Chapter 8.4, Problem 118E

a.

To determine

To adjust: the given model so that $t=0$ corresponds to 2005 rather than 2000.

Answer to Problem 118E

  $g\left(t\right)=-0.039t^2+0.91t+23.225$ for $5\le t\le 18$ .

Explanation of Solution

Given:

The function corresponds to $t=0\left(2000\right)$ is given by:

  $f\left(t\right)=-0.039t^2+1.30t+17.7$

Such that $0\le t\le 13$

Given graph:

  

Consider the given model.

  $f\left(t\right)=-0.039t^2+1.30t+17.7$

Now, adjust the given model so that $t=0$ corresponds to 2005 rather than 2000.

Shift the graph of $f\left(t\right)$ by 5 units to the left.

  $\begin{array}{c}g\left(t\right)=f\left(t+5\right)\\ =-0.039{\left(t+5\right)}^2+1.30\left(t+5\right)+17.7\\ =-0.039\left(t^2+10t+25\right)+1.30t+6.5+17.7\\ =-0.039t^2-0.39t+1.30t-0.975+6.5+17.7\\ =-0.039t^2+0.91t+23.225\end{array}$

Hence, $g\left(t\right)=-0.039t^2+0.91t+23.225$ such that $5\le t\le 18$ .

b.

To determine

To graph: the function $f\left(t\right)$ and $g\left(t\right)$ in same viewing window.

Explanation of Solution

Given:

The function corresponds to $t=0\left(2000\right)$ is given by:

  $f\left(t\right)=-0.039t^2+1.30t+17.7$

Such that $0\le t\le 13$

And

  $g\left(t\right)=-0.039t^2+0.91t+23.225$ such that $5\le t\le 18$ .

Construct the graph of $f\left(t\right)$ and $g\left(t\right)$ in same viewing window as shown:

  

The graph $f\left(t\right)$ is shifted horizontally by 5 units to obtain the graph of $g\left(t\right)$ .