## Precalculus (6th Edition) Blitzer

(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.