Question

Given the set points K+1:(-1,8),(1,0) and (2,5), find the quadratic polynomial interpolate

Answer 1

The interpolating polynomial is
`p(x) = 1-4x+3x^2`.

Explanation:

We want to find a quadratic polynomial
`p(x)` such that
`p(-1)=8`,
`p(1)=0` and
`p(2)=5`. In order to do this let us write
`p(x) = a_0+a_1x+a_3x^2`.

Now, evaluating the polynomial in the points -1, 1 and 2 we get

`\begin{cases} 8 = p(-1) &= a_0-a_1+a_2\\ 0 = p(1) &= a_0+a_1+a_2\\ 5 = p(2) &= a_0+2a_1+4a_2\end{cases}`

This relations give us a linear system of equations:

`\begin{cases} 8 &= a_0-a_1+a_2\\ 0 &= a_0+a_1+a_2\\ 5&= a_0+2a_1+4a_2\end{cases}`

where the
`a_0`,
`a_1` and
`a_2` are the unknowns.

The augmented matrix of the system is

`\begin{pmatrix}1 & -1 & 1 & 8\\ 1 & 1 & 1 & 0\\ 1 & 2 & 4 & 5\end{pmatrix}`

In this matrix it is easy to eliminate the 1's of the first column and get

`\begin{pmatrix} 1 & -1 & 1 & 8\\ 0 & 2 & 0 & -8\\ 0 & 3 & 3 & -3\end{pmatrix}`

From this matrix we can find the values of each unknown. Notice that the second row gives us
`2a_2=-8` that yields
`a_1=-4`.

Then, the third row means
`3a_1+3a_2=-3` that gives
`-12+3a_2=-3`. So,
`a_2=3`.

Finally, the first row is
`a_0-a_1+a_2=8` and substituting is
`a_0+7=8` that yields
`a_0=1`.

Therefore, the interpolating polynomial is

`p(x) = 1-4x+3x^2`.