Just needed some guidance to see if my approach is correct. The following is how I proceeded to do this problem. What seems to be confusing me is that it is in the potential form and not $\dot{x} = ...$. From the equation I can tell that this is a saddle node bifurcation.
$$x' = r- x^2$$
Since we know that $x' = -\frac{dV}{dx}$, we can integrate the system to solve for the potential. $$ V = \frac{x^3}{3} -rx + C$$ We can let $C= 0$. Therefore: $$ V = \frac{x^3}{3} - rx$$
The following graphs are the graphs for values of $r > 0 , r < 0$ and $r = 0$.
The question asks to include bifurcation values (r critical), how would I go about finding those? From my understanding I know the bifurcation occurs when the two curves $x^3/3$ and $-rx$ are tangent to each other. So I can take the derivative of both of them then solve for x then plug that into the equation for V to solve for r. Is this correct?