 Solution

Draw the downward-opening nullcline $y=x(2-x)$, meeting the axes at $x=0,2$, and the line $y=kx$. Their positive intersection is $P=(2-k,k(2-k))$. Use the sign rulesAbove both curves the arrows point northwest; below both they point southeast. For $0<x<2-k$, the parabola lies above the line, so arrows between them point northeast. For $2-k<x<2$, their order reverses, so arrows between them point southwest. For $x>2$, $\dot{x}<0$ throughout the nonnegative quadrant, with the vertical direction still determined by the line. Mark the origin as an unstable node and $P$ as a saddle, and trace sample trajectories consistent with those directions.

The quadrant is not positively invariant: on $x=0,y>0$, $\dot{x}=-y<0$; on $y=0,x>0$, $\dot{y}=-kx<0$. Trajectories reaching either axis there leave the plotted region. The phase portrait describes the portions inside the quadrant, not a claim that all forward trajectories remain inside it.