= 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 write
$$
y=a_1p+a_2x^2+O(x^3,xp,p^2),\qquad
z=b_1p+b_2x^2+O(x^3,xp,p^2).
$$
The <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 gives
$$
-qa_1-b_1=0,\quad 1-qa_2-b_2=0,\qquad
a_1+1-b_1=0,\quad a_2-b_2=0.
$$
Since $q+1\ne0$, their solution is
$$
a_1=-\frac1{q+1},\quad a_2=\frac1{q+1},\qquad
b_1=\frac q{q+1},\quad b_2=\frac1{q+1}.
$$
Hence the reduced <normal form> is
$$
\boxed{\dot x=\frac{x^2-p}{q+1}+O(x^3,xp,p^2).}
$$
Both the quadratic and unfolding coefficients are nonzero, so \b[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.
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