Question

Find b₀, b_1, and b₂  to fit the second degree parabola =b₀+b₁ x+b₂ x² for the following data:

x	1	2	3	4
y	1.7	1.8	2.3	3.2

		b0 = -2, b1=-0.5, b2 = -0.2
		b0 =2, b1= -0.5, b2 = 0.2
		b0 =2, b1=0.5, b2 =0.2
		b0 =2, b1=0.5, b2 = -0.2

Answer 1

Please note x2 = x²

                                 X3 = x³

And so on.

The equation is y=a+bx+cx2 and the normal equations are

∑y=an+b∑x+c∑x2

∑xy=a∑x+b∑x2+c∑x3

∑x2y=a∑x2+b∑x3+c∑x4


The values are calculated using the following table

x	y	x2	x3	x4	x⋅y	x2⋅y
1	1.7	1	1	1	1.7	1.7
2	1.8	4	8	16	3.6	7.2
3	2.3	9	27	81	6.9	20.7
4	3.2	16	64	256	12.8	51.2
---	---	---	---	---	---	---
∑x=10	∑y=9	∑x2=30	∑x3=100	∑x4=354	∑x⋅y=25	∑x2⋅y=80.8

Substituting these values in the normal equations

4a+10b+30c=9

10a+30b+100c=25

30a+100b+354c=80.8


Solving these 3 equations,
Total Equations are 3

4a+10b+30c=9→(1)

10a+30b+100c=25→(2)

30a+100b+354c=80.8→(3)

Select the equations (1) and (2), and eliminate the variable a.

4a+10b+30c= 9	×5→			20a	+	50b	+	150c	=	45
		−
10a+30b+100c = 25	×2→			20a	+	60b	+	200c	=	50
					-	10b	-	50c	=	-5	→ (4)

Select the equations (1) and (3), and eliminate the variable a.

4a+10b+30c        =     9	×15→			60a	+	150b	+	450c	=	135
		−
30a+100b+354c =   80.8	× 2→			60a	+	200b	+	708c	=	161.6
					-	50b	-	258c	=	-26.6	→ (5)

Select the equations (4) and (5), and eliminate the variable b.

-10b-50c =     -5	× 5→				-	50b	-	250c	=	-25
		−
-50b-258c= -26.6	× 1→				-	50b	-	258c	=	-26.6
								8c	=	1.6	→ (6)

Now use back substitution method
From (6)
8c=1.6

⇒c=1.68=0.2

From (4)
-10b-50c=-5

⇒-10b-50(0.2)=-5

⇒-10b-10=-5

⇒-10b=-5+10=5

⇒b=5-10=-0.5

From (1)
4a+10b+30c=9

⇒4a+10(-0.5)+30(0.2)=9

⇒4a+1=9

⇒4a=9-1=8

⇒a=84=2

Solution using the Elimination Method.
a=2,b=-0.5,c=0.2

Now substituting these values in the equation is y=a+bx+cx2, we get

y = 2 - 0.5 x + 0.2 x²

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