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:
[tex]y=72000+5040\cdot t[/tex]
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:
[tex]\begin{gathered} y(0)=72000 \\ y(1)=1.06\cdot y(0)=1.06\cdot72000 \\ y(2)=1.06\cdot y(1)=1.06\cdot1.06\cdot72000=1.06^2\cdot72000 \\ \Rightarrow y(t)=72000\cdot1.06^t \end{gathered}[/tex]
We now can complete the table for the first offer as:
[tex]\begin{gathered} y(1)=72000+5040\cdot1=72000+5040=77040 \\ y(5)=72000+5040\cdot5=72000+25200=97200 \\ y(10)=72000+5040\cdot10=72000+50400=122400 \\ y(15)=72000+5040\cdot15=72000+75600=147600 \\ y(20)=72000+5040\cdot20=72000+100800=172800 \end{gathered}[/tex]
Now, we calculate for the second offer:
[tex]\begin{gathered} y(1)=72000\cdot1.06^1=76320 \\ y(5)=72000\cdot1.06^5\approx96352 \\ y(10)=72000\cdot1.06^{10}\approx128941 \\ y(15)=72000\cdot1.06^{15}\approx172552 \\ y(20)=72000\cdot1.06^{20}\approx230914 \end{gathered}[/tex]
