a) Sketch the phase line for the differential equation

dy/dt=1/((y-2)(y+1))
and discuss the behavior of the solution with initial condition y(0)=1/2
b) Apply analytic techniques to the initial-value problem
dy/dt=1/((y-2)(y+1))), y(0)=1/2
and compare your results with your discussion in part (a).

I couldn't get the equilibrium points for the equation so I did the phase line without them, and everything above 2 and below -1 was positive and between 2 and -1 is negative.
When y(0)=1/2, the solution is negative but what happens when it gets to 2 or -1 if they are not equilibrium points?