Fixed points have $y=0$ and $x^2-bx-a=0$. They coalesce on $a=-b^2/4$, the saddle-node curve. Away from it, $x_\pm=(b\pm\sqrt{b^2+4a})/2$; the Jacobian has trace $-1$ and <determinant> $2x-b$, so $x_+$ is stable and $x_-$ a saddle. At $(a,b)=(0,0)$ the branches $x=0$ and $x=b$ cross in a transcritical bifurcation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 31B / b / Solution

 Solution (source code)