= Solution
Intersect the <nullclines> $y=x(2-x)$ and $y=kx$. Their intersection equation is $x(2-k-x)=0$, giving the origin and $(2-k,k(2-k))$. The <Jacobian matrix> is
$$
M(x,y)=\begin{pmatrix}2-2x&-1\\-k&1\end{pmatrix}.
$$
At the origin, $\operatorname{tr}M=3$, $\det M=2-k$ and
$$
\lambda_\pm=\frac{3\pm\sqrt{1+4k}}2.
$$
For \b[$k>2$], only the origin is in the nonnegative quadrant, and it is a \b[saddle] because its determinant is negative. For \b[$0<k<2$], the origin is an \b[unstable node], with two distinct positive <eigenvalues>, and there is a second equilibrium in the open quadrant. At that second equilibrium, $\det M=k-2<0$, so it is a \b[saddle]. These signs give <hyperbolic equilibria>, so the nonlinear classifications follow from the <linearization stability theorem>. The boundary value $k=2$ is excluded from the two stated cases.
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