To determine To find: a. Each real zero and its multiplicity. Answer a. Real zeros = − 1 2 , − 1 2 , − 4 , − 4 , − 4 _ . The multiplicity of the polynomial = − 1 2 ( Multiplicity = 2 ) _ . The multiplicity of the polynomial = − 4 ( Multiplicity = 3 ) _ . Explanation Given: f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 = − 2 ( x 2 + x + 1 4 ) ( x 3 + 12 x 2 + 48 x + 64 ) = − 2 ( x 5 + 13 x 4 + 241 x 3 4 + 115 x 2 + 76 x + 16 ) The real zeros of the polynomial f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 . a. − 1 2 , − 1 2 , − 4 , − 4 , − 4 The multiplicity of the polynomial = − 1 2 (Multiplicity = 2 ). The multiplicity of the polynomial = − 4 . (Multiplicity = 3 ). To determine To find: b. To determine whether the graph crosses or touches the x -axes at each x -intercept . Answer b. Zero − 1 2 has even multiplicity . Therefore the graph touches the x -axis _ . Zero − 4 has even multiplicity . Therefore the graph crosses the x -axis _ . Explanation Given: f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 = − 2 ( x 2 + x + 1 4 ) ( x 3 + 12 x 2 + 48 x + 64 ) = − 2 ( x 5 + 13 x 4 + 241 x 3 4 + 115 x 2 + 76 x + 16 ) The real zeros of the polynomial f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 . b. Zero − 1 2 has even multiplicity. Therefore the graph touches the x -axis . Zero − 4 has odd multiplicity. Therefore the graph crosses the x -axis . To determine To find: c. To determine the maximum number of turning points on the graph. Answer c. The maximum number of turning points = ( n − 1 ) = 4 _ . Explanation Given: f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 = − 2 ( x 2 + x + 1 4 ) ( x 3 + 12 x 2 + 48 x + 64 ) = − 2 ( x 5 + 13 x 4 + 241 x 3 4 + 115 x 2 + 76 x + 16 ) The real zeros of the polynomial f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 . c. The degree of f ( x ) = n = 5 . The maximum number of turning points = ( n − 1 ) = 4 . To determine To find: d. To determine the end behavior (power function that the graph of f resembles for large values of x ). Answer d. For large values of x in the positive or negative direction value of f approaches positive infinity lim x → ∞ f ( x ) = − ∞ _ . Explanation Given: f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 = − 2 ( x 2 + x + 1 4 ) ( x 3 + 12 x 2 + 48 x + 64 ) = − 2 ( x 5 + 13 x 4 + 241 x 3 4 + 115 x 2 + 76 x + 16 ) The real zeros of the polynomial f ( x ) = − 2 ( x + 1 2 ) 2 ( x + 4 ) 3 . d. f ( x ) = − 2 ( x 5 + 13 x 4 + 241 x 3 4 + 115 x 2 + 76 x + 16 ) The polynomial f ( x ) = a n x n . Here a n = − 2 < 0 . For large values of x in the positive direction or negative direction value of f approaches negative infinity lim x → ∞ f ( x ) = − ∞ . Unbounded in the negative direction.

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Textbook Question

Chapter 4.1, Problem 53AYU

In Problems 57-68, for each polynomial function:

(a) List each real zero and its multiplicity.

(b) Determine whether the graph crosses or touches the $x -axis$ at each $x -intercept$

(c) Determine the maximum number of turning points on the graph.

(d) Determine the end behavior; that is, find the power function that the graph of $f$ resembles for large values of $| x |$ .

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$

Expert Solution

To determine

To find:

a. Each real zero and its multiplicity.

Answer to Problem 53AYU

a. $Real zeros = - \frac{1}{2}, - \frac{1}{2}, - 4, - 4, - 4_$ .

$The multiplicity of the polynomial = - \frac{1}{2} (Multiplicity = 2)_$ .

$The multiplicity of the polynomial = - 4 (Multiplicity = 3)_$ .

Explanation of Solution

Given:

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3} = - 2 (x^{2} + x + \frac{1}{4}) (x^{3} + 12 x^{2} + 48 x + 64)$

$= - 2 (x^{5} + 13 x^{4} + \frac{241 x^{3}}{4} + 115 x^{2} + 76 x + 16)$

The real zeros of the polynomial $f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$ .

a. $- \frac{1}{2}, - \frac{1}{2}, - 4, - 4, - 4$

The multiplicity of the polynomial $= - \frac{1}{2}$ (Multiplicity $= 2$ ).

The multiplicity of the polynomial $= - 4$ . (Multiplicity $= 3$ ).

Expert Solution

To determine

To find:

b. To determine whether the graph crosses or touches the $x -axes$ at each $x -intercept$ .

Answer to Problem 53AYU

b. $Zero - \frac{1}{2} has even multiplicity . Therefore the graph touches the x -axis_$ .

$Zero - 4 has even multiplicity . Therefore the graph crosses the x -axis_$ .

Explanation of Solution

Given:

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3} = - 2 (x^{2} + x + \frac{1}{4}) (x^{3} + 12 x^{2} + 48 x + 64)$

$= - 2 (x^{5} + 13 x^{4} + \frac{241 x^{3}}{4} + 115 x^{2} + 76 x + 16)$

The real zeros of the polynomial $f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$ .

b. Zero $- \frac{1}{2}$ has even multiplicity. Therefore the graph touches the $x -axis$ .

Zero $- 4$ has odd multiplicity. Therefore the graph crosses the $x -axis$ .

Expert Solution

To determine

To find:

c. To determine the maximum number of turning points on the graph.

Answer to Problem 53AYU

c. $The maximum number of turning points = (n - 1) = 4_$ .

Explanation of Solution

Given:

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3} = - 2 (x^{2} + x + \frac{1}{4}) (x^{3} + 12 x^{2} + 48 x + 64)$

$= - 2 (x^{5} + 13 x^{4} + \frac{241 x^{3}}{4} + 115 x^{2} + 76 x + 16)$

The real zeros of the polynomial $f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$ .

c. The degree of $f (x) = n = 5$ .

The maximum number of turning points $= (n - 1) = 4$ .

Expert Solution

To determine

To find:

d. To determine the end behavior (power function that the graph of $f$ resembles for large values of $x$ ).

Answer to Problem 53AYU

d. $For large values of x in the positive or negative direction value of f approaches positive infinity \lim_{x \to \infty} f (x) = - \infty_$ .

Explanation of Solution

Given:

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$

$f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3} = - 2 (x^{2} + x + \frac{1}{4}) (x^{3} + 12 x^{2} + 48 x + 64)$

$= - 2 (x^{5} + 13 x^{4} + \frac{241 x^{3}}{4} + 115 x^{2} + 76 x + 16)$

The real zeros of the polynomial $f (x) = - 2 {(x + \frac{1}{2})}^{2} {(x + 4)}^{3}$ .

d. $f (x) = - 2 (x^{5} + 13 x^{4} + \frac{241 x^{3}}{4} + 115 x^{2} + 76 x + 16)$

The polynomial $f (x) = a_{n} x^{n}$ .

Here $a_{n} = - 2 < 0$ .

For large values of $x$ in the positive direction or negative direction value of $f$ approaches negative infinity $\lim_{x \to \infty} f (x) = - \infty$ .

Unbounded in the negative direction.

Chapter 4.1, Problem 52AYU

Chapter 4.1, Problem 54AYU

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