Bifurcation flow field? assuming I have
$$x'=5+mx+2x^2$$
how would I find the flow field of this bifurcation with the changing variable m?
EDIT: 
Take for example 
http://www.math.colostate.edu/~shipman/47/volume3b2011/M640_MunozAlicea.pdf
At page 2 there is a very strange looking diagram that is a parabola upside down. I don't see how he would draw that. I understand that the axes are shift but like what is the thought process? How would I also draw a flow diagram etc... please help!
 A: The first thing we do is to set $x'=0$ and find the critical points. This yields:
$$x'=5+mx+2x^2 = 0 \implies x=\frac{1}{4} \left(-\sqrt{m^2-40}-m\right), x=\frac{1}{4} \left(+\sqrt{m^2-40}-m\right)$$
Now, we can do a plot in the $xm-plane$ for the first root, yielding:

Doing a similar $xm-plane$ plot for the second root yields:

The major difference between these and the ones you reference in the book is that you typically find where the points are stable and unstable and use a dashed line to designate where the system is unstable and the solid line stable. I will leave you to do some of the work.
Now, if we were to draw a direction field plot of the system while varying $m$, you can observe that we transition through a change. The RHS and LHS behave the same for the two roots above. Here is an animated direction field plot.

Lastly, it is worth mentioning that we can find a closed form solution to this system as:
$$x(t)\to \frac{1}{2} \left(\sqrt{20-m^2} \tan \left(\frac{1}{2} \left(c_1 \sqrt{20-m^2}+\sqrt{20-m^2} t\right)\right)-m\right)$$
Try plotting $x$ as a function of $t$ for different initial conditions and what do you notice.
Update
We can also write $x' = 5 + mx + 2x^2, ~ y' = -y$ and do a phase portrait since you asked for that animation. I will let you do the analysis. Here is the parametrized ($ -15 \le m \le 15$) phase portrait.

