigg|_{x=k/a}=-k<0$$ so $x=\frac{k}{a}$ is a maximum. For large $x$, the term $-\frac{ax^3}{3}$ dominates, so $U(x)$ decreases and becomes negative. Therefore the graph starts at a minimum at the origin, rises to a maximum, then falls and crosses the axis later. This matches option $D$.