Chapter 13.3, Problem 28E

To determine

To find:The least squares regression parabola for the given points and verify the equation using a graphing utility.

Answer to Problem 28E

The equation is $y=-\frac{5}{16}{x}^2+\frac{3}{20}x+\frac{37}{16}$ and the graph is shown in Figure 1.

Explanation of Solution

Given:

The points are $\left(-3,-1\right)$ , $\left(-1,2\right)$ , $\left(1,2\right)$ and $\left(3,0\right)$ .

Calculation:

Let the general equation of the least squares regression parabola be $f\left(x\right)=a{x}^2+bx+c$ where the following equations are used to determine the unknowns a, b and c:

  $\begin{array}{c}cn+b{\displaystyle \sum _{i=1}^n{x}_i}+a{\displaystyle \sum _{i=1}^n{x}_i{}^2}={\displaystyle \sum _{i=1}^n{y}_i}\\ c{\displaystyle \sum _{i=1}^n{x}_i}+b{\displaystyle \sum _{i=1}^n{x}_i{}^2}+a{\displaystyle \sum _{i=1}^n{x}_i{}^3}={\displaystyle \sum _{i=1}^n{x}_i{y}_i}\\ c{\displaystyle \sum _{i=1}^n{x}_i{}^2}+b{\displaystyle \sum _{i=1}^n{x}_i{}^3}+a{\displaystyle \sum _{i=1}^n{x}_i{}^4}={\displaystyle \sum _{i=1}^n{x}_i{}^2{y}_i}\end{array}$

…(1)

In order to determine value of the coefficients of a, b and c in the set of equations (1), construct a table as shown below:

x	y	$x^2$	$x^3$	$x^4$	$xy$	$x^{2}y$
-3	-1	9	-27	81	3	-9
-1	2	1	-1	1	-2	2
1	2	1	1	1	2	2
3	0	9	27	81	0	0
${\displaystyle \sum _{i=1}^3{x}_i}=0$	${\displaystyle \sum _{i=1}^3{y}_i}=3$	${\displaystyle \sum _{i=1}^3{x}_i{}^2}=20$	${\displaystyle \sum _{i=1}^3{x}_i{}^3}=0$	${\displaystyle \sum _{i=1}^3{x}_i{}^4}=164$	${\displaystyle \sum _{i=1}^3{x}_i{y}_i}=3$	${\displaystyle \sum _{i=1}^3{x}_i{}^2{y}_i}=-5$

Table 1

Substitute the values from the above table into the set of equations (1).

  $\begin{array}{c}4c+0b+20a=3\\ 0c+20b+0a=3\\ 20c+0b+164a=-5\end{array}$

Simplify the set of equations to determine the values of a, b and c.

  $\begin{array}{c}a=-\frac{5}{16}\\ b=\frac{3}{20}\\ c=\frac{37}{16}\end{array}$

So, the equation of the least squares regression parabola becomes,

  $\begin{array}{c}f\left(x\right)=\left(-\frac{5}{16}\right)x^2+\left(\frac{3}{20}\right)x+\frac{37}{16}\\ y=-\frac{5}{16}{x}^2+\frac{3}{20}x+\frac{37}{16}\end{array}$

In order to verify the equation $y=-\frac{5}{16}{x}^2+\frac{3}{20}x+\frac{37}{16}$ , plot the points and the equation in a graphing utility. The graph is shown below:

  

Figure 1