Consider the differential equation $\frac{dy}{dt}=ry(1-\frac{y}{a})(1-2by+y^2)$ with $0<a<b$. I wish to answer the two following questions:
(1) Find the steady states for the model
(2) Plot $\frac{dy}{dt}$ versus $y$ for the cases (i) $b<1$ and (ii) b-$\sqrt{b^2-1}>a$.
My attempt: (a) A steady state occurs when $\frac{dy}{dt}=0$. Therefore we can look at the factors of $\frac{dy}{dt}$ to see that $y=a$ is a steady state solution. For the factor $1-2by+y^2$, by the quadratic formula we get $y=b\pm\sqrt{b^2-1}$. If $b<1$ then the only steady state is $y=a$, if $b=1$ then the two steady states are $y=a$ and $y=b$, and if $b>1$ then the 3 steady states are $y=a, y=b\pm\sqrt{b^2-1}$.
(b) I'm a little confused at this point - since $b<1$ can we say that the factor $1-2by+y^2\approx 1+y^2$, and then $\frac{dy}{dt}\approx ry(1-y)(1+y^2)$? And for (ii) I'm not sure what to do.
In part (ii), you're simply supposed to sketch the graph of the function $f(y)=ry(1-\frac{y}{a})(1-2by+y^2)$. And the result will depend on how many zeros the function has, and how they are situated relative to each other. And that, in turn, you can determine by taking the conditions into account (you basically already wrote the answer in your question). You're not really supposed to make any approximations here, I think. On the other hand, one is mainly interested in the sign (and the zeros) of $f(y)$, so perhaps a rather crude sketch would be acceptable (qualitative rather than quantitative).