Fit a curve $y=ax+bx_2$ for the data:

x	1	2	3	4	5	6
y	2.51	5.82	9.93	14.84	20.55	27.06

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written 7.8 years ago by

• modified 7.8 years ago

x	$x^2$	$x^3$	$x^4$	y	$xy$	$x^2y$
1	1	1	1	2.51	2.51	2.51
2	4	8	16	5.82	11.64	23.28
3	9	27	81	9.93	29.79	89.37
4	16	64	256	14.84	59.36	237.44
5	25	125	625	20.55	102.75	513.75
6	36	216	1296	27.06	62.36	974.16
$\sum x=21$	$ \sum x^2=91$	$\sum x^3=441$	$\sum x^4=2275$	$ \sum y=80.71$	$\sum xy=368.41$	$\sum x^2y=1840.51$

$ \\ \; \\ \; \\ \sum y \;=\; a\sum x + b \sum x^2 + c \sum x^2 + cN \; \; \; \therefore 80.71=21a+91b+6c \; \; \; \; \ldots (i) \\ \; \\ \; \\ \sum xy \;=\; a \sum x^2 + b \sum x^3 + c \sum x \; \; \; \therefore 91a+441b+21c \; \; \; \; \ldots (ii) \\ \; \\ \; \\ \sum x^2y \;=\; a \sum x^3 + b \sum x^4 + c \sum x^2 \; \; \; \therefore 1840.51 = 441a+2275b+91c \; \; \; \; \ldots (iii) $

Solving equations (i), (ii), (iii) simultaneously, we get

a=2.11 , b=0.4 , c=0

$ \therefore y=2.11x+0.4x^2 $ This is an equation of parabola.

Note: This is another method where we didn’t convert into X=x-3.5 (in this case). We can solve by previous method also.

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