Chapter 4.3, Problem 22PPSE

Solution

a.

To Complete: The given table.

Number of points, $x$	$2$	$3$	$4$	$5$
Number of segments, $y$	1	3	$6$	$10$

Given:

Number of points, $x$	$2$	$3$
Number of segments, $y$	1	3

Explanation:

Complete the table by plotting points and draw segments (not collinear).

Number of points, $x$	$2$	$3$	$4$	$5$
Number of segments, $y$	1	3

  

From above diagram, number of segments by connecting the $4$ points form a quadrilateral polygon.

That the $6$ segments in $4$ points that are not collinear.

  $\left(x,y\right)\to \left(4\text{ points, }6\text{ segments}\right)$

  

The the number of segments by connecting the $5$ points that form a pentagon.

Thus, $10$ segments in $5$ points that are not collinear.

  $\left(x,y\right)\to \left(5\text{ points, }10\text{ segments}\right)$

Completing the table:

Number of points, $x$	$2$	$3$	$4$	$5$
Number of segments, $y$	1	3	$6$	$10$

b.

To Write: The quadratic model for the data.

The quadratic model for the data is $y=\frac{1}{2}{x}^2-\frac{1}{2}x$

Given:

Number of points, $x$	$2$	$3$
Number of segments, $y$	1	3

Explanation:

The standard form of quadratic model is $y=a{x}^2+bx+c$ :

From part $\left(a\right)$ :

Number of points, $x$	$2$	$3$	$4$	$5$
Number of segments, $y$	1	3	$6$	$10$

Substitute the first point $\left(2,1\right)\to \left(x,y\right)$

  $\begin{array}{l}y=a{x}^2+bx+c\\ 1=a{\left(2\right)}^2+b\left(2\right)+c\\ 1=4a+2b+c\to \left(1\right)\end{array}$

Substitute the first point $\left(3,3\right)\to \left(x,y\right)$

  $\begin{array}{l}y=a{x}^2+bx+c\\ 3=a{\left(3\right)}^2+b\left(3\right)+c\\ 3=9a+3b+c\to \left(2\right)\end{array}$

Substitute the third point $\left(4,6\right)\to \left(x,y\right)$

  $\begin{array}{l}y=a{x}^2+bx+c\\ 6=a{\left(4\right)}^2+b\left(4\right)+c\\ 6=16a+4b+c\end{array}$

Substitute the third point $\left(5,10\right)\to \left(x,y\right)$

  $\begin{array}{c}y=a{x}^2+bx+c\\ 10=a{\left(5\right)}^2+b\left(5\right)+c\\ 10=25a+5b+c\end{array}$

Multiply equation $1$ with $-1$ :

  $-1=-4a-2b-c\to \left(1.1\right)$

Solve equation $\left(1.1\right)$ and $\left(2\right)$ :

  $\begin{array}{l}\underset{¯}{\begin{array}{l}-1=-4a-2b-c\\ +3=9a+3b+c\end{array}}\\ 2=5a+b\to \left(5\right)\end{array}$

Multiply equation $3$ with $-1$ :

  $-6=-16a-4b-c\to \left(3.1\right)$

Solve equation $\left(3.1\right)$ and $\left(4\right)$ :

  $\begin{array}{l}-6=-16a-4b-c\\ \underset{¯}{10=25a+5b+c}\\ 4=9a+b\to \left(6\right)\end{array}$

Multiply equation $5$ with $-1$ :

  $-2=-5a-b\to \left(5.1\right)$

Solve equation $\left(5.1\right)$ and $\left(6\right)$ :

  $\begin{array}{l}c-2=-5a-b\\ \underset{¯}{4=9a+b}\\ 2=4a\\ a=\frac{1}{2}\end{array}$

Substitute $a=\frac{1}{2}$ in equation $\left(6\right)$ to find $b$ :

  $\begin{array}{c}4=9a+b\\ 4=9\cdot \frac{1}{2}+b\\ 4=\frac{9}{2}+b\\ -\frac{1}{2}=b\end{array}$

Substitute $a=\frac{1}{2}$ , and $b=-\frac{1}{2}$ in equation $1$ :

  $\begin{array}{c}1=4a+2b+c\\ 1=4\cdot \frac{1}{2}+2\cdot -\frac{1}{2}+c\\ 1=2-1+c\\ 0=c\end{array}$

Thus, plug $a=\frac{1}{2},b=-\frac{1}{2}$ and $c=0$ in $y=a{x}^2+bx+c$ :

  $y=\frac{1}{2}{x}^2-\frac{1}{2}x$

c.

To Predict: The number of segments that can be drawn using $10$ points.

There are $45$ segments in $10$ points.

Given:

Number of points, $x$	$2$	$3$
Number of segments, $y$	1	3

Explanation:

From part $\left(b\right)$ ,

  $y=\frac{1}{2}{x}^2-\frac{1}{2}x$

For predicting number of segments with $10$ points consider $x=10$ :

Substitute $x=10$ in $y=\frac{1}{2}{x}^2-\frac{1}{2}x$ L

  $\begin{array}{c}y=\frac{1}{2}{x}^2-\frac{1}{2}x\\ =\frac{1}{2}{\left(10\right)}^2-\frac{1}{2}\left(10\right)\\ =50-5\\ =45\end{array}$

Thus, there are $45$ segments in $10$ points.