Chapter 1.4, Problem 66E

(a)

To determine

To find: The transformation of the parent function $f\left(t\right)=\sqrt{t}$ .

Answer to Problem 66E

The transformation of the parent function $f\left(t\right)=\sqrt{t}$ is shifting of function by $3.7$ units to the left and stretching it vertically by a factor $1.9$ .

Explanation of Solution

Given information:

The depreciation $D$ (in millions of dollars) of the $\text{WD}-40$ company assets from $2009$ through $2013$ can be approximated by the function $D\left(t\right)=1.9\sqrt{t+3.7}$ where $t=0$ represents $2009$ .

Calculation:

Shift the graph of the parent function $f\left(t\right)=\sqrt{t}$ by $3.7$ units to the left, the function transforms to $\sqrt{t+3.7}$ .

Now stretch the graph of function $\sqrt{t+3.7}$ vertically by a factor $1.9$ , the function transforms to the depreciation function $D\left(t\right)=1.9\sqrt{t+3.7}$ .

Therefore, the transformation of the parent function $f\left(t\right)=\sqrt{t}$ is shifting of function by $3.7$ units to the left and stretching it vertically by a factor $1.9$ .

(b)

To determine

To graph: The model of the depreciation over the interval $0\le t\le 4$ .

Explanation of Solution

Given information:

The depreciation $D$ (in millions of dollars) of the $\text{WD}-40$ company assets from $2009$ through $2013$ can be approximated by the function $D\left(t\right)=1.9\sqrt{t+3.7}$ where $t=0$ represents $2009$ .

Graph:

To graph a function $D\left(t\right)=1.9\sqrt{t+3.7}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=1.9\sqrt{t+3.7}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{1.9}\boxed{(}\boxed{\text{X}}\boxed{+}\boxed{3.7}\boxed{)}\boxed{^}\boxed{(}\boxed{1}\boxed{÷}\boxed{2}\boxed{)}$ .

The display will show the equation,

  ${\text{Y}}_{\text{1}}=1.9\left(\text{X}+3.7\right)^\left(1/2\right)$

Press the $\boxed{\text{WINDOW}}$ key and adjust the window by $\left[0,4\right]$ by $\left[-1,7\right]$ .

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given quadratic equation in standard window as shown in Figure (1).

  

Figure (1)

(c)

To determine

To find: The year when the depreciation of $\text{WD}-40$ assets be approximately $6$ million dollars.

Answer to Problem 66E

The year when the depreciation of $\text{WD}-40$ assets be approximately $6$ million dollars is $2015$ .

Explanation of Solution

Given information:

The depreciation $D$ (in millions of dollars) of the $\text{WD}-40$ company assets from $2009$ through $2013$ can be approximated by the function $D\left(t\right)=1.9\sqrt{t+3.7}$ where $t=0$ represents $2009$ .

Calculation:

Substitute $6$ for $D\left(t\right)$ in the function of depreciation.

  $\begin{array}{c}D\left(t\right)=1.9\sqrt{t+3.7}\\ 6=1.9\sqrt{t+3.7}\\ \sqrt{t+3.7}=\frac{6}{1.9}\\ \sqrt{t+3.7}=3.157\end{array}$

Further simplify,

  $\begin{array}{c}\sqrt{t+3.7}=3.157\\ {\left(\sqrt{t+3.7}\right)}^2={\left(3.157\right)}^2\\ t+3.7=9.97\\ t=6.27\end{array}$

The value of depreciation will be $6$ million dollars at year corresponding to $t=6$ , as $t=0$ represents $2009$ so, $t=6$ represents $2015$ .

Therefore, the year when the depreciation of $\text{WD}-40$ assets be approximately $6$ million dollars is $2015$ .

(d)

To determine

To find: The new function such that $t=0$ represents $2011$ .

Answer to Problem 66E

The new function such that $t=0$ represents $2011$ is $D\left(t\right)=1.9\sqrt{t+5.7}$ .

Explanation of Solution

Given information:

The depreciation $D$ (in millions of dollars) of the $\text{WD}-40$ company assets from $2009$ through $2013$ can be approximated by the function $D\left(t\right)=1.9\sqrt{t+3.7}$ where $t=0$ represents $2009$ .

Calculation:

As in the new depreciation function $t=0$ represents $2011$ . So, $t=-2$ represents the year $2009$ .

This means to shift the graph of the depreciation function by $2$ units to the left.

  $\begin{array}{c}D\left(t\right)=1.9\sqrt{t+3.7}\\ D\left(t\right)=1.9\sqrt{\left(t+2\right)+3.7}\\ D\left(t\right)=1.9\sqrt{t+5.7}\end{array}$

Therefore, the new function such that $t=0$ represents $2011$ is $D\left(t\right)=1.9\sqrt{t+5.7}$ .