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anonymous

  • one year ago

determine the polynomial function whose graph passes through the given points (-1,-1) (0,0) (1,1) (2,4)

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  1. anonymous
    • one year ago
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    @amistre64

  2. amistre64
    • one year ago
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    we have 4 reference points, what do you consider we can do with them?

  3. amistre64
    • one year ago
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    4 equations, with 4 unknowns ... can we do matrix stuff yet?

  4. anonymous
    • one year ago
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    ax^2 + bx +c ?

  5. amistre64
    • one year ago
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    thats only 3 unknowns :) lets go another degree

  6. anonymous
    • one year ago
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    oooohhhh ax^3 +bx^2+cx

  7. amistre64
    • one year ago
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    ax^3 + bx^2 + cx + d = y

  8. amistre64
    • one year ago
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    but yeah, since 0,0, then d=0 by default

  9. anonymous
    • one year ago
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    so then |dw:1444009179944:dw|

  10. anonymous
    • one year ago
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    is this right then

  11. amistre64
    • one year ago
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    the 0 0 0 0 is not required, but yeah, its fine. then row reduce it to echelon form rref{{(-1)^3,(-1)^2,(-1)^1,(-1)},{(1)^3,(1)^2,(1)^1,(1)},{(2)^3,(2)^2,(2)^1,(4)}}

  12. amistre64
    • one year ago
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    for checking .. if the link posts properly http://www.wolframalpha.com/input/?i=rref {{%28-1%29^3%2C%28-1%29^2%2C%28-1%29^1%2C%28-1%29}%2C{%281%29^3%2C%281%29^2%2C%281%29^1%2C%281%29}%2C{%282%29^3%2C%282%29^2%2C%282%29^1%2C%284%29}}

  13. anonymous
    • one year ago
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    so my matrix table is right then

  14. amistre64
    • one year ago
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    yes

  15. amistre64
    • one year ago
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    only thing that may be off is that you dont have your column for d, but d=0 due to the point 0,0

  16. anonymous
    • one year ago
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    but i have 4 column

  17. amistre64
    • one year ago
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    |dw:1444009601928:dw|

  18. amistre64
    • one year ago
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    your last column is a y column .. you had no d variable column

  19. amistre64
    • one year ago
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    |dw:1444009738187:dw|

  20. amistre64
    • one year ago
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    |dw:1444009782699:dw|

  21. amistre64
    • one year ago
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    do you follow that?

  22. anonymous
    • one year ago
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    nah really

  23. amistre64
    • one year ago
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    each column is an unknown, and the last is the y ... we have 4 unknown so we need 4 columns, and a 5th one for y

  24. amistre64
    • one year ago
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    IF we can already determine a variable, then it is reduced to 3 unknown ... 3 columns and a 4th for y

  25. amistre64
    • one year ago
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    ax^3 + bx^2 + cx + d = y -1a + 1b + -1c + 1d = -1 0a + 0b + 0c + 1d = 0 1a + 1b + 1c + 1d = 1 8a + 4b + 2c + 1d = -1

  26. amistre64
    • one year ago
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    do you see that d=0 already? if not we can simply work the matrix with these columns as is

  27. anonymous
    • one year ago
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    yes

  28. anonymous
    • one year ago
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    alright thanks ill go ahead and work on it now

  29. amistre64
    • one year ago
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    ok

  30. anonymous
    • one year ago
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    so now how do i graph it

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