Carissa15
  • Carissa15
Find the interpolating polynomial p(x)=a0+a1x+a2x^2 for the data (1,12),(2,15), (3,16). Can anyone help me with this question please?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
you will get 3 equations in three unknowns if you plug in
Carissa15
  • Carissa15
I do not understand how to even attempt this question :-(
anonymous
  • anonymous
12=a0+a1(1)+a2(1)^2 15=a0+a1(2)+a2(2)^2 16=a0+a1(3)+a2(3)^2

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More answers

anonymous
  • anonymous
Do you see what i did?
Carissa15
  • Carissa15
oh ok, just by plugging in the values given for x and the result. Then what needs to be done with the equations?
anonymous
  • anonymous
solve for a0, a1, a2, then plug them into the original one to get the result
anonymous
  • anonymous
12=a0+a1(1)+a2(1)^2 15=a0+a1(2)+a2(2)^2 16=a0+a1(3)+a2(3)^2 simplified ; 12=a0+a1*1+a2*1 15=a0+a1*2+a2*4 16=a0+a1*3+a2*9 we have 3 equations in 3 unknowns, a0, a1, a2
Carissa15
  • Carissa15
solve for the value of a in each equation and then "check" that value in each equation?
anonymous
  • anonymous
this is like solving 12=x+1y+1z 15=x+2y+4z 16=x+3y+4z
Carissa15
  • Carissa15
oh ok, so it will be very similar to my previous linear equations, solve for each unknown and sub back into original equation to see whether the equation has 1, many or no solutions?
ganeshie8
  • ganeshie8
Hey!
Carissa15
  • Carissa15
hey
ganeshie8
  • ganeshie8
There is an easier way to find the interpolation polynomial, wana learn ?
Carissa15
  • Carissa15
sure, that would be great
ganeshie8
  • ganeshie8
so you want to find a quadratic that passes through the points (1,12),(2,15), (3,16)
ganeshie8
  • ganeshie8
Look at below quadratic, notice anything interesting ? \[\dfrac{(x-2)(x-3)}{(1-2)(1-3)}\cdot 12\]
ganeshie8
  • ganeshie8
what is its value when \(x=1\) ?
Carissa15
  • Carissa15
12?
ganeshie8
  • ganeshie8
Yes, what do you get when you plugin \(x=2\) or \(x = 3\) ?
misty1212
  • misty1212
this looks interesting, don't know it @ganeshie8 got a source for this i can read?
ganeshie8
  • ganeshie8
one sec, i have a really nice short pdf on this...
misty1212
  • misty1212
cool thnx
Carissa15
  • Carissa15
I got 0
ganeshie8
  • ganeshie8
here it is http://www2.lawrence.edu/fast/GREGGJ/Math420/Section_3_1.pdf @misty1212
ganeshie8
  • ganeshie8
@Carissa15 rigjt, so we have just cooked up a quadratic that passes through \((1, 12)\) but produces \(0\) at two other points that we are interested in
misty1212
  • misty1212
great thanks!
ganeshie8
  • ganeshie8
Here is the final quadratic that passes through \((1,12),(2,15), (3,16)\) : \[\dfrac{(x-2)(x-3)}{(1-2)(1-3)}\cdot 12 + \dfrac{(x-1)(x-3)}{(2-1)(2-3)}\cdot 15 + \dfrac{(x-1)(x-2)}{(3-1)(3-2)}\cdot 16\] see if you can figure out why/how it works.. :)
Carissa15
  • Carissa15
each og the equations equal 1 and therefore are only multiplied by the 12,15 and 16. giving total 43?
ganeshie8
  • ganeshie8
you don't have teamviewer?
Carissa15
  • Carissa15
no, i don't :(

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