D midpoint of EC -----------------> FD parallel to AC and FD=AC/2=14/2=7
<span>2-EB=EA E midpoint of AB </span>
<span>DB=DC D midpoint BC ...............> ED=AC/2=2 </span>
<span>3-T midpoint of SR </span>
<span>U midpoint of QR ---------> TU = QS/2 </span>
<span>QS=2 TU = 4.4 </span>
<span>4- The same steps SR=2 UV=9 </span>
<span>5-N midpoint of KM </span>
<span>O midpoint of ML </span>
<span>* NO parallel to Kl</span>
We can see that the graph touches
without crossing the x-axis (i.e. it is a double solution), and then there's another zero at
(this time it's a crossing zero, so a single solution).
This leads, up to multiple, to the polynomial

If we impose the passing through
we have

So, the polynomial is

Finally, to solve
, simply look at the graph, searching for the points, where the graph is below the x-axis. You can see that this happens only if
, so that's the solution to your question.
At sea level, the pressure is 1013, that's when the altitude is at 0, sea level, let's see

so, the inital amount is 1013, when t = 0,

now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.