# Use the appropriate Lagrange interpolating polynomials to find the cubic polynomial whose graph passes through the given points. (2, 1), (3, 1), (4, −3), (5, 0).

Question
Polynomial graphs
Use the appropriate Lagrange interpolating polynomials to find the cubic polynomial whose graph passes through the given points. $$\displaystyle{\left({2},\ {1}\right)},\ {\left({3},\ {1}\right)},\ {\left({4},\ −{3}\right)},\ {\left({5},\ {0}\right)}$$.

2020-10-29
Step 1 With $$\displaystyle{x}_{{{1}}}={2},\ {x}_{{{2}}}={3},\ \text{and}\ {x}_{{{4}}}={5}$$, the Langrange interpolating polynomials give: $$\displaystyle{p}_{{{1}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{2}}}\right)}{\left\lbrace{x}\ -\ {x}_{{{3}}}\right)}{\left({x}\ -\ {x}_{{{4}}}\right)}}}{{{\left({x}_{{{1}}}\ -\ {x}_{{{2}}}\right)}{\left({x}_{{{1}}}\ -\ {x}_{{{3}}}\right)}{\left({x}_{{{1}}}\ -\ {x}_{{{4}}}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}\ -\ {3}\right)}{\left({x}\ -\ {4}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({2}\ -\ {3}\right)}{\left({2}\ -\ {4}\right)}{\left({2}\ -\ {5}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {7}{x}\ +\ {12}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left(-{1}\right)}{\left(-{2}\right)}{\left(-{3}\right)}}}}$$
$$\displaystyle=\ -\ {\frac{{{x}^{{{3}}}\ -\ {5}{x}^{{{2}}}\ -\ {7}{x}^{{{2}}}\ +\ {35}{x}\ +\ {12}{x}\ -\ {60}}}{{{6}}}}$$
$$\displaystyle{\color{red}{=\ -\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ +\ {2}{x}^{{{2}}}\ -\ {\frac{{{47}}}{{{6}}}}{x}\ +\ {10}}}$$
$$\displaystyle{p}_{{{2}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{1}}}\right)}{\left({x}\ -\ {x}_{{{3}}}\right)}{\left({x}\ -\ {x}_{{{4}}}\right)}}}{{{\left({x}_{{{2}}}\ -\ {x}_{{{1}}}\right)}{\left({x}_{{{2}}}\ -\ {x}_{{{3}}}\right)}{\left({x}_{{{2}}}\ -\ {x}_{{{4}}}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}\ -\ {2}\right)}{\left({x}\ -\ {4}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({3}\ -\ {2}\right)}{\left({3}\ -\ {4}\right)}{\left({3}\ -\ {5}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {6}{x}\ +\ {8}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({1}{\left(-{1}\right)}{\left(-{2}\right)}\right)}}}}$$
$$\displaystyle=\ {\frac{{{x}^{{{3}}}\ -\ {11}{x}^{{{2}}}\ +\ {38}{x}\ -\ {40}}}{{{2}}}}$$
$$\displaystyle{\color{red}{=\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ -\ {\frac{{{11}}}{{{2}}}}{x}^{{{2}}}\ +\ {19}{x}\ -\ {20}}}$$ Step 2 $$\displaystyle{p}_{{{3}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{1}}}\right)}{\left({x}\ -\ {x}_{{{2}}}\right)}{\left({x}\ -\ {x}_{{{4}}}\right)}}}{{{\left({x}_{{{3}}}\ -\ {x}_{{{1}}}\right)}{\left({x}_{{{3}}}\ -\ {x}_{{{2}}}\right)}{\left({x}_{{{3}}}\ -\ {x}_{{{4}}}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}\ -\ {2}\right)}{\left({x}\ -\ {3}\right)}{\left({x}\ -\ {5}\right)}}}{{{\left({4}\ -\ {2}\right)}{\left({4}\ -\ {3}\right)}{\left({4}\ -\ {5}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {5}{x}\ +\ {6}\right)}{\left({x}\ -\ {5}\right)}}}{{{2}\ \cdot\ {1}{\left(-{1}\right)}}}}$$
