a. Explanation Result used: Real zeros and x-intercepts of a function : A real number r is a zero of a polynomial function f ( x ) if and only if it is an x -intercept of f . Real zero of even multiplicity : If a real number r is a zero of even multiplicity of a polynomial function f ( x ) , then the graph of f touches the x -axis at r . Real zero of odd multiplicity : If a real number r is a zero of odd multiplicity of a polynomial function f ( x ) , then the graph of f crosses the x -axis at r . Maximum number of turning points : The maximum number of turning points on the graph of a polynomial function of degree n is n − 1 . The end behaviour of the graph of a polynomial function : The power function that resembles the end behaviour of a polynomial function f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 with a n ≠ 0 is the function y = a n x n . Calculation: Consider the given function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 with 0 < a < b < c . Substitute the value f ( x ) = 0 in f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 and obtain the x -intercepts of the function f as follows. 0 = − a x 2 ( x − b ) ( x + c ) 2 ⇒ − a x 2 = 0 , x − b = 0 , ( x + c ) 2 = 0 ⇒ x 2 = 0 , x = b , x + c = 0 ⇒ x = 0 , x = b , x = − c Thus, the x -intercepts of the function f are ( 0 , 0 ) , ( b , 0 ) and ( − c , 0 ) . Hence, the real zeros of the function f are x = 0 , x = b , x = − c . Note that, the multiplicity of the root x = 0 is 2, the multiplicity of the root x = b is 1 and the multiplicity of the root x = − c is 2. Since the zeros x = 0 and x = − c are of even multiplicity, the graph of f touches the x -axis at the points x = 0 and x = − c . Since the zeros x = b are of odd multiplicity, the graph of f crosses the x -axis at the point x = b . Note that, the degree of the function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 is 5. Hence, the function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 has at most 5 − 1 = 4 turning points. Rewrite the function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 as follows. f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 = − a x 3 + a b x 2 + ( x 2 + 2 c x + c 2 ) = − a x 3 + ( a b + 1 ) x 2 + 2 c x + c 2 Therefore, the power function that resembles the end behaviour of the polynomial function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 is y = − a x 3 . Also, since the values 0 < a < b < c , the values − c < − b < − a < 0 . For any value of x < − c , the value − a x 2 < 0 , x − b < 0 and ( x + c ) 2 > 0 b. To determine The intervals in which the function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 has a local maximum value. c. To determine The numbers where the function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 has a local minimum value. d. To determine The interval where the function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 is such that f ( x ) < 0 . e. To determine The interval where the function value f ( − x − 4 ) < 0 . f. To determine Whether the function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 is increasing, decreasing or neither on the interval ( − ∞ , − c ] .

11th Edition

ISBN: 9780135163078

Author: Michael Sullivan

Publisher: PEARSON

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Question

Chapter 5.2, Problem 49AYU

To determine

To graph: The function f(x)=−ax2(x−b)(x+c)2 with the values 0<a<b<c.

To determine

The intervals in which the function f(x)=−ax2(x−b)(x+c)2 has a local maximum value.

To determine

The numbers where the function f(x)=−ax2(x−b)(x+c)2 has a local minimum value.

To determine

The interval where the function f(x)=−ax2(x−b)(x+c)2 is such that f(x)<0.

To determine

The interval where the function value f(−x−4)<0.

To determine

Whether the function f(x)=−ax2(x−b)(x+c)2 is increasing, decreasing or neither on the interval (−∞,−c].

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Chapter 5.2, Problem 48AYU

Chapter 5.2, Problem 50AYU

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Chapter 5 Solutions

Algebra And Trigonometry (11th Edition)

Ch. 5.1 - Prob. 1AYU Ch. 5.1 - Prob. 2AYU Ch. 5.1 - Prob. 3AYU Ch. 5.1 - Prob. 4AYU Ch. 5.1 - Prob. 5AYU Ch. 5.1 - Prob. 6AYU Ch. 5.1 - Prob. 7AYU Ch. 5.1 - Prob. 8AYU Ch. 5.1 - Prob. 9AYU Ch. 5.1 - Prob. 10AYU

Knowledge Booster

To graph: The function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 with the values 0 &lt; a &lt; b &lt; c .

To graph: The function f(x)=−ax2(x−b)(x+c)2 with the values 0<a<b<c.

The intervals in which the function f(x)=−ax2(x−b)(x+c)2 has a local maximum value.

The numbers where the function f(x)=−ax2(x−b)(x+c)2 has a local minimum value.

The interval where the function f(x)=−ax2(x−b)(x+c)2 is such that f(x)<0.

The interval where the function value f(−x−4)<0.

Whether the function f(x)=−ax2(x−b)(x+c)2 is increasing, decreasing or neither on the interval (−∞,−c].

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Chapter 5 Solutions

To graph: The function f ( x ) = − a x 2 ( x − b ) ( x + c ) 2 with the values 0 < a < b < c .