In order to see which function is a more appropiated model for the data recorded by Rosa we can plot the graphs of the four functions in the same grid where the data points are plotted.
As you can see in options A and D we have linear functions whereas in options B and C we have exponential functions. Let's start with these two.
Let's choose some random x-values and find their respective y-values using each of the exponential functions. For example we can take x=0, x=3, x=7 and x=10. Then we find their respective y-values with B(x) (the function of option B) and C(x) (the funciton of option C).
We start with B:
[tex]\begin{gathered} B(x)=2.0\cdot(1.4)^x \\ x=0\Rightarrow y=2.0\cdot(1.4)^0=2.0\Rightarrow(0,2.0) \\ x=3\Rightarrow y=2.0\cdot(1.4)^3=5.49\Rightarrow(3,5.49) \\ x=7\Rightarrow y=2.0\cdot(1.4)^7=21.08\Rightarrow(7,21.08) \\ x=10\Rightarrow y=2.0\cdot(1.4)^{10}=57.85\Rightarrow(10,57.85) \end{gathered}[/tex]So the function in B passes through those points. The y-values of the last two are very high so the graph of B(x) clearly diverges from the data points so it seems it's not an appropiate model.
Then we continue with the function in option C:
[tex]\begin{gathered} C(x)=2.2\cdot(1.1)^x \\ x=0\Rightarrow y=2.2\cdot(1.1)^0=2.2\Rightarrow(0,2.2) \\ x=3\operatorname{\Rightarrow}y=2.2(1.1)^3=2.93\operatorname{\Rightarrow}(3,2.93) \\ x=7\operatorname{\Rightarrow}y=2.2(1.1)^7=4.29\operatorname{\Rightarrow}(7,4.29) \\ x=10\operatorname{\Rightarrow}y=2.2(1.1)^{10}=5.71\operatorname{\Rightarrow}(10,5.71) \end{gathered}[/tex]So the graph of C(x) must pass through those 4 points.
For the linear functions the calculations are easier. In order to graph a line we just need two of its points so we just find the y-values corresponding to x=0 and x=10. We start with option A:
[tex][/tex]