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Fit a second degree parabolic curve to the following data:

Data :

x 1 2 3 4 5 6 7 8 9
y 2 6 7 8 10 11 11 10 9
0
1

We write X such that $\sum X=0$ . Here N=9 (odd)

x $X_i$ $Y_i$ $X_i^2$ $X_i^3$ $X_i^4$ $X_iY_i$ $X_i^2Y_i$
1 -4 2 16 -64 256 -8 32
2 -3 6 9 -27 81 -8 54
3 -2 7 4 -8 16 -14 28
4 -1 8 1 -1 1 -8 8
5 0 10 0 0 0 0 0
6 1 11 1 1 1 11 11
7 2 11 4 8 16 22 44
8 3 10 9 27 81 30 90
9 4 9 16 64 256 36 144
N=9 $\sum X_i=0$ $\sum Y_i = 74$ $\sum X_i^2=60$ $\sum X_i^3=0$ $\sum X_i^4=708$ $\sum X_iY_i=51$ $\sum X_i^2Y_i=411$

{Here, we made the equation } We use the method called “the method of least squares”.

The Equation of parabola is $y=a+bx+cx^2$

Hence, the normal equations are

$\sum Y_i \;=\; Na + b \sum X_i + c \sum X_i^2 \\ \; \\ \; \\ \sum X_i Y_i \;=\; a \sum X_i + b \sum X_i^2 + c \sum X_i^3 \\ \; \\ \; \\ \sum X_i^2Y_i \;=\; a \sum X_i^2 + b \sum X_i^3 + c \sum X_i^4$

[The principle of least squares states that the parabola should be such that the distances of the given points from the parabola measured along the y axis must be minimum].

$\therefore 74=9a+b(0)+60c \; \; \therefore 9a+60c=74 \; \; \ldots (i) \\ \; \\ 51=a(0)+60b+0c \; \; \therefore 60b=51 \; \; \; \therefore b \;=\; \dfrac{51}{60} \;=\; 0.85 \\ \; \\ 411=60a+0b+708c \; \; \therefore 60a+708c=411 \; \; \; \; \ldots (ii)$

Solving (i) and (ii) simultaneously, we get
a=10.004 , c=-0.267

The Equation of parabola is therefore,

$y=10.004+0.85X-0.267X^2 \\ \; \\ =10.004+0.85(x-5)-0.267(x-5)^2 \\ \; \\ =10.004+0.85x-4.25-0.267(x^2-10x+25) \\ \; \\ =10.004+0.85x-4.25-0.267x^2+2.67x-6.675 \\ \; \\ \therefore y \;=\; =-0.921+3.52x-0.267x^2$

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