# Find the quadratic function that is best fit for f(x) defined by the table below

Find the quadratic function that is best fit for f(x) defined by the table below

$\begin{array}{|cc|}\hline X& f\left(x\right)\\ 0& 0\\ 2& 401\\ 4& 1598\\ 6& 3595\\ 8& 6407\\ 10& 10,009\\ \hline\end{array}$

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Step 1 Let the quadratic equation be $f\left(x\right)=a{x}^{2}+bx+c$ Since there are 3 unlnowns a, b, c Let us use the least square method for the curve fit

Step 2

$\begin{array}{|ccccccc|}\hline X& y& xy& {x}^{2}& {x}^{2}y& {x}^{3}& {x}^{4}\\ 0& 0& 0& 0& 0& 0& 0\\ 2& 401& 802& 4& 1604& 8& 16\\ 4& 1598& 6392& 16& 25568& 64& 256\\ 6& 3595& 21570& 36& 129420& 216& 1296\\ 8& 6407& 51256& 64& 410048& 512& 4096\\ 10& 10009& 100090& 100& 1000900& 1000& 10000\\ \sum x=30& \sum y=22010& \sum xy=180110& \sum {x}^{2}=220& \sum {x}^{2}y=1567540& \sum {x}^{3}=1800& \sum {x}^{4}=1\\ \hline\end{array}$

Step 3 number of data $=n=6$ $\sum y=a\sum {x}^{2}+b\sum x+nc$
$22010=220a+30b+6c\dots .\left(1\right)$
$\sum xy=a\sum {x}^{3}+b\sum {x}^{2}+c\sum x$
$180110=1800a+220b+30c\dots .\left(2\right)$
$\sum {x}^{2}y=a\sum {x}^{4}+b\sum {x}^{3}+c\sum {x}^{2}$
$1567540=15664a+1800b+220c\dots .\left(3\right)$ Step 4 Solve equation (1), (2) and (3) $a=100.29$
$b=2.04$
$c=1.25$
$y=100.29{x}^{2}-2.04x+1.25$