 Solution

Adjoin $\dot{p}=0$ and use an extended centre manifold with local coordinates $(x,p)$. Count $p$ as order $x^2$, as appropriate to a steady-state fold, and writeThe centre-manifold invariance equation differentiates these graphs along $\dot{x}=y$. The derivatives $y_{x}y$ and $z_{x}y$ first contribute at weighted order three. Comparing the coefficients of $p$ and $x^2$ in the right sides therefore givesSince $q+1\ne 0$, their solution isHence the reduced normal form isBoth the quadratic and unfolding coefficients are nonzero, so the steady-state bifurcation is a saddle-node, with equilibria on the $p>0$ side. The exact steady equations confirm this: $y=0$, $z=x^2$, and $p=z+z^3=x^2+x^6$, giving two nearby equilibria for $p>0$ and none for $p<0$.

For $q>-1$ the transverse modes are attracting; the negative-$x$ branch is locally attracting and the positive-$x$ branch has one unstable direction. For $q<-1$, one transverse eigenvalue is already positive, so neither full equilibrium is asymptotically stable, although the reduced fold is still present with reversed flow orientation. The parameter $q=-1$ requires a higher-dimensional degeneracy analysis and is outside the stated assumption.