This is part of Strogatz exercise $3.2.3:$
This is the process by which I found the bifurcation point/points for $\dot x=x-rx(1-x)$:
By the method of tangential intersection we have: $$x=rx(1-x)$$ $$and$$ $$\frac{d}{dx}(x)=\frac{d}{dx}(rx(1-x)).$$
From $\frac{d}{dx}(x)=\frac{d}{dx}(rx(1-x))$ we get $1=r-2rx$ and therefore $x=\frac{r-1}{2r}$.
Substituting $x=\frac{r-1}{2r}$ into $\dot x=x-rx(1-x)$ we get:
\begin{align} 0 & =\frac{r-1}{2r}-\frac{r-1}{2}(1-\frac{r-1}{2r})\\ & = \frac{r-1}{2r}-\frac{r-1}{2}(\frac{r+1}{2r})\\ & = \frac{r-1}{2r}-\frac{r^2-1}{4r}\\ & = \frac{-r^2+2r-1}{4r} \end{align}
Therefore $r^2-2r+1=0$. Therefore $r=1$ with algebraic multiplicity $2$.
With the help of Mathematica I sketched the graphs of tangential intersection at $r = 0$, $r=1$, and $r=2$ which you can see in below.
Now here is my question: Since the algebraic multiplicity of the bifurcation point is $2$, what is the geometric significance of it, if there is any?
Thank you in advance.