How to approach transistor-based exercises?
Your schematic has an applied input voltage and a base resistor that might be better replaced with a current source for analysis purposes. That doesn't really change the question or the curve much, but it may help thinking about it.



simulate this circuit &ndash; Schematic created using CircuitLab
On the left is the schematic I'm suggesting helps simplify the analysis, a bit. On the right is the first thing I want you to consider. On the right, in case where I've disconnected the collectors of both BJTs, the current \$I_1\$ must flow through both base-emitter junctions, equally. So, we'd predict that \$V_{\text{BE}_1}=V_{\text{BE}_2}\$. But what happens when we reconnect them back into the situation shown on the left?
Well, in this case, all of the collector current in \$Q_1\$ is added to \$I_1\$ to form \$Q_1\$'s emitter current. That means that \$I_{\text{E}_1}=\left(\beta_1+1\right)\cdot I_1\$, now. That's a lot bigger than \$I_1\$. So \$Q_2\$'s base current will be on the order of \$\beta\$ times as much as \$Q_1\$'s base current. Since \$\beta\$ is at least 100, this means that there are two orders of magnitude difference between the collector currents, with \$Q_2\$'s collector current being about \$\beta\$ more than \$Q_1\$'s.
Since there's about \$60\:\text{mV}\$ difference in \$V_\text{BE}\$ for each factor of 10 difference in currents, and since there are (at least) two factors of 10 difference here, we'd expect that \$V_{\text{BE}_2}\ge V_{\text{BE}_1}+120\:\text{mV}\$. So as soon as the BJTs are connected back up (moving now from the right-hand schematic back to the left-hand one), their base-emitter voltages are no longer the same but will instead be different by about this much. At least, at first.
The other conclusion here is that most of the current will be sunk by \$Q_2\$ because it has so much more base current to work with. In fact, we'd expect about \$\beta\$ times as much. Which, for all intents and purposes is all of it. So that also now means that \$R_1\$ and \$R_2\$ simply form a voltage divider with the only question being, &quot;What's \$Q_2\$'s collector voltage?&quot;
Well, that's not hard to work out at first. The collector current for \$Q_2\$, at least at first anyway, will be \$I_{\text{C}_2}=I_1\cdot\left(\beta_1+1\right)\cdot\beta_2\$. So long as \$I_{\text{C}_2}\$ is itself tiny compared to \$R_1+R_2\$, then there won't be much of a voltage drop across them. So long as that voltage drop doesn't exceed some tens of millivolts, \$V_u\$ will look about the same -- namely, \$V_u\approx +V\$. That's the first part of your curve and it exists until A is reached.
Now, what exactly happens to make that start to change? Well, one thing is that \$V_\text{DROP}=I_{\text{C}_2}\cdot\left(R_1+R_2\right)\$ must be large enough to measure, now. Let's say this is of greater magnitude than \$100\:\text{mV}\$. What's the behavior now as we continue to increase \$I_1\$?
This is the point to stop and think and also think back, too. Prior to this point, we could change \$I_1\$ by orders of magnitude -- not just linear changes -- and we'd see no difference at all. But throughout this time, we've been increasing \$V_{\text{BE}_1}\$ and \$V_{\text{BE}_2}\$ linearly, by \$60\:\text{mV}\$ for each decade increase in \$I_1\$. And as \$I_1=\frac{I_{\text{C}_2}}{I_1\cdot\left(\beta_1+1\right)\cdot\beta_2}\$, it's pretty certain that there isn't much of a voltage drop across your circuit's \$R_3\$, either. So the input voltage, \$V_i\$, has been tracking \$V_{\text{BE}_1}+V_{\text{BE}_2}\$ right up to this point, or near it. So at first, with linear increases in \$V_i\$ there's really been no change in \$V_\text{DROP}\$. Again, this is why that first stretch of \$V_u\$, so far, has looked flat.
But now we are right at the point where a linear change in \$I_1\$ is going to directly impact \$V_\text{DROP}\$ also in a linear way. You can see this because of \$V_\text{DROP}=I_1\cdot\left[\left(\beta_1+1\right)\cdot\beta_2\cdot\left(R_1+R_2\right)\right]\$. So what happens this time when you change \$I_1\$ by exactly one decade (order of magnitude?) Isn't the voltage drop now going to change also by one order of magnitude??
So we are now entering an area where if we only just noticed that \$V_u\$ started changing by its first \$V_\text{DROP}=100\:\text{mV}\$ (enough to start seeing a change at \$V_u\$), then one decade's change in \$I_1\$ will imply also one decade change in \$V_\text{DROP}\$ so that \$V_\text{DROP}=1\:\text{V}\$, now. And the next decade of change in \$I_1\$ will see \$V_\text{DROP}=10\:\text{V}\$!

Note: This was happening before, too. For each change of \$V_i\$ by
about 2 base-emitter drops (\$120 \:\text{mV}\$), \$V_\text{DROP}\$
was changing by about an order of magnitude. But early on this means
\$V_\text{DROP}=1\:\mu\text{V}\$, then
\$V_\text{DROP}=10\:\mu\text{V}\$, then
\$V_\text{DROP}=100\:\mu\text{V}\$, and so on. Those changes, while
orders of magnitude between them, were still too small for us to care
about in our linear world. For much of the time, these values are
smaller than the accuracy errors of our meters to measure them! It's
only when these order of magnitude changes transgress into our ability
to measure them easily that they show up, visibly. So it may look
sudden. But it really wasn't if you change your perspective.

However, these two decades we are now looking closely at have only meant two more \$60\:\text{mV}\$ changes in the base-emitter junction voltages of \$Q_1\$ and \$Q_2\$ (plus perhaps now just a little added voltage drop across \$R_3\$ to reach \$V_i\$.) So now you see \$V_u\$ dropping like a rock while \$V_i\$ is changing at about the same rate (maybe just slightly faster than before.)
Now, what about the next phase? Is there one? Well, yes, there must be. Obviously, \$V_c\$ can't go below \$0\:\text{V}\$. So at some point \$V_\text{DROP}\$ (keep in mind this is defined earlier as the drop for both resistors) can't get any bigger. Clearly, \$V_\text{DROP}\lt +V\$ in all scenarios! When \$V_\text{DROP}\$ has smashed \$Q_2\$'s collector (\$V_c\$) down about as low as it can go, then our prior assumption that the current in \$R_1\$ and \$R_2\$ is about equal to \$Q_2\$'s collector current starts to fail.
At this point, \$V_c\$ isn't going to be changing much, anymore. It's gone almost as close to ground as possible. The only remaining thing that can change is \$V_u\$ as \$V_i\$ (and \$I_1\$) increases still further. And it is here that \$Q_1\$'s collector current can, in fact, continue to increase. This does drive more current into \$Q_2\$'s base. But no longer can this cause \$Q_2\$ to sink more current. Worse, as \$V_u\$ declines further because of the increasing level of \$Q_1\$'s collector current, there's more voltage drop across \$R_1\$ and, because of the declining \$V_u\$, less current in \$R_2\$. So, in point of fact, \$Q_2\$'s collector current may actually begin to decline! Even with more base current. \$Q_2\$ is now going into deeper saturation.
Think about the process and see if you can come up with the answers you need.