fit a parabola, by the method of least squares, to the following data: x: 1,2,3,4,5 , y: 5,12,26,60,97

Solution:

$ \begin{array}{|c|c|c|c|c|c|c|} \hline Sr. & x & y & x^{2} & x^{3} & x^{4} & x y & x^{2} y \\ \hline 1 & 1 & 5 & 1 & 1 & 1 & 5 & 5 \\ \hline 2 & 2 & 12 & 4 & 8 & 16 & 24 & 48 \\ \hline 3 & 3 & 26 & 9 & 27 & 81 & 78 & 234 \\ \hline 4 & 4 & 60 & 16 & 64 & 256 & 64 & 960 \\ \hline 5 & 5 & 97 & 25 & 125 & 625 & 485 & 2425 \\ \hline \Sigma Total & 15 & 200 & 55 & 225 & 979 & 832 & 3672 \\ \hline \end{array} $

let $y=a x^{2}+bx+c$ be the best fit

Then normal equations are,

$$ \begin{array}{I} a \Sigma x^{2}+b \Sigma x+ c\Sigma N=\Sigma y \\ a \Sigma x^{3}+b \Sigma x^{2}+c \Sigma x=\Sigma xy \\ a \Sigma x^{4} + b \Sigma x^{3} +c \Sigma x^{2}= \Sigma x^{2}y \end{array} $$

Hence solving normal equations, it becemes

$$ \begin{array}{l} 55 a+15 b+5 c=200 \\ 225 a+55 b+15 c=832 \\ 979 a+225 b+55 c=3672 \end{array} $$

On solving above equations, we get

$$ a=5.72, b=-11.09, c=10.40 $$

Hence the porabola is,

$$ y=5.72 x^{2}-11.09 x=10.40 $$