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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)
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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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3A 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: \[\Large \left( {\begin{array}{*{20}{c}} 1&{  1}&1&{  1} \\ 1&1&1&1 \\ {16}&{  8}&4&{  2} \\ {81}&{27}&9&3 \end{array}\;\;\begin{array}{*{20}{c}} 1 \\ 1 \\ 1 \\ 1 \end{array}} \right)\quad \left( {\begin{array}{*{20}{c}} a \\ b \\ c \\ \begin{gathered} d \hfill \\ e \hfill \\ \end{gathered} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {  1} \\ 5 \\ 2 \\ {67} \end{array}} \right)\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0what if OP doesn't know matrices? These days Linear Algebra is taught at either a very well known private high school or public university

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I think that she knows algebra of matrices, since it is the only method that I know to solve her problem @UsukiDoll

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3applying the "Gauss Elimination Method" we get: the subsequent equvalent system: \[\Large \left( {\begin{array}{*{20}{c}} 1&{  1}&1&{  1} \\ 0&2&0&2 \\ 0&0&{  12}&{10} \\ 0&0&0&6 \end{array}\;\;\left. {\begin{array}{*{20}{c}} 1 \\ 0 \\ {  15} \\ {10} \end{array}} \right\;\begin{array}{*{20}{c}} {  1} \\ 6 \\ {  6} \\ {50} \end{array}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3as we can see there are infinity solutions, which are depending on one real parameter

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3In 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3oops.. square system...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3using the polynomial of third grade, I got this system: \[\Large \left\{ \begin{gathered}  1 =  a + b  c + d \hfill \\ 5 = a + b + c + d \hfill \\ 2 =  a + 4b  2c + d \hfill \\ 67 = 27a + 9b + 3c + d \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3or, using matrices: \[\Large \left( {\begin{array}{*{20}{c}} {  1}&1&{  1}&1 \\ 1&1&1&1 \\ {  8}&4&{  2}&1 \\ {27}&9&3&1 \end{array}\left. {\begin{array}{*{20}{c}} {} \\ {} \\ {} \\ {} \end{array}} \right\;\begin{array}{*{20}{c}} {  1} \\ 5 \\ 2 \\ {67} \end{array}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3again, I apply the "Gauss Elimination Method" and I get: \[\Large \left( {\begin{array}{*{20}{c}} {  1}&1&{  1}&1 \\ 0&2&0&2 \\ 0&0&6&{  3} \\ 0&0&0&{  20} \end{array}\left. {\begin{array}{*{20}{c}} {} \\ {} \\ {} \\ {} \end{array}} \right\;\begin{array}{*{20}{c}} {  1} \\ 4 \\ 2 \\ {  40} \end{array}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now 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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