In order to find the exponential regression we are going to select some values of the given data.
An special value is when x=0.
On the table we can see that when x=0 then y=9
Replacing x by 0 in the given choices, we have that:
[tex]\begin{gathered} A\text{.} \\ y=8.04\cdot0.98^x \\ \downarrow \\ y=8.04\cdot0.98^0=8.04\cdot1 \\ =8.04 \end{gathered}[/tex][tex]\begin{gathered} B\text{.} \\ y=3.02\cdot3.67^x \\ \downarrow \\ y=3.02\cdot3.67^0=3.02\cdot1 \\ =3.02 \end{gathered}[/tex][tex]\begin{gathered} C\text{.} \\ y=6.61\cdot1.55^x \\ \downarrow \\ y=6.61\cdot1.55^0=6.61\cdot1 \\ =6.61 \end{gathered}[/tex][tex]\begin{gathered} D\text{.} \\ y=2.27\cdot2.09^x \\ \downarrow \\ y=2.27\cdot2.09^0=2.27\cdot1 \\ =2.27 \end{gathered}[/tex]Observing the results we have that the two choices with closer results to 9 are A (with 8.04) and C (with 6.61)
Now, we are going to select two additional values from the table in order to find which is the best answer: A or C.
Let's take x=1.
When x = 1, then y=10.
Replacing on the equation A we have:
[tex]\begin{gathered} A\text{.} \\ y=8.04\cdot0.98^x \\ \downarrow \\ y=8.04\cdot0.98^1=8.04\cdot0.98 \\ =7.879 \end{gathered}[/tex]and for the equation C:
[tex]\begin{gathered} C\text{.} \\ y=6.61\cdot1.55^x \\ \downarrow \\ y=6.61\cdot1.55^1=6.61\cdot1.55 \\ =10.2455 \end{gathered}[/tex]For x=1, the nearest result is from the equation C.
Let's verify what happens when x=2.
When x=2 then y=16. Replacing on the equation A we have:
[tex]\begin{gathered} A\text{.} \\ y=8.04\cdot0.98^x \\ \downarrow \\ y=8.04\cdot0.98^2 \\ =7.7216 \end{gathered}[/tex]and for the equation C:
[tex]\begin{gathered} C\text{.} \\ y=6.61\cdot1.55^x \\ \downarrow \\ y=6.61\cdot1.55^2 \\ =15.88 \end{gathered}[/tex]Again, for x=2, the nearest result is from the equation C.
Then, we can conclude that the best candidate is equation C.
We could try other values of x to double check:
[tex]\begin{gathered} C\text{.} \\ y=6.61\cdot1.55^x \\ \downarrow \\ y=6.61\cdot1.55^2 \\ =15.88 \end{gathered}[/tex]