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Expert-verified Found in: Page 19 ### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974 # Find the polynomial of degree 4 whose graph goes through the points ${\mathbf{\left(}}{\mathbf{1}}{\mathbf{,}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{,}}{\mathbf{\left(}}{\mathbf{2}}{\mathbf{,}}{\mathbf{-}}{\mathbf{1}}{\mathbf{\right)}}{\mathbf{,}}{\mathbf{\left(}}{\mathbf{3}}{\mathbf{,}}{\mathbf{-}}{\mathbf{59}}{\mathbf{\right)}}{\mathbf{,}}{\mathbf{\left(}}{\mathbf{-}}{\mathbf{1}}{\mathbf{,}}{\mathbf{5}}{\mathbf{\right)}}$, and${\mathbf{\left(}}{\mathbf{-}}{\mathbf{2}}{\mathbf{,}}{\mathbf{-}}{\mathbf{29}}{\mathbf{\right)}}$. Graph this polynomial.

The graphical representation of the cubic polynomial $f\left(t\right)=1-5t+4{t}^{2}+3{t}^{3}-2{t}^{4}$ is, See the step by step solution

## Step 1: Consider the points and form the equations

The equation is, $f\left(t\right)=a+bt+c{t}^{2}+d{t}^{3}+e{t}^{4}$

Substitute the points in the above equation to form the equations.

$\left(1,1\right):\phantom{\rule{0ex}{0ex}}f\left(1\right)=a+b\left(1\right)+c{\left(1\right)}^{2}+d{\left(1\right)}^{3}+e{\left(1\right)}^{4}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}a+b+c+d+e=1\dots \dots \left(1\right)$

$\begin{array}{l}\left(2,-1\right):\end{array}\phantom{\rule{0ex}{0ex}}f\left(2\right)=a+b\left(2\right)+c{\left(2\right)}^{2}+d{\left(2\right)}^{3}+e{\left(2\right)}^{4}\phantom{\rule{0ex}{0ex}}=1$

$a+2b+4c+8d+16e=-1\dots \dots \left(2\right)$

$\left(3,-59\right)\phantom{\rule{0ex}{0ex}}f\left(3\right)=a+b\left(3\right)+c{\left(3\right)}^{2}+d{\left(3\right)}^{3}+e{\left(3\right)}^{4}\phantom{\rule{0ex}{0ex}}=-59$

$a+3b+9c+27d+81e=-59\dots \dots \left(3\right)$

$\left(-1,5\right):\phantom{\rule{0ex}{0ex}}f\left(-1\right)=a+b\left(-1\right)+c{\left(-1\right)}^{2}+d{\left(-1\right)}^{3}+e{\left(-1\right)}^{4}\phantom{\rule{0ex}{0ex}}=5$

$a-b+c-d+e=5\dots \dots \left(4\right)$

$\left(-2,-29\right):\phantom{\rule{0ex}{0ex}}f\left(-2\right)=a+b\left(-2\right)+c{\left(-2\right)}^{2}+d{\left(-2\right)}^{3}+e{\left(-2\right)}^{4}\phantom{\rule{0ex}{0ex}}=-29$

$a-2b+4c--8d+16e=-29\dots \dots \left(5\right)$

## Step 2: Consider the matrix

Re-write the equations in terms of a matrix.

$\left|\begin{array}{c}a+b+c+d+e=1\\ a+2b+4c+8d+16e=-1\\ a+3b+9c+27d+81e=-59\\ a-b+c-d+e=5\\ a-2b+4c-8d+16e=-29\end{array}\right|$

The matrix form is,

$\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 2& 4& 8& 16& -1\\ 1& 3& 9& 27& 81& -59\\ 1& -1& 1& -1& 1& -1\\ 1& -2& 4& -8& 16& -29\end{array}\right]$

## Step 3: Solve the matrix

Consider the matrix.

$\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 2& 4& 8& 16& -1\\ 1& 3& 9& 27& 81& -59\\ 1& -1& 1& -1& 1& -1\\ 1& -2& 4& -8& 16& -29\end{array}\right]$

Using row Echelon form to reduce the matrix.

$\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 1& 2& 4& 8& 16& -1\\ 1& 3& 9& 27& 81& -59\\ 1& -1& 1& -1& 1& -1\\ 1& -2& 4& -8& 16& -29\end{array}\right]=\left[\begin{array}{cccccc}1& 1& 1& 1& 1& 1\\ 0& -3& 3& -9& 15& -30\\ 0& 0& 10& 20& 90& -80\\ 0& 0& 0& 8& 8& 2\\ 0& 0& 0& 0& -12& 21\end{array}\right]\phantom{\rule{0ex}{0ex}}=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 1\\ 0& 1& 0& 0& 0& -5\\ 0& 0& 1& 0& 0& 4\\ 0& 0& 0& 1& 0& 3\\ 0& 0& 0& 0& 1& -2\end{array}\right]$

The values are, $a=1,b=-5,c=4,d=3,e=-2$.

Therefore, the polynomial is,

$f\left(t\right)=a+bt+c{t}^{2}+d{t}^{3}+e{t}^{4}\phantom{\rule{0ex}{0ex}}=1-5t+4{t}^{2}+3{t}^{2}-2{t}^{4}$

## Step 4: Sketch the graph for the obtained polynomial

The graphical representation is,  ### Want to see more solutions like these? 