Need help determining bifurcation points So I have to find the bifurcation points of the system: $\dot{x}=(ax-x^3+x^5)(x-a+2)$, where $a\in\mathbb{R}$ is a parameter.
Attempt:
I know that a bifurcation point is the point, where there is a change in stability or number of fixed points.
I have tried visualising the graph, and have come to the conclusion, that there are: 
4 fixed points for $a\leq 0$.
6 fixed points for $0<a\leq 0.2$. 
2 fixed points for $0.2<a<2$. 
1 fixed point for $a=2$
2 fixed points for $2<a$.
The change in stability happens at the same time as the number of fixed points changes.
From what I have learned, I'm pretty sure that one bifurcation point is $(a,x)=(2,0)$, and I think that a transcritical bifurcation happens at this point.
I think there is another bifurcation point, when we go from 4 to 6 to 2 points. I just don't know exactly what that point is? $a=0$? $a=0.2$? It confueses me, that the change seems to happen before and after the interval $0<a\leq 0.2$. Normally the change should happen at a single point?
All help is appreciated!
 A: Lets take this step-by-step:

*

*Find the fixed points as a function of our parameters (e.g. $a$).

*Investigate the changes of the locations as we vary the parameters

*Determine the stability of the fixed points.

*Investigate the nature of these fixed points as we vary the parameters.

The fixed points:
$$ x_c = 0, a-2, \pm \sqrt{\frac{1 \pm \sqrt{1 - 4a}}{2}} $$
When $a < 0$, we have four solutions. Our first (pitchfork) bifurcation occurs at $a = 0$ where two new solutions emerge. We now have six solutions for $0 < a < 1/4$, where we reach our next double (saddle-node) bifurcation and lose two solutions. Down to only two fixed points, one more (transcritical) bifurcation occurs at $a = 2$.
Stability can be found by linearizing about the critical points in the usual way.

A: One extra tip that augments the prior answer. The easy way to locate the fixed points is to first use the equation $\dot x=0$  by setting each factor equal to zero, and then (the sneaky part) plot the solution set for each factor by expressing $a$ as a function of $x$.
(This is much easier than solving for $x$ as a function of $a$.)
In your example the first factor give the equation (i)  $a=x+2$.
The second factor gives either (ii) $a= x^2- x^4$  or (iii) $x=0$ with no restriction on $a$. Plot all these solutions on a single picture and this will reveal how the number of solutions varies as you adjust the variable $a$.
In the graph below (i) is blue, (iii) is the faint green vertical axis, and (ii) is the quartic.
Horizontal lines correspond to setting $a$ constant. As you vary the height ( the value of $a$), the number of solutions changes.
