How to find orignal equations of type y=ax2+bx+c. Given 3 coordinate points? (40,30)(60,28)(20,25)

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How to find orignal equations of type y=ax2+bx+c. Given 3 coordinate points?  (40,30)(60,28)(20,25)

Maritza Mccall · Accepted Answer

Your first problem is in the title. The function you&#039;re looking for is y=ax2+bx+c. And you have 3 points on that parabola. So just plug them in.1600a+40b+c=303600a+60b+c=28400a+20b+c=25

trnovitom06 · Accepted Answer

in general, if a polynomial ∑i=1naixi goes through the n points (xj,yj) (different x-coordinates) then  (x1nx1n−1⋯1...xnnxnn−1⋯1)(anan−1...a0)=(y1...yn)

RizerMix · Accepted Answer

Perhaps an easier way will be to use the Lagrange Interpolation Formula. Given n points such that no two of them lies on a same vertical line, the formula guarantee the existence of a polynomial P(x) such that P(x) passes through these n points and the degree of P(x) is at most n−1. Since 3 points are given, by using the formula, the polynomial formed will be at most degree of 2. Here is how the formula works: Suppose the 3 points are p1=(x1=40,y1=30),p2=(x2=60,y2=28),p3=(x3=20,y3=25) Now, form 3 polynomial R1(x),R2(x),R3(x) such that R1(x1)=1,R1(x2)=R1(x3)=0, R2(x2)=1,R2(x1)=R2(x3)=0 and R3(x3)=1,R3(x1)=R3(x2)=0. It is easy to see that R1(x)=(x−60)(x−20)(40−60)(40−20)=−x2−80x+1200400 R2(x)=(x−40)(x−20)(60−40)(60−20)=x2−60x+800800 R3(x)=(x−40)(x−60)(20−40)(20−60)=x2−100x+2400800 This P(x) that we want to find is P(x)=30R1(x)+28R2(x)+25R3(x) And the rest are just algebra. Generally, given n points, p1,p2,⋯,pn, such that pi=(xi,yi), a polynomial P(x) that passes through all these points is P(x)=∑i=0nyi∏j≠ix−xjxi−xj

How to find orignal equations of type y=ax^{2}+bx+c. Given 3

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