Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

Find the polynomial of degree 4 whose graph goes through the points $(1, 1), (2, - 1), (3, - 59), (- 1, 5)$ , and $(- 2, - 29)$ . Graph this polynomial.

The graphical representation of the cubic polynomial $f (t) = 1 - 5 t + 4 t^{2} + 3 t^{3} - 2 t^{4}$ is,

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Consider the points and form the equations

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

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

$(1, 1) : f (1) = a + b (1) + c {(1)}^{2} + d {(1)}^{3} + e {(1)}^{4} = 1 a + b + c + d + e = 1 \dots \dots (1)$

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

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

$(3, - 59) f (3) = a + b (3) + c {(3)}^{2} + d {(3)}^{3} + e {(3)}^{4} = - 59$

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

$(- 1, 5) : f (- 1) = a + b (- 1) + c {(- 1)}^{2} + d {(- 1)}^{3} + e {(- 1)}^{4} = 5$

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

$(- 2, - 29) : f (- 2) = a + b (- 2) + c {(- 2)}^{2} + d {(- 2)}^{3} + e {(- 2)}^{4} = - 29$

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

Step 2: Consider the matrix

Re-write the equations in terms of a matrix.

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

The matrix form is,

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

Step 3: Solve the matrix

Consider the matrix.

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

Using row Echelon form to reduce the matrix.

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

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

Therefore, the polynomial is,

$f (t) = a + b t + c t^{2} + d t^{3} + e t^{4} = 1 - 5 t + 4 t^{2} + 3 t^{2} - 2 t^{4}$

Step 4: Sketch the graph for the obtained polynomial

The graphical representation is,

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Linear Algebra With Applications

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Short Answer

Find the polynomial of degree 4 whose graph goes through the points $(1, 1), (2, - 1), (3, - 59), (- 1, 5)$ , and $(- 2, - 29)$ . Graph this polynomial.

Step by Step Solution

Step 1: Consider the points and form the equations

Step 2: Consider the matrix

Step 3: Solve the matrix

Step 4: Sketch the graph for the obtained polynomial

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Linear Algebra With Applications

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Short Answer

Find the polynomial of degree 4 whose graph goes through the points (1,1),(2,-1),(3,-59),(-1,5), and(-2,-29). Graph this polynomial.

Step by Step Solution

Step 1: Consider the points and form the equations

Step 2: Consider the matrix

Step 3: Solve the matrix

Step 4: Sketch the graph for the obtained polynomial

Most popular questions for Math Textbooks

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Find the polynomial of degree 4 whose graph goes through the points $(1, 1), (2, - 1), (3, - 59), (- 1, 5)$ , and $(- 2, - 29)$ . Graph this polynomial.