##### Hint

Online, you have three branches of equilibria that you can manipulate so they form the bifurcation diagram. Two branches are curves that come together at the bifurcation point (the red point). The third branch is a line.

Move the red point to indicate the location of the bifurcation point. Move the two points on the adjoining curves to indicate how the equilibria change as a function of the parameter $\alpha$. Move two points on the line so that the line indicates how the third equilibrium changes as a function of the parameter $\alpha$. You can click each branch of equilibria to switch between a solid line (indicating a stable equilibrium) and a dashed line (indicating an unstable equilibrium).

To put the bifurcation point in the right spot, look at the graph with $\alpha=\alpha^*$ and determine the value of $v$, let's call it $v^*$, where the graph of $f$ is just touching the horizontal axis. That value $v=v^*$ is the location of the equilibrium at the bifurcation point. Note that on the bifurcation diagram, the location of $v$ is flipped so that it is on the vertical axis. The horizontal axis is for the parameter $\alpha$. Therefore, move the red point indicating the bifurcation to the coordinates $(\alpha^*, v^*)$.

Next, focus on the equilibria that come from the branch of the graph of $f$ where the bifurcation occurs. Those equilibria exist only for $\alpha$ either greater than or less than $\alpha^*$, so the two curves coming of the red point should either both go to the right or both go to the left. To pin down their exact locations, look at one of the graphs of $f$ where these extra equilibria exist and note the values of $v$ of those equilibria. Then, move the two points on curves so their *vertical* coordinates equal those values of $v$; their horizontal coordinates should be the value of $\alpha$ of the graph that you consulted.

The third equilibrium that exists in all graphs of $f$ corresponds to the remaining line on the bifurcation diagram. To correctly position the line, look up the values of $v$ of that equilibrium on two of the graphs of $f$. Position the two points so their first coordinates are the values of $\alpha$ of the graphs you consulted and their second coordinates are the values of $v$ that you found from the graph.

The last step is to get the stability of the branches of equilibria correct. Determine the stability of the equilibria from the graphs (either by thinking of the direction in which $v$ would move on either side of the equilibrium or by noting the sign of the slope of $f$ at each equilibrium and using the stability theorem). Then, click each curve in the bifurcation diagram so that stable equilibria are represented by solid curves and unstable equilibria are represented by dashed curves.

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