Why does a transcritical bifurcation occur when $ab = 1$? From Nonlinear Dynamics and Chaos by Strogatz:

Show that the first-order system $\dot{x} = x(1-x^{2})-a(1-e^{-bx})$ undergoes a transcritical bifurcation at $x=0$ when the parameters $a,b$ satisfy a certain equation, to be determined. (This equation defines a bifurcation curve in the $(a,b)$ parameter space.) Then find an approximate formula for the fixed point that bifurcates from $x=0$, assuming that the parameters are close to the bifurcation curve.
Solution: Note that $x=0$ is a fixed point for all $(a,b)$. This makes it plausible that the fixed point will birfurcate transcritically, if it bifurcates at all. For small $x$, we find
\begin{align*}
1-e^{-bx} &amp;= 1-[1-bx+\dfrac{1}{2}b^{2}x^{2} + O(x^{3})] \\
&amp;= bx - \dfrac{1}{2}b^{2}x^{2} + O(x^{3})
\end{align*}
and so
\begin{align*}
\dot{x} &amp;= x - a(bx - \dfrac{1}{2}b^{2}x^{2})+O(x^{3}) \\
&amp;=(1-ab)x + (\dfrac{1}{2}ab^{2})x^{2}+O(x^{3}).
\end{align*}
Hence a transcritical bifurcation occurs when $ab=1$; this is the equation for the bifurcation curve. The nonzero fixed point is given by the solution of $1-ab+(\dfrac{1}{2}ab^{2})x \approx 0,$ that is $$x^{*} \approx \dfrac{2(ab-1)}{(ab)^{2}}.$$

After looking through the solution given in my textbook, I still don't understand why a transcritical bifurcation occurs when $ab=1$?
The transcritical bifurcation occurs when $r=0$ in the normal form $\dot{y}=ry\pm y^2$. You can notice that if you rescale $x$ and $t$ to eliminate the quadratic term you will obtain the transcritical normal form with $r=(1-ab)/\sqrt{2a}b$. The zero of $r$ are given by $ab=1$ as shown in the textbook.
The transcritical bifurcation occurs when $r=0$ in the normal form $\dot{y}=ry\pm y^2$. You can notice that if you rescale $x$ and $t$ to eliminate the quadratic term you will obtain the transcritical normal form with $r=(1-ab)/\sqrt{2a}b$. The zero of $r$ are given by $ab=1$ as shown in the textbook.