= Solution
In physical <concentration> coordinates, the two <nullclines> are
$$
\boxed{\dot C=0:\ C=\frac{E_0S}{S+K_M},\quad K_M=\frac{k_{-1}+k_2}{k_1};\qquad
\dot S=0:\ C=\frac{E_0S}{S+K_S},\quad K_S=\frac{k_{-1}}{k_1}.}
$$
Since $K_S<K_M$, the <substrate> <nullcline> lies above the complex <nullcline> for $S>0$. Starting at $(S_0,0)$, the complex rises and <substrate> falls; with $E_0/S_0\ll1$, this fast segment is nearly vertical when the axes use their respective <concentration> scales. The <trajectory> then follows the complex quasi-steady manifold toward the origin. On its slow descending portion it is slightly above the complex <nullcline> so that $\dot C<0$; the leading outer curve itself sets that <derivative> to zero only at zeroth order.
\Image[/past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2008-enzyme-phase-plane.png]
{title=Enzyme kinetics in scaled concentration coordinates: rapid complex buildup followed by slow substrate depletion}
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