Rearrange the data as follows:
x values y values xdiff ydiff slope=ydif/xdiff
x1=612 2052 =y1
x2=693 2779=y2 81 727 8.9753
x3=721 2170=y3 28 -609 -12.7747
It is possible to fit (i) a line of best fit of the form y=mx+c , but here as the y difference/xdifference is too much diffrent the slope is not constant and so the linear relationship is not suitable. (ii) a prarabola of 2nd degree of the form y=ax^2+bx+c between x and y.So, I chose the second form to fit here that establish the relation between x and y.
Since there are 3 pairs of entries we can find a solution of the form y=ax^2+bx+c. Then we get 3 equations if we substitute x=xi and y=yi for i=1,2 and 3 and i is a suffix to x or y.
axi^2+bxi+ c = yi for i =1,2 and 3
Solve the 3 simultaneous equations and determine the values of a,b and c. Substituting the values for xi's and yi's we get:
a(612)^2+b(612)+c=2052 (1)
a(693)^2+b(693)+c= 2779 (2)
a(727)^2+b(727)+c=2170 (3)
(2)-(1) eliminates c and (3)-(2) also eliminates c. We get 2 equation with 2 unkwons in a,b to be determined.:
a{693^2-612^2)+b(693-612)=2779-2052
a(721^2-693^2)+b(721-693)=2170-2779
Equations reduces to after dividing the former by (693-612) and the latter by (721-693):
1305a+b=8.97508642 (4)
1414a+b=-21.75 (5)
(5)-(4) eliminates b , giving an equation with only a to be determined:
109a=30.72508642
a=-0.281881526
Substiture a=-0.281881526 in equation (5) to get b and
b=-21.75-1414(-0.281881526)=376.83047789
Having known a and b ,substitute the values of them in any of the original equations flagged at (1 ) or (2) or (3), to obtain c. So you will get the requred relation once you know all a,b and c determined:
From equatio (1)
c= 2052-(a*612^2+b(612)
=2052-{-0.281881526*612^2+376.83047789*612}
=-122991.2181
Therefore the relation required is:
y=(-0.281881526)x^2+(376.83047789)x-122991.2181.
Hope this helps.
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