Chapter 9.5, Problem 91E

a.

To determine

To rewrite: The model in term of $g\left(t\right)$

Answer to Problem 91E

  $g\left(t\right)=-0.009t^2+0.96t+23.025$

Explanation of Solution

Given:The amount $f\left(t\right)$ (in billions of dollars) of child support collected in the United Stated from 2002 through 2009 can be approximated by the model

  $f\left(t\right)=-0.009t^2+1.05t+18.0,2\le t<9$

Where t represents the year, with $t=2$ corresponding to 2002.

In question $t=2$ corresponding to 2007 rather than 2002. So, shift graph five unit left to get $g\left(t\right)=f\left(t+5\right)$ .

  $\begin{array}{l}g\left(t\right)=f\left(t+5\right)\\ g\left(t\right)=-0.009{\left(t+5\right)}^2+1.05\left(t+5\right)+18.0\\ g\left(t\right)=-0.009\left(t^2+10t+25\right)+1.05t+5.25+18.0\\ g\left(t\right)=-0.009t^2-0.09t-0.225+1.05t+5.25+18.0\\ g\left(t\right)=-0.009t^2+0.96t+23.025\end{array}$

b.

To determine

To graph: The $g\left(t\right)\text{ and }f\left(t\right)$ on same window.

Answer to Problem 91E

  

Explanation of Solution

Given: The amount $f\left(t\right)$ (in billions of dollars) of child support collected in the United Stated from 2002 through 2009 can be approximated by the model

  $f\left(t\right)=-0.009t^2+1.05t+18.0,2\le t<9$ and $g\left(t\right)=-0.009t^2+0.96t+23.025$

Now, graph $f\left(t\right)=-0.009t^2+1.05t+18.0,2\le t<9$ and $g\left(t\right)=-0.009t^2+0.96t+23.025$ on same graphing window.

  

c.

To determine

To find: The year in which collections exceeds $25 billion.

Answer to Problem 91E

  $2007$

Explanation of Solution

Given: The graph of $f\left(t\right)=-0.009t^2+1.05t+18.0,2\le t<9$ and $g\left(t\right)=-0.009t^2+0.96t+23.025$

  

First draw a horizontal line, $y=25$

Line will cut both graph.

For graph $f\left(t\right)$ ,

  $\begin{array}{c}f\left(t\right)=25\\ -0.009t^2+1.05t+18.0=25\\ t\approx 7\end{array}$

Since, $t=2$ represents 2002.

For $t=7$ represents 2007.