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

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.