Answer
Linear model: $\qquad F =65t-60$
Exponential model: $\qquad F= 97.2(1.20)^{t}$
The exponential model is more appropriate.
Work Step by Step
A linear model has the form: $F=mt+b$
Given the points: $(4,200)$ and $(10,590)$, we find the slope:
$m=\displaystyle \frac{F_{2}-F_{1}}{t_{2}-t_{1}}=\frac{590-200}{10-4}=65$
So, $F=65t+b$
To find b, insert $(t=4,F=200)$
$b=200-(65)(4)=-60$
Linear model: $\qquad F =65t-60$
The differences of successive terms in the table are not constant.
They are increasing, so our linear model will not be a good fit.
Exponential models have the form: $F=Ab^{t}$
Substitute $(t=4,F=200)\qquad 200=Ab^{4},$
substitute $(t=10,F=590)\qquad 590=Ab^{10}.$
Divide the second equation by the first:
$\displaystyle \begin{aligned}\frac{590}{200}&=\displaystyle \frac{b^{10}}{b^{4}}=b^{6}\\2.95&=b^{6}\displaystyle \\b&=(2.95)^{16}\displaystyle \approx 1.19758\approx 1.20\end{aligned}$
Substitute the value of $b$ in $200=Ab^{4}$
$\displaystyle \begin{aligned}200&\displaystyle \approx A(1.19758)^{4}\approx 2.0569A,\text{so}\\A&\displaystyle \approx\frac{200}{2.0569}\approx 97.2\end{aligned}$
Exponential model: $\qquad F= 97.2(1.20)^{t}$
The successive ratios of the values of $F$ in the table are roughly $(1.20)^{2}\approx 1.44$, and, knowing $F(0)$, we would multiply it with $1.20$ to obtain $F(1)$,
and then again with $1.20$ to obtain $F(2)$
so, $F(2)\approx 100(1.20)^{2}\approx 97.2(1.20)^{2}$.
Thus, we will take the exponential model to be more appropriate.
