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Carissa15

  • one year ago

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?

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  1. anonymous
    • one year ago
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    you will get 3 equations in three unknowns if you plug in

  2. Carissa15
    • one year ago
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    I do not understand how to even attempt this question :-(

  3. anonymous
    • one year ago
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    12=a0+a1(1)+a2(1)^2 15=a0+a1(2)+a2(2)^2 16=a0+a1(3)+a2(3)^2

  4. anonymous
    • one year ago
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    Do you see what i did?

  5. Carissa15
    • one year ago
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    oh ok, just by plugging in the values given for x and the result. Then what needs to be done with the equations?

  6. anonymous
    • one year ago
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    solve for a0, a1, a2, then plug them into the original one to get the result

  7. anonymous
    • one year ago
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    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

  8. Carissa15
    • one year ago
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    solve for the value of a in each equation and then "check" that value in each equation?

  9. anonymous
    • one year ago
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    this is like solving 12=x+1y+1z 15=x+2y+4z 16=x+3y+4z

  10. Carissa15
    • one year ago
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    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?

  11. ganeshie8
    • one year ago
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    Hey!

  12. Carissa15
    • one year ago
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    hey

  13. ganeshie8
    • one year ago
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    There is an easier way to find the interpolation polynomial, wana learn ?

  14. Carissa15
    • one year ago
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    sure, that would be great

  15. ganeshie8
    • one year ago
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    so you want to find a quadratic that passes through the points (1,12),(2,15), (3,16)

  16. ganeshie8
    • one year ago
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    Look at below quadratic, notice anything interesting ? \[\dfrac{(x-2)(x-3)}{(1-2)(1-3)}\cdot 12\]

  17. ganeshie8
    • one year ago
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    what is its value when \(x=1\) ?

  18. Carissa15
    • one year ago
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    12?

  19. ganeshie8
    • one year ago
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    Yes, what do you get when you plugin \(x=2\) or \(x = 3\) ?

  20. misty1212
    • one year ago
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    this looks interesting, don't know it @ganeshie8 got a source for this i can read?

  21. ganeshie8
    • one year ago
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    one sec, i have a really nice short pdf on this...

  22. misty1212
    • one year ago
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    cool thnx

  23. Carissa15
    • one year ago
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    I got 0

  24. ganeshie8
    • one year ago
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    here it is http://www2.lawrence.edu/fast/GREGGJ/Math420/Section_3_1.pdf @misty1212

  25. ganeshie8
    • one year ago
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    @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

  26. misty1212
    • one year ago
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    great thanks!

  27. ganeshie8
    • one year ago
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    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.. :)

  28. Carissa15
    • one year ago
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    each og the equations equal 1 and therefore are only multiplied by the 12,15 and 16. giving total 43?

  29. ganeshie8
    • one year ago
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    you don't have teamviewer?

  30. Carissa15
    • one year ago
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    no, i don't :(

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