Find the cubic polynomial whose graph passes through the points ( -1, -1), (0, 1), (1, 3), (4, -1).

Find the cubic polynomial whose graph passes through the points ( -1, -1), (0, 1), (1, 3), (4, -1).

Question
Polynomial graphs
asked 2021-02-14
Find the cubic polynomial whose graph passes through the points \(\displaystyle{\left(-{1},-{1}\right)},{\left({0},{1}\right)},{\left({1},{3}\right)},{\left({4},-{1}\right)}.\)

Answers (1)

2021-02-15

Step 1
Denote the interpolating cubic polynomial by
\(\displaystyle{p}{\left({x}\right)}={a}_{{{0}}}+{a}_{{{1}}}{x}+{a}_{{{2}}}{x}^{{{2}}}+{a}_{{{3}}}{x}^{{{3}}}\)
Let's substitute the given points in the equation of the polynomial. The points \(\displaystyle{\left(-{1},-{1}\right)},{\left({0},{1}\right)},{\left({1},{3}\right)}{\quad\text{and}\quad}{\left({4},-{1}\right)}\) thus have to satisfy.
\(\displaystyle-{1}={a}_{{{0}}}-{a}_{{{1}}}+{a}_{{{2}}}-{a}_{{{3}}}\)
\(\displaystyle{1}={a}_{{{0}}}\)
\(\displaystyle{3}={a}_{{{0}}}+{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}\)
\(\displaystyle-{1}={a}_{{{0}}}+{4}{a}_{{{1}}}+{16}{a}_{{{2}}}+{64}{a}_{{{3}}}\)
First, let's substitute \(\displaystyle{a}_{{{0}}}={1}\) into the three remaining equations. That way we reduced the original system to a system of three equations with three unknowns:
\(\displaystyle-{a}_{{{1}}}+{a}_{{{2}}}-{a}_{{{3}}}=-{2}\)
\(\displaystyle{a}_{{{1}}}+{a}_{{{2}}}+{a}_{{{3}}}={2}\)
\(\displaystyle{4}{a}_{{{1}}}+{16}{a}_{{{2}}}+{64}{a}_{{{3}}}=-{2}\)
Let's solve this system using Gauss-Jordan elimination. The augmented matrix of the system is.
\(\begin{bmatrix}-1 & 1 & -1 & -2 \\1 & 1 & 1 & 2\\ 4 & 16 & 64 & -2 \end{bmatrix}\)
Step 2
Multiply the first row by -1.
\(\begin{bmatrix}1 & -1 & 1 & 2 \\1 & 1 & 1 & 2\\ 4 & 16 & 64 & -2 \end{bmatrix}\)
Add -1 times the first row to the second row, Add -4 times the first row to the third row.
\(\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 2 & 0 & 0\\ 0 & 20 & 60 & -10 \end{bmatrix}\)
Multiply the second row by \(\displaystyle\frac{{1}}{{2}}\)
\(\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 1 & 0 & 0\\ 0 & 20 & 60 & -10 \end{bmatrix}\)
Add -20 times the second row to the third row.
\(\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 1 & 0 & 0\\ 0 & 0 & 60 & -10 \end{bmatrix}\)
Multiply the third row by \(\displaystyle\frac{{1}}{{60}}\).
\(\begin{bmatrix}1 & -1 & 1 & 2 \\0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1/6 \end{bmatrix}\)
Add -1 times the third row to the first row.
\(\begin{bmatrix}1 & -1 & 0 & 13/6 \\0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1/6 \end{bmatrix}\)
Add 1 times the second row to the first row.
\(\begin{bmatrix}1 & 0 & 0 & 13/6 \\0 & 1 & 0 & 0\\ 0 & 0 & 1 & -1/6 \end{bmatrix}\)
The solution is thus
\(=\frac{13}{6}, a_{2}=0,a_{3}=-\frac{1}{6}\)
Along with \(\displaystyle{a}_{{{0}}}={1}\), the interpolating polynomial is
\(\displaystyle{p}{\left({x}\right)}={1}+{\frac{{{13}}}{{{6}}}}{x}-{\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\)

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