Consider the function f(x)=4sin(π2(x−3))+8. State the amplitude A, period P, and midline. State the phase shift and vertical translation. In the full period [0, P], state the maximum and minimum y-values and their corresponding x-values. Enter the exact answers. Amplitude: A= Period: P=Midline: y= The phase shift is _____.The vertical translation is_______ Hints for the maximum and minimum values of f(x):The maximum value of y=sin(x) is y=1 and the corresponding x values are x=π2 and multiples of 2π less than and more than this x value. You may want to solve π2(x−3)=π2.The minimum value of y=sin(x) is y=−1 and the corresponding x values are x=3π2 and multiples of 2π less than and more than this x value. You may want to solve π2(x−3)=3π2.If you get a value for x that is less than 0, you could add multiples of P to get into the next cycles.If you get a value for x that is more than P, you could subtract multiples of P to get into the previous cycles. For x in the interval [0, P], the maximum y-value and corresponding x-value is at:X=Y=For x in the interval [0, P], the minimum y-value and corresponding x-value is at: X=Y=

we have f(x)=4sin(2π(x-3))+8Amplitude A=cofficient of sin(2π(x-3))=4period P=2πcofficient of…

Answered: Consider the function…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Consider the function f(x)=4sin(π2(x−3))+8. State the amplitude A, period P, and midline. State the phase shift and vertical translation. In the full period [0, P], state the maximum and minimum y-values and their corresponding x-values.

Enter the exact answers.

Amplitude: A=

Period: P=

Midline: y=

The phase shift is _____

The vertical translation is_______

Hints for the maximum and minimum values of f(x):

The maximum value of y=sin(x) is y=1 and the corresponding x values are x=π2 and multiples of 2π less than and more than this x value. You may want to solve π2(x−3)=π2.
The minimum value of y=sin(x) is y=−1 and the corresponding x values are x=3π2 and multiples of 2π less than and more than this x value. You may want to solve π2(x−3)=3π2.
If you get a value for x that is less than 0, you could add multiples of P to get into the next cycles.
If you get a value for x that is more than P, you could subtract multiples of P to get into the previous cycles.

For x in the interval [0, P], the maximum y-value and corresponding x-value is at:

For x in the interval [0, P], the minimum y-value and corresponding x-value is at:

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