Q32E Find the polynomial of degree 3 ... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 1: Q32E (page 19)

Find the polynomial of degree 3 [a polynomial of the form $f (t) = a + b t + c t^{2} + d t^{3}$ ] whose graph goes through the points $(0, 1), (1, 0), (- 1, 0)$ and $(2, - 15)$ . Sketch the graph of this cubic.

Short Answer

Expert verified

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

Step by step solution

Consider the points and form the equations

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

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

$(0, 1) : F (0) = a + b (0) + c {(0)}^{2} + d {(0)}^{3} = 1 a = 1 . . . . . . (1)$

$(1, 0) : f (1) = a + b (1) + c {(1)}^{2} + d {(1)}^{3} = 0 a + b + c + d = 0 . . . . . . . (2)$

(- 1, 0) : f (- 1) = a + b (- 1) + c {(- 1)}^{2} + d {(- 1)}^{3} a - b + c - d = 0 . . . . . . (3)

$(2, - 15) : f (2) = a + b (2) + c {(2)}^{2} + d {(2)}^{3} a + 2 b + 4 c + 8 d = - 15 . . . . . . . (4)$

Consider the matrix

Re-write the equations in terms of a matrix.

$|\begin{matrix} a = 1 \\ a + b + c + d = 0 \\ a - b + c - d = 0 \\ a + 2 b + 4 c + 8 d = - 15 \end{matrix}|$

The matrix form is,

$[\begin{matrix} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ 1 & - 1 & 1 & - 1 & 0 \\ 1 & 2 & 4 & 8 & - 15 \end{matrix}]$

Solve the matrix

Consider the matrix.

$[\begin{matrix} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ 1 & - 1 & 1 & - 1 & 0 \\ 1 & 2 & 4 & 8 & - 15 \end{matrix}]$

Using row Echelon form to reduce the matrix.

$[\begin{matrix} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \\ 1 & - 1 & 1 & - 1 & 0 \\ 1 & 2 & 4 & 8 & - 15 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 2 & 4 & 8 & - 16 \\ 0 & 0 & 3 & 3 & - 9 \\ 0 & 0 & 0 & - 2 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 2 \\ 0 & 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & - 1 & - 2 \end{matrix}]$

The values are, $a = 1, b = 2, c = - 1, d = - 2$

Therefore, the polynomial is,

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

Sketch the graph.

The graphical representation is,

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91影视

Find the polynomial of degree 3 [a polynomial of the form $f (t) = a + b t + c t^{2} + d t^{3}$ ] whose graph goes through the points $(0, 1), (1, 0), (- 1, 0)$ and $(2, - 15)$ . Sketch the graph of this cubic.

Short Answer

Step by step solution

Consider the points and form the equations

Consider the matrix

Solve the matrix

Sketch the graph.

One App. One Place for Learning.

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91影视

Find the polynomial of degree 3 [a polynomial of the formf(t)=a+bt+ct2+dt3] whose graph goes through the points (0,1),(1,0),(-1,0)and (2,-15). Sketch the graph of this cubic.

Short Answer

Step by step solution

Consider the points and form the equations

Consider the matrix

Solve the matrix

Sketch the graph.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Calculus

Logic and Functions

Decision Maths

Mechanics Maths

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Find the polynomial of degree 3 [a polynomial of the form $f (t) = a + b t + c t^{2} + d t^{3}$ ] whose graph goes through the points $(0, 1), (1, 0), (- 1, 0)$ and $(2, - 15)$ . Sketch the graph of this cubic.