SOLUTION: what is the amplitude, period, and phase shift of f(x)= -3sin(4x-pi)-5?

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Question 1156186: what is the amplitude, period, and phase shift of f(x)= -3sin(4x-pi)-5?
Found 3 solutions by Edwin McCravy, MathLover1, ikleyn:
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!

For f%28x%29=Asin%28Bx%2BC%29%2BD or f%28x%29=Acos%28Bx%2BC%29%2BD

matrix%281%2C4%2Cabs%28A%29%2Cis%2Cthe%2Camplitude%29

matrix%281%2C4%2C2pi%2FB%2Cis%2Cthe%2Cperiod%29

matrix%281%2C9%2CC%2FB%2C+with%2C+its%2C+sign%2C+changed%2C+is%2C+the%2C+phase%2C+shift%29

matrix%281%2C5%2CD%2Cis%2Cthe%2Cvertical%2Cshift%29

Edwin

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

what is the amplitude, period, and phase shift of
f%28x%29=+-3sin%284x-pi%29-5
y=asin%28bx-c%29%2Bd
where
a= the amplitude
b= the number of times the graph will repeat itself in the “normal” period, which is 2pi for sin
c= the horizontal shift or phase shift of the graph, when c is subtracted from x, it shifts to the right
d= the vertical shift of the graph (up or down)
in your case:
amplitude: -3
period: b=2pi%2F4=pi%2F2
phase shift to the right: c%2Fb=+pi%2F4
vertical shift: -5 (5 units down)

Answer by ikleyn(52903) About Me  (Show Source):
You can put this solution on YOUR website!
.

1)  Amplitude is ALWAYS positive value and NEVER is negative value, by the definition.


    So, in this problem the amplitude is |-3| = 3.


        NOT  -3  (!) (!) (!)




2)  The parent function to measure the shift in this problem  IS NOT -3sin(4x) - 5, as @Mathlover MISTAKENLY thinks 

     and determines the phase shift incorrectly.


    To determine the shift, the given function should be presented wit POSITIVE amplitude

        -3sn(4x-pi) - 5 = 3sin(4x) - 5

    which shows that the phase shift is 0 (zero, ZERO) in this case.


See the plot below.



    


    Plot y = -3*sin(4x-3.14) - 5


Thus regarding the amplitