Figure 2 here represents the bifurcation diagram for the equation $\dot{x}=\mu - x^2.$ Clearly for $\mu < 0$ there are no critical points. Since $Df(x)=-2x$ the trajectory is stable for $x >0$ and unstable for $x <0,$ depicted by the solid and dashed lines respectively. I know if a critical point is stable arrows move towards, and away if it is unstable. Can someone explain me how the phase portraits have been drawn for the three instances. Thank you.
We check the dynamical system for equilibria in the typical way (indicating with $x_e$ the equilibria and assuming $x_e \in \mathbb{R}$)
$$\dot x_e=0 \Leftrightarrow x_e^2=\mu \Leftrightarrow x_e \in\begin{cases} \{-\sqrt{\mu},\sqrt{\mu} \} &,\mu>0 \\ \{0\} &,\mu=0 \\ \{\} &,\mu<0\end{cases}$$
We already see that for $\mu>0$ we have two equilirbia, for $\mu=0$ one and for $\mu<0$ none, as seen in figure 1. The stability is left. As you computed correctly $Df(x)=-2x$ therefore
$$Df(x_e)=\begin{cases} \mp2 \sqrt{\mu} &,\mu>0\\ 0&,\mu=0 \end{cases}$$
Therefore in the case $\mu>0$ we get that $-\sqrt{\mu}$ is stable and $\sqrt{\mu}$ is unstable. We see in figure one that the trajectories move to right equilibrium $\sqrt{\mu}$ and move away $-\sqrt{\mu}$, exactly like we computed.
For $\mu=0$ we just have $x'=-x^2<0$ for $x\neq 0$ so we always move to the left if we follow one trajectory - except for $x=0$ which stays in its point since it is a equilibrium. The same holds for $x'=\mu-x^2<0$ for all $x \in \mathbb{R}$.
Now figure 2 takes all this into a bifurcation diagram. You see that for $\mu<0$ there is no 'graph' (no equilibrium) and for $\mu=0$ only one point (one equilibrium). For $\mu>0$ we know have this nice bifurcation indicating $\sqrt{\mu}$ and $-\sqrt{\mu}$.
All in all you just draw the function $\mu=x^2=:f(x)$ (setting $\dot x=0$ in the differential equation). To see this just tilt your head a bit and look at the $x$-$\mu$-diagram and not the $\mu$-$x$-diagram.