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# 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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