Answer:
The interpolating polynomial is
.
Step-by-step explanation:
We want to find a quadratic polynomial
such that
,
and
. In order to do this let us write
.
Now, evaluating the polynomial in the points -1, 1 and 2 we get

This relations give us a linear system of equations:

where the
,
and
are the unknowns.
The augmented matrix of the system is

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

From this matrix we can find the values of each unknown. Notice that the second row gives us
that yields
.
Then, the third row means
that gives
. So,
.
Finally, the first row is
and substituting is
that yields
.
Therefore, the interpolating polynomial is
.