Question

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

Answer 1

Answer:

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

Step-by-step 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

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