I have to find bifurcation for the system $\dot{X} = AX$ where $A = \begin{pmatrix} -2 & \frac{1}{4} \\ -1 & \mu \end{pmatrix}$. where $X =\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$
My attempt: I first try to find eigenvalues of the matrix $A$ by calculating $det(A-\lambda I) = 0$ I get a quadratic $$ \lambda^2 + \lambda(2-\mu) + (\frac{1}{4}-\mu) =0$$ Solving which I get $$\lambda = \frac{-(2-\mu)\pm \sqrt{(\mu+1)(\mu+3)}}{2}$$ How should I proceed further ? I am stuck here.
As pointed out by Moo, using Wolfram we can plot $\lambda$ versus $\mu$ and observe possibility of bifurcation(change of the stability of the fixed points as the parameter varies),here the parameter is $\mu$.
In this dynamical system we have two fixed points whose stability can be decided from the sign of real part of the eigenvalues.
If the real part of the eigenvalue corresponding to a fixed point is negative then that fixed point is stable else it is unstable.
So from the plot you can observe for which parameter there is change of stability of the two fixed points and the point where such a change of stability occurs is the bifurcation point.
As from the plot fig(1) and fig(2) we can see that in fig $(1)$ the real part of this eigenvalue is always negative for all set of $\mu$ so this fixed point remains stable forever but in fig$(2)$ we see that the second fixed point changes its stability as we see that as $\mu $ varies the real part of the eigenvalue changes its sign!
Hence the bifurcation!
Also this could be a good reference.