Question

What polynomial has a graph that passes through the given points?

(-2, 2) (-1, -1) (1, 5) (3, 67)

Answer 1

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:
\[\Large \left( {\begin{array}{*{20}{c}}
1&{ - 1}&1&{ - 1} \\
1&1&1&1 \\
{16}&{ - 8}&4&{ - 2} \\
{81}&{27}&9&3
\end{array}\;\;\begin{array}{*{20}{c}}
1 \\
1 \\
1 \\
1
\end{array}} \right)\quad \left( {\begin{array}{*{20}{c}}
a \\
b \\
c \\
\begin{gathered}
d \hfill \\
e \hfill \\
\end{gathered}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 1} \\
5 \\
2 \\
{67}
\end{array}} \right)\]

Answer 2

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

Answer 3

I think that she knows algebra of matrices, since it is the only method that I know to solve her problem @UsukiDoll

Answer 4

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

Answer 5

as we can see there are infinity solutions, which are depending on one real parameter

Answer 6

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

Answer 7

requested*

Answer 8

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

Answer 9

oops.. square system...

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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