Chapter 4.6, Problem 3DC

Solution

To Draw: a scatter plot of given points and also find the power model of the given point.

  $y=1906\cdot {\left(x\right)}^{-0.098}$ .

Given: points are given in $\left(x,y\right)$ as $\left(1,1906\right)$ , $\left(2,1832\right)$ , $\left(3,1762\right)$ , $\left(4,1692\right)$ ,and $\left(5,1627\right)$ .

Concept used:

(1) If a function $y=f\left(x\right)$ passes through a point $\left(x_0,y_0\right)$ , then the point $\left(x_0,y_0\right)$ satisfies the equation of the function, that is, $y_0=f\left(x_0\right)$ .

(2) A power model or function is defined as:

  $y=a\cdot {(x)}^b$ ; where $a\ne 0$ and $b\ge 0$ .

(3) Equation of a line passing through the given two points as like $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

Then, $\left(y-y_1\right)=m\left(x-x_1\right)$ .

Here, $m$ is the slope. And $m$ is as follows:

  $m=\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

The table of data pairs $\left(x,y\right)$ written as;

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ y& 1906& 1832& 1762& 1692& 1627\end{array}$

The Scatter graph of $\left(x,y\right)$ is given as;

  

Now, the general power function form is $y=a\cdot {\left(x\right)}^b$ ;

Here, choose two points $\left(1,1906\right)$ and $\left(5,1627\right)$ then the equation of the line;

  $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Now, substitute the value;

  $\begin{array}{c}y-1906=\frac{1627-1906}{5-1}\left(x-1\right)\\ y-1906=\frac{-279}{4}\left(x-1\right)\\ y-1906=69.75\left(x-1\right)\\ y=69.75x-69.75+1906\end{array}$

Thus,

  $y=69.75x+1836.25$

This is a required equation of a line between given two points.

Now, the general power function of the model form is $y=a\cdot {\left(x\right)}^b$   ....... (1)

The point $\left(1,1906\right)$ passes through the line. Then, the above equation (1) is as follows:

  $1906=a{\left(1\right)}^b$

  $a=\frac{1906}{{\left(1\right)}^b}$  ....... (2)

Again similarly the point $\left(5,1627\right)$ passes through the line. Then, the above equation (1) is as follows:

  $1627=a{\left(5\right)}^b$  ....... (3)

Now substitute the value of $a$ from equation (2) in equation (3) becomes;

  $\begin{array}{c}1627=\frac{1906}{{(1)}^b}\cdot {\left(5\right)}^b\\ \frac{1627}{1906}={\left(\frac{5}{1}\right)}^b\\ 0.854={\left(5\right)}^b\end{array}$

Take on both sides logarithm with base $\left(5\right)$ ;

  $\begin{array}{c}{\log }_{5}0.854={\log }_5{\left(5\right)}^b\\ -0.098=b{\log }_{5}5\left[{\log }_a{m}^n=n{\log }_{a}m\right]\\ b=-0.098\left[{\log }_{a}a=1\right]\end{array}$

Substitute the value of $b=-0.098$ in the above equation (2);

  $\begin{array}{c}a=\frac{1906}{{\left(1\right)}^{-0.098}}\\ a=1906\cdot {\left(1\right)}^{0.098}\\ a=1906\left[{\left(1\right)}^n=1;n\ge 1\right]\end{array}$

Therefore, substitute the value of $a$ and $b$ in equation (1);

  $y=1906\cdot {\left(x\right)}^{-0.098}$

This is a power function of the model of given data pairs.

Conclusion:

Hence, the required power function of the model that satisfies the given points is $y=1906\cdot {\left(x\right)}^{-0.098}$ .