Explanation Given Information: The provided function is y = 15 sin ( 2 3 x − 120 ° ) . Formula Used: For a given function y = A s c t ( B x + Q ) , where s c t = sin / cos / tan Here, A = Amplitude , Period = 360 B and Phase = Q B Calculation: Consider the provided function: y = 15 sin ( 2 3 x − 120 ° ) Now, compare the provided function y = 15 sin ( 2 3 x − 120 ° ) with the function y = A s c t ( B x + Q ) . So, it can be concluded that, Amplitude = 15 And, period = 360 2 3 = 360 ⋅ 3 2 = 540 And, phase = − 120 2 3 = − 180 Thus, The amplitude is 15 , period is 540 ° and phase is 180 ° right n the given function y = 15 sin ( 2 3 x − 120 ) . Graph: Consider the provided function: y = 15 sin ( 2 3 x − 120 ) Now, obtain the values of y by assuming corresponding values of x as: Substitute x = 0 in the function y = 15 sin ( 2 3 x − 120 ) . y = 15 cos ( 2 3 ⋅ 0 − 120 ) = 15 cos ( 120 ) = − 15 ⋅ 0.5 = − 7.5 Substitute x = 30 in the function y = 15 sin ( 2 3 x − 120 ) . y = 15 cos ( 2 3 ⋅ 30 − 120 ) = 15 cos ( − 100 ) = − 15 ⋅ 0.1736 = − 2.6047 Substitute x = 60 in the function y = 15 sin ( 2 3 x − 120 ) . y = 15 cos ( 2 3 ⋅ 60 − 120 ) = 15 cos ( − 80 ) = 15 ⋅ 0

12th Edition

ISBN: 9781337630580

Author: Dale Ewen

Publisher: Cengage Learning

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Chapter 14.2, Problem 24E

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To calculate: The period and amplitude of the given expression y=15sin(23x−120°) and graph the provided function.

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Chapter 14.2, Problem 23E

Chapter 14.3, Problem 1E

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Chapter 14 Solutions

Elementary Technical Mathematics

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The period and amplitude of the given expression y = 15 sin ( 2 3 x − 120 ° ) and graph the provided function.

Solution Summary: The author calculates the period and amplitude of the given expression y=15mathrmsin(23x-120°).

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To calculate: The period and amplitude of the given expression y=15sin(23x−120°) and graph the provided function.

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Chapter 14 Solutions