Explanation Given Information: The provided function is y = ( x + 2 x − 3 ) 2 . Formula used: A vertical asymptote of a rational function h ( x ) = f ( x ) g ( x ) is x = a where g ( a ) = 0 and f ( a ) ≠ 0 . A horizontal asymptote of a rational function h ( x ) = f ( x ) g ( x ) is, Step 1. A line y = 0 if the degree of the numerator is less than the degree of the denominator. Step 2. The line y = a n b n ratio of the leading coefficients if the degree of the numerator is equal to the degree of the denominator. Step 3. Does not exist the degree of the numerator is greater than the degree of the denominator. To find relative maxima and minima of a function, 1. Set the first derivative of the function to zero f ' ( x ) = 0 , to find the critical values of the function. 2. Substitute the critical values into f ( x ) and calculate the critical points. 3. Calculate the sign of the function to the left and right of the critical values. If f ′ ( x ) < 0 and f ′ ( x ) > 0 on the left and right side respectively of a critical value, then the point is relative minimum. If f ′ ( x ) > 0 and f ′ ( x ) < 0 on the left and right side respectively of a critical value, then the point is relative maximum. Calculation: Consider the provided function, y = ( x + 2 x − 3 ) 2 Recall that a vertical asymptote of a rational function h ( x ) = f ( x ) g ( x ) is x = a where, g ( a ) = 0 and f ( a ) ≠ 0 . Set the denominator x of the function equal to zero. ( x − 3 ) 3 = 0 x − 3 = 0 x − 3 + 3 = 0 + 3 x = 3 Substitute 3 for x in the numerator. x + 2 = 3 + 2 = 5 ≠ 0 Thus, the vertical asymptote is the line x = 3 . Degree of the numerator is 2 and denominator is 2 of the function y = ( x + 2 x − 3 ) 2 . Since, the degree of the numerator is same as the denominator. Thus, the horizontal asymptote is ratio of the leading coefficients. y = 1 1 = 1 Now, calculate the first derivative of the function y = ( x + 2 x − 3 ) 2 . y ' = 2 ( x + 2 x − 3 ) × 1 × ( x − 3 ) − 1 × ( x + 2 ) ( x − 3 ) 2 = 2 ( x + 2 x − 3 ) × x − 3 − x − 2 ( x − 3 ) 2 = 2 ( x + 2 x − 3 ) × − 5 ( x − 3 ) 2 = − 10 ( x + 2 ) ( x − 3 ) 3 Now, to obtain the critical values, set y ' = 0 as, − 10 ( x + 2 ) ( x − 3 ) 3 = 0 − 10 ( x + 2 ) = 0 Thus, x + 2 = 0

Mathematical Applications for the Management, Life, and Social Sciences

11th Edition

ISBN: 9781305108042

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 10.5, Problem 16E

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To calculate: The horizontal and vertical for the provided function y=(x+2x−3)2 and sketch its graph.

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Chapter 10.5, Problem 15E

Chapter 10.5, Problem 17E

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Mathematical Applications for the Management, Life, and Social Sciences

Ch. 10.1 - 1. The x- values of critical points are found...Ch. 10.1 - Prob. 2CP Ch. 10.1 - Prob. 3CP Ch. 10.1 - Prob. 4CP Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - 3. Use the graph of in Problem 1 to identify at...Ch. 10.1 - 4. Use the graph of in Problem 2 to identify at...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...

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The horizontal and vertical for the provided function y = ( x + 2 x − 3 ) 2 and sketch its graph.

Solution Summary: The author explains how to calculate the vertical and horizontal asymptotes of a rational function.

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To calculate: The horizontal and vertical for the provided function y=(x+2x−3)2 and sketch its graph.

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Chapter 10 Solutions