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The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its derivatives are $V^{\prime }=6x(1-x)$ and $V^{\prime \prime }=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.

Put $v=\dot{x}$. Newton's second law gives the phase plane system $\dot{x}=v$, $\dot{v}=-V^{\prime }=6x(x-1)$. Its conserved mechanical energy and trajectories areThe linearization at $(0,0)$ has eigenvalues $\pm i\sqrt{6}$, and the surrounding energy curves are closed, so it is a center equilibrium. At $(1,0)$ the eigenvalues are $\pm \sqrt{6}$, giving a saddle equilibrium. The figure shows the cubic potential barrier phase portrait; arrows point right when $v>0$ and left when $v<0$.

For $0<E<1$, there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.

For $E=1$, the factorization $1-V(x)=(x-1{)}^2(2x+1)$ gives the separatrixThe branch with $-1/2\le x<1$ is a homoclinic orbit: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an equilibrium trajectory.

For $E>1$, the particle can cross the barrier. There is one left turning point below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For $E<0$, only a right-hand open component exists, with its turning point beyond $3/2$. At $E=0$, the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.