$$\displaystyle=\ {\frac{{{x}^{{{3}}}\ -\ {10}{x}^{{{2}}}\ +\ {31}{x}\ -\ {30}}}{{{2}}}}$$
$$\displaystyle{\color{red}{=\ {\frac{{{1}}}{{{2}}}}{x}^{{{3}}}\ +\ {5}{x}^{{{2}}}\ -\ {\frac{{{31}}}{{{2}}}}{x}\ +\ {15}}}$$
$$\displaystyle{p}_{{{4}}}{\left({x}\right)}=\ {\frac{{{\left({x}\ -\ {x}_{{{1}}}\right)}{\left({x}\ -\ {x}_{{{2}}}\right)}{\left({x}\ -\ {x}_{{{3}}}\right)}}}{{{\left({x}_{{{4}}}\ -\ {x}_{{{1}}}\right)}{\left({x}_{{{4}}}\ -\ {x}_{{{2}}}\right)}{\left({x}_{{{4}}}\ -\ {x}_{{{3}}}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}\ -\ {2}\right)}{\left({x}\ -\ {3}\right)}{\left({x}\ -\ {4}\right)}}}{{{\left({5}\ -\ {2}\right)}{\left({5}\ -\ {3}\right)}{\left({5}\ -\ {4}\right)}}}}$$
$$\displaystyle=\ {\frac{{{\left({x}^{{{2}}}\ -\ {5}{x}\ +\ {6}\right)}{\left({x}\ -\ {4}\right)}}}{{{3}\ \cdot\ {2}\ \cdot\ {1}}}}$$
$$\displaystyle=\ {\frac{{{x}^{{{3}}}\ -\ {9}{x}^{{{2}}}\ +\ {26}{x}\ -\ {24}}}{{{6}}}}$$
$$\displaystyle{\color{red}{=\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ -\ {\frac{{{3}}}{{{2}}}}{x}^{{{2}}}\ +\ {\frac{{{13}}}{{{3}}}}{x}\ -\ {4}}}$$ Step 3 With $$\displaystyle{y}_{{{1}}}={1},\ {y}_{{{2}}}=\ -{1},\ {y}_{{{3}}}={3}\ {\quad\text{and}\quad}\ {y}_{{{4}}}={0},$$ Formula (9) gives: $$\displaystyle{p}{\left({x}\right)}={1}{\left(-\ {\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ +\ {2}{x}^{{{2}}}\ -\ {\frac{{{47}}}{{{6}}}}{x}\ +\ {10}\right)}-\ {1}{\left({\frac{{{1}}}{{{2}}}}{x}^{{{3}}}\ -\ {\frac{{{11}}}{{{2}}}}{x}^{{{2}}}\ +\ {19}{x}\ -\ {20}\right)}$$
$$\displaystyle+\ {3}{\left(-\ {\frac{{{1}}}{{{2}}}}{x}^{{{3}}}\ +\ {5}{x}^{{{2}}}\ -\ {\frac{{{31}}}{{{2}}}}{x}\ +\ {15}\right)}+\ {0}{\left({\frac{{{1}}}{{{6}}}}{x}^{{{3}}}\ -\ {\frac{{{3}}}{{{2}}}}{x}^{{{2}}}\ +\ {\frac{{{13}}}{{{3}}}}{x}\ -\ {4}\right)}$$
$$\displaystyle={x}^{{{3}}}{\left(-\ {\frac{{{1}}}{{{6}}}}\ -\ {\frac{{{1}}}{{{2}}}}\ -\ {\frac{{{3}}}{{{2}}}}\right)}\ +\ {x}^{{{2}}}{\left({2}\ +\ {\frac{{{11}}}{{{2}}}}\ +\ {15}\right)}\ +\ {x}{\left(-\ {\frac{{{47}}}{{{6}}}}\ -\ {19}\ -\ {\frac{{{92}}}{{{2}}}}\right)}\ +\ {10}\ -\ {20}\ +\ {45}$$
$$\displaystyle{\color{red}{=\ {\frac{{{11}}}{{{2}}}}{x}^{{{3}}}\ -\ {\frac{{{37}}}{{{2}}}}{x}^{{{2}}}\ +\ {\frac{{{173}}}{{{3}}}}{x}\ -\ {55}}}$$

### Relevant Questions

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