The first offer is a linear function.
It starts at y = 72000 for t = 0 and increases 5040 per year, so the slope of the line is m = 5040.
We can model the salary as:

The second offer increases proportionally, so it is an exponential growth.
It increases 6% per year, so the salary at year t is 1.06 times the salary of year (t-1).
We can express it as:

We now can complete the table for the first offer as:

Now, we calculate for the second offer:
