Chapter 7.3, Problem 91E

To determine

To calculate: Solve the given system for $a,b$ and $c$ to find the least squares regression parabola for the points and confirm the result with the help of graphing utility.

Answer to Problem 91E

The value of $a,b,c$ are $-\frac{5}{24},-\frac{3}{10},\frac{41}{6}$ respectively and the equation of regression parabola is $y=-\frac{5}{24}{x}^2-\frac{3}{10}x+\frac{41}{6}$ and verify the graph with figure(1).

Explanation of Solution

Given information:

The given system of equations

  $\begin{array}{c}4c+40a=19\\ 40b=-12\\ 40c+544a=160\end{array}$ .

Formula used:

Solve the given system of equations to get the values of $a,b,c$ and substitute in the formula $a{x}^2+bx+c$ to get the least squares parabola.

Calculation:

As per given system of equations

  $\begin{array}{c}4c+40a=19\text{}\text{}\text{}\text{}\text{}\dots \left(1\right)\\ 40b=-12\dots \left(2\right)\\ 40c+544a=160\dots \left(3\right)\end{array}$ .

Solve equation $\left(2\right)$

  $\begin{array}{c}40b=-12\\ b=\frac{-12}{40}\\ b=-0.3\end{array}$ .

Solve equation $\left(1\right)$

  $\begin{array}{c}4c+40a=19\\ 4c=19-40a\\ c=\frac{19-40a}{4}\end{array}$ .

Substitute the value of $c$ in equation $\left(3\right)$

  $\begin{array}{c}40\left(\frac{19-40a}{4}\right)+544a=160\\ 10\left(19-40a\right)+544a=160\\ 190-400a+544a=160\\ 190+144a=160\\ \end{array}$

Further simplify

  $\begin{array}{c}144a=-30\\ a=\frac{-30}{144}\\ a=-\frac{5}{24}\end{array}$ .

Substitute the value of $a$ in equation $\left(1\right)$

  $\begin{array}{l}4c+40\left(\frac{-5}{24}\right)=19\\ 4c-\frac{25}{3}=19\\ 4c=19+\frac{25}{3}\\ 4c=\frac{82}{3}\\ c=\frac{41}{6}\end{array}$ .

Therefore the value of $a,b,c$ are $-\frac{5}{24},-\frac{3}{10},\frac{41}{6}$ respectively.

Therefore the least squares regression parabola is

  $y=-\frac{5}{24}{x}^2-\frac{3}{10}x+\frac{41}{6}$ .

Verify the graph with parabola $y=-\frac{5}{24}{x}^2-\frac{3}{10}x+\frac{41}{6}$

  

Figure(1)

Therefore, both the graphs gives same values at each points.