Chapter 1.4, Problem 65E

a.

To determine

To identify:

The transformation of parent function $f\left(t\right)=t^2$ .

Answer to Problem 65E

The transformation of parent function $f\left(t\right)=t^2$ .is as below:

1. Horizontal shift of $26.17$ units to right.
2. Vertical shrink of $0.03$ units.
3. Reflection about the $t-$ axis.
4. Vertical shrink of $126.5$ units upwards.

Explanation of Solution

Give information:

Given:

  $N\left(t\right)=0.03{\left(t-26.17\right)}^2+126.5$ .

Calculation:

The numbers $N\left(inmillions\right)$ of house hold from $2000$ to $2013$ are given.

Model for the given data is ,

  $N\left(t\right)=0.03{\left(t-26.17\right)}^2+126.5$

Where

  $t=0$ represents $2000$ .

The parent function of the given function $f\left(t\right)=t^2$ ,

To get the graph of $y=f\left(x-c\right)$ , shifts graph of $y=f\left(x\right)$ a distance $c$ units to right.

To get the graph of $y=cf\left(x\right)$ , shrink graph of $y=f\left(x\right)$ vertically by a factor of $0<c<1$ .

To get the graph of $y=-f\left(x\right)$ , reflect the graph of $y=f\left(x\right)$ about $x-$ axis.

To get the graph of $y=f\left(x\right)+c$ , shift the graph of $y=f\left(x\right)$ a distance $c$ units upwards.

Thus, transformation of $N\left(t\right)=-0.03{\left(t-26.17\right)}^2+126.5$ from $f\left(t\right)=t^2$ is obtained by,

1. Horizontal shift of $26.17$ units to right.
2. Vertical shrink of $0.03$ units.
3. Reflection about the $t-$ axis.
4. Vertical shrink of $126.5$ units upwards.

b.

To determine

To Graph:

Model and given data in the same viewing window.

Explanation of Solution

Give information:

Given:

  $N\left(t\right)=0.03{\left(t-26.17\right)}^2+126.5$ .

Graph:

  

Interpretation:

Graph model and given data points using graphing utility is as above.

c.

To determine

To calculate:

Function so that $t=0$ represents $2009$ .

Answer to Problem 65E

Required function is $G\left(t\right)=-0.03{\left(t-17.17\right)}^2+126.5$

Explanation of Solution

Give information:

Given:

  $N\left(t\right)=0.03{\left(t-26.17\right)}^2+126.5$ .

Calculation:

If $t=0$ represents $2009$ , then to fit the data there will be a horizontal shift of $9units$ to the left and so the new functioncan be get from $N\left(t\right)$ as,

  $\begin{array}{l}G\left(t\right)=N\left(t+9\right)\\ G\left(t\right)=-0.03{\left(t-26.17\right)}^2+126.5\\ G\left(t\right)=-0.03{\left(t-17.17\right)}^2+126.5\end{array}$

Required new function is,

  $G\left(t\right)=-0.03{\left(t-17.17\right)}^2+126.5$