Answer
See the full explanation below.
Work Step by Step
(a)
The value of the number of discharges for ${{x}_{1}}=0\text{ }$ is:
$\begin{align}
& f\left( x \right)=1.1{{x}^{3}}-35{{x}^{2}}+264x+557 \\
& f\left( 0 \right)=1.1{{\left( 0 \right)}^{3}}-35{{\left( 0 \right)}^{2}}+264\left( 0 \right)+557 \\
& f\left( 0 \right)=557 \\
\end{align}$
The value of the number of discharges for ${{x}_{2}}=4$ is
$\begin{align}
& f\left( x \right)=1.1{{x}^{3}}-35{{x}^{2}}+264x+557 \\
& f\left( 4 \right)=1.1{{\left( 4 \right)}^{3}}-35{{\left( 4 \right)}^{2}}+264\left( 4 \right)+557 \\
& f\left( 4 \right)=70.4-560+1056=557 \\
& f\left( 4 \right)=1123.4 \\
\end{align}$
Substitute the values $\left( {{x}_{2}},{{x}_{1}} \right)=\left( 4,0 \right)$ and $\left( f\left( {{x}_{2}} \right),f\left( {{x}_{1}} \right) \right)=\left( 1123.4,557 \right)$ to get the slope:
$\begin{align}
& m=\frac{f\left( {{x}_{2}} \right)-f\left( {{x}_{1}} \right)}{{{x}_{2}}-{{x}_{1}}} \\
& =\frac{1123.4-557}{4-0} \\
& =\frac{566.4}{4} \\
& \approx 142
\end{align}$
Therefore, the slope of the secant line is $142$ from ${{x}_{1}}=0\text{ to }{{x}_{2}}=4$.
(b)
The difference between the slope determines whether it is underestimates or overestimates.
The difference in slope is
$\begin{align}
& {{m}_{2}}-{{m}_{1}}=142-137 \\
& \Delta m=5
\end{align}$
Hence, the slope overestimates by $5$ discharges per year.
