= Solution
The <fixed point> zero has multiplier $F'(0)=\mu$, so it is attracting for $|\mu|<1$ and repelling for $|\mu|>1$. Other real <fixed points> satisfy
$$
\boxed{ax^2+bx+\mu-1=0,\qquad x=\frac{-b\pm\sqrt{b^2+4a(1-\mu)}}{2a}.}
$$
Their multiplier is $M=1+bx+2ax^2$, and they are attracting exactly when $-2<bx+2ax^2<0$. At equality $M=\pm1$ the linear test is inconclusive. For $b\ne0$ two branches meet at $x_s=-b/(2a)$, $\mu_s=1+b^2/(4a)$, a generic <saddle-node bifurcation>. At $\mu=1$, the nonzero branch crossing zero gives a <transcritical bifurcation> when $b\ne0$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-4-fixed-points.png]
{title=Representative fixed-point diagrams with solid attracting and dashed repelling branches}
The three sketches use $(a,b)=(1,0),(1,1/2),(-1,-1/2)$ respectively; case (iii) therefore illustrates the subcritical-flip choice.
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