Answer
Linear model: $\qquad I =5.88t-24.6$
Exponential model: $\qquad I= 1.87(1.34)^{t}$
The exponential model is more appropriate.
Work Step by Step
A linear model has the form: $I=mt+b$
Given the points: $(6,10.7)$ and $(10,34.2)$, we find the slope:
$m=\displaystyle \frac{I_{2}-I_{1}}{t_{2}-t_{1}}=\frac{34.2-10.7}{10-6}=\frac{23.5}{4}=5.88$
So, $I=5.88t+b$
To find b, insert $(t=6,I=10.7)$
$b=10.7-5.88(6)=-24.6$
Linear model: $\qquad I =5.88t-24.6$
Exponential models have the form: $I=Ab^{t}$
Substitute $(t=6,F=10.7)\qquad 10.7=Ab^{6},$
substitute $(t=10,F=34.2)\qquad 34.2=Ab^{10}.$
Divide the second equation by the first:
$\begin{aligned}
\displaystyle \frac{34.2}{10.7}&=\displaystyle \frac{b^{10}}{b^{6}}=b^{4}\\3.19626168224&=b^{4}\displaystyle \\b&=(3.19626168224)^{1/4}\displaystyle \approx 1.34\end{aligned}$
Substitute the value of $b$ in $10.7=Ab^{6}$
$\displaystyle \begin{aligned}10.7&\displaystyle \approx A(1.34)^{6},\text{so}\\A&\displaystyle \approx\frac{10.7}{(3.19626168224)^{6}}\approx 1.87\end{aligned}$
Exponential model: $\qquad I= 1.87(1.34)^{t}$
The differences between successive terms are increasing.
Since they are not roughly constant, the linear model will not be appropriate.
The ratios between successive terms are in the range $1.27$ to $1.40$, so we can say roughly, $1.34$, which is why we take the exponential model to be appropriate.
