bifurcation in differential equations I have an equation
$$
y' = (2+y)(k-y^2)
$$
and I am asked to find the equilibrium solutions and bifurcation values for all values of $k$.
My approach was that... there exists $2$ equlibrium solutions for $k>0$, and $1$ equilibrium soltuion for $k = 0$ and $0$ equilibrium solution for $k < 0$.
After differentiating the function to
$$
f' = -4y - 3y^2
$$
I do not know how to continue. Do I plug in $\sqrt k$ and $\sqrt {-k}$, which is the equilibrium solution to the differentiated function...?
Thanks.

Hints:
To find the critical points, we set up and solve $y' = 0$, hence:
$$y' = (2+y)(k-y^2) = 0 \implies y = -2, ~\pm ~\sqrt{k}$$
Now that we have these critical points, how do they behave for various values of $k$?
Can you determine how to find the bifurcation points from this?
Look at the direction field plot and what happens to  number of critical points for $k = 4, 1, 0, -1$:



Another point of interest is $k = 4$, as look at the direction field plot (collapses to two equilibrium points):

