Hello,
Here's what I'm thinking - mightn't be spot on but here we go...
Can someone explain to me why the graph looks the way it does (and not flipped the other way, e.g positive first peak)
Imagine the solenoid as a battery (similar to generators), in that a current is being
produced. So, like in a generator, (or in the little bit between two battery terminals) the conventional current goes from negative to positive. Other things outside of this will have current flowing in the normal, conventional way from positive to negative. These 'other things' could be light globes, etc. In this case, it's a voltmeter.
Now looking at the current that the magnet induces in the solenoid. As the south pole moves towards the top of the solenoid, a south pole is induced to oppose the change (Lenz's Law). Using the right-hand grip rule, the conventional current moves from the top to the bottom of the solenoid. In between the terminals of the voltmeter/ datalogger is what we're looking for, as that's when the current leaves the 'battery' and enters the circuit. Hence -ve to +ve. The first blip is in the negative direction to reflect this. The second shows the current moving from the bottom of the solenoid to the top, so from the datalogger's POV, current is going from +ve to -ve.
Also, why is the (inflexion?) point where it changes polarity not instantaneous and is dragged out over a longer time?
From what I'm guessing, we're supposed to assume that the solenoid of considerable length. In which case, there will be some time when the magnet is completely inside the solenoid ( and still falling down). It won't be inducing any current during this time, so the voltmeter won't pick it up. I'm not sure on the specifics of why this is the case, though. Perhaps it's because while it's inside, a south pole wants to be generated at the top as the north pole at the top of the magnet leaves, it - but at the same time a south pole wants to be generated at the bottom of the solenoid to repel the magnet's south pole as it accelerates towards it (therefore while there is an EMF, no net current is produced)? Please don't take my word for it, though.
Hopefully something was helpful, regardless