 # Find the quadratic function that is best fit for f(x) vakirnarhh 2021-12-03 Answered
Find the quadratic function that is best fit for f(x) defined by the table below
$\begin{array}{}X& f\left(x\right)\\ 0& 0\\ 2& 401\\ 4& 1598\\ 6& 3595\\ 8& 6407\\ 10& 10,009\end{array}$
The quadratic function is y=.
(Type an equation using as the variable. Round to two decimals places as needed)
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Step 1
Let the quadratic equation be
$f\left(x\right)=a{x}^{2}+bx+c$
Since there are 3 unknowns a, b, c.
Let us use the least square method for the curve fit
Step 2
$\begin{array}{}x& y& xy& {x}^{2}& {x}^{2}y& {x}^{3}& {x}^{4}\\ 0& 0& 0& 0& 0& 0& 0\\ 2& 401& 802& 4& 1604& 8& 16\\ 4& 1598& 6392& 16& 25568& 64& 256& \\ 6& 3595& 21570& 36& 129420& 216& 1296\\ 8& 6407& 51256& 64& 410048& 512& 4096\\ 10& 10009& 100090& 100& 1000900& 1000& 10000\\ \sum x=30& \sum y=22010& \sum xy=180110& \sum {x}^{2}=220& \sum {x}^{2}y=1567540& \sum {x}^{3}=1800& {x}^{4}=15664\end{array}$
Step 3
number of data $=n=6$
Step 2
$\sum y=a\sum {x}^{2}+b\sum x+nc$
$22010=220a+30b+6c$....(1)
$\sum xy=a\sum {x}^{3}+b\sum {x}^{2}+c\sum x$
$180110=1800a+220b+30c$....(2)
$\sum {x}^{2}y=a\sum {x}^{4}+b\sum {x}^{3}+c\sum {x}^{2}$
$1567540=15664a+1800b+220c$....(3)
Step 4
Solve equation (1), (2) and (3)
$a=100.29$
$b=-2.04$
$c=1.25$
$y=100.29{x}^{2}-2.04x+1.25$