Chapter 7.3, Problem 67E

To determine

To find: The least squares regression parabola for the given points.

Answer to Problem 67E

The values of $a,b,\text{and c}$ in the given figure are $1,-1,\text{and 0}$ , respectively.

Explanation of Solution

Given information:

The data points are given in the graph below:

  

For the above data points, the sums have been evaluated and that results to the following system of linear equations.

  $\{\begin{array}{c}4c+9b+29a=20\\ 9c+29b+99a=70\\ 29c+99b+353a=254\end{array}$

Concept used:

One way to find the least squares regression parabola $y=a{x}^2+bx+c$ for a set of points $\left(x_1,y_1\right),\left(x_2,y_2\right),\mathrm{...},\left(x_n,y_n\right)$ is by solving the system below for $a,b,\text{and }c.$

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

Calculations:

The steps of solving the system of equations using Row-Echelon Form and Back substitutions are as follows:

Subtracting the first equation from the second equation to produce a new first equation, we get

  $\{\begin{array}{c}5c+20b+70a=50\\ 9c+29b+99a=70\\ 29c+99b+353a=254\end{array}$

Now dividing the first equation by 5 to produce a new first equation, we get

  $\{\begin{array}{c}c+4b+14a=10\\ 9c+29b+99a=70\\ 29c+99b+353a=254\end{array}$

Adding $-9$ times the first equation to the second equation to produce a new second equation and adding $-29$ times the first equation to the third equation to produce a new third equation, we get

  $\{\begin{array}{c}c+4b+14a=10\\ -7b-27a=-20\\ -17b-53a=-36\end{array}$

Now, adding 17 times the second equation to $-7$ times the third equation to produce a new third equation, we get

  $\{\begin{array}{c}c+4b+14a=10\\ -7b-27a=-20\\ -88a=-88\end{array}$

Solving $-88a=-88$ for $a$ , we obtain $a=1$ .

Substituting this value in the second equation, we get

  $\begin{array}{l}-7b-27(1)=-20\\ \Rightarrow -7b-27=-20\\ \Rightarrow -7b=27-20=7\\ \Rightarrow b=-1\end{array}$

Now, substituting the above finding values in the first equation, we get

  $\begin{array}{l}c+4(-1)+14(1)=10\\ \Rightarrow c-4+14=10\\ \Rightarrow c=0\end{array}$

Conclusion:

Therefore, the values of $a,b,\text{and c}$ in the given figure are $1,-1,\text{and 0}$ , respectively.

To verify these results, we will be using Desmos graphing utility.

  

Hence, the results we got are correct.