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anonymous

  • one year ago

What polynomial has a graph that passes through the given points? (-2, 2) (-1, -1) (1, 5) (3, 67)

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  1. Michele_Laino
    • one year ago
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    A possible solution is a polynomial of fourth grade, namely: \[\Large y = a{x^4} + b{x^3} + c{x^2} + dx + e\] where, a,b , c, d, and e are real coeeficient. Now using your data we can write this algebraic system: \[\Large \left\{ \begin{gathered} - 1 = a - b + c - d + e \hfill \\ 5 = a + b + c + d + e \hfill \\ 2 = 16a - 8b + 4c - 2d + e \hfill \\ 67 = 81a + 27b + 9c + 3d + e \hfill \\ \end{gathered} \right.\] or using matrices: \[\Large \left( {\begin{array}{*{20}{c}} 1&{ - 1}&1&{ - 1} \\ 1&1&1&1 \\ {16}&{ - 8}&4&{ - 2} \\ {81}&{27}&9&3 \end{array}\;\;\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \\ 1 \end{array}} \right)\quad \left( {\begin{array}{*{20}{c}} a \\ b \\ c \\ \begin{gathered} d \hfill \\ e \hfill \\ \end{gathered} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - 1} \\ 5 \\ 2 \\ {67} \end{array}} \right)\]

  2. UsukiDoll
    • one year ago
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    what if OP doesn't know matrices? These days Linear Algebra is taught at either a very well known private high school or public university

  3. Michele_Laino
    • one year ago
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    I think that she knows algebra of matrices, since it is the only method that I know to solve her problem @UsukiDoll

  4. Michele_Laino
    • one year ago
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    applying the "Gauss Elimination Method" we get: the subsequent equvalent system: \[\Large \left( {\begin{array}{*{20}{c}} 1&{ - 1}&1&{ - 1} \\ 0&2&0&2 \\ 0&0&{ - 12}&{10} \\ 0&0&0&6 \end{array}\;\;\left. {\begin{array}{*{20}{c}} 1 \\ 0 \\ { - 15} \\ {10} \end{array}} \right|\;\begin{array}{*{20}{c}} { - 1} \\ 6 \\ { - 6} \\ {50} \end{array}} \right)\]

  5. Michele_Laino
    • one year ago
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    as we can see there are infinity solutions, which are depending on one real parameter

  6. Michele_Laino
    • one year ago
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    I think that I have made an error, since the reqiested polynoimial is a polynomial of third grade, like below: \[\Large y = a{x^3} + b{x^2} + cx + d\] In that case more simpler methods can be applied in order to solve the corresponding algebraic system

  7. Michele_Laino
    • one year ago
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    requested*

  8. Michele_Laino
    • one year ago
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    In that case we have a squre system, and if the matrix of coefficient is a nonsingular matrix, then we have one and only one solution for coefficients a, b, c, d

  9. Michele_Laino
    • one year ago
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    oops.. square system...

  10. Michele_Laino
    • one year ago
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    using the polynomial of third grade, I got this system: \[\Large \left\{ \begin{gathered} - 1 = - a + b - c + d \hfill \\ 5 = a + b + c + d \hfill \\ 2 = - a + 4b - 2c + d \hfill \\ 67 = 27a + 9b + 3c + d \hfill \\ \end{gathered} \right.\]

  11. Michele_Laino
    • one year ago
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    or, using matrices: \[\Large \left( {\begin{array}{*{20}{c}} { - 1}&1&{ - 1}&1 \\ 1&1&1&1 \\ { - 8}&4&{ - 2}&1 \\ {27}&9&3&1 \end{array}\left. {\begin{array}{*{20}{c}} {} \\ {} \\ {} \\ {} \end{array}} \right|\;\begin{array}{*{20}{c}} { - 1} \\ 5 \\ 2 \\ {67} \end{array}} \right)\]

  12. Michele_Laino
    • one year ago
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    again, I apply the "Gauss Elimination Method" and I get: \[\Large \left( {\begin{array}{*{20}{c}} { - 1}&1&{ - 1}&1 \\ 0&2&0&2 \\ 0&0&6&{ - 3} \\ 0&0&0&{ - 20} \end{array}\left. {\begin{array}{*{20}{c}} {} \\ {} \\ {} \\ {} \end{array}} \right|\;\begin{array}{*{20}{c}} { - 1} \\ 4 \\ 2 \\ { - 40} \end{array}} \right)\]

  13. Michele_Laino
    • one year ago
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    now being a square system, and being the matrix of coefficients of our system a nonsingular matrix, then we can state that there is one and only one solution

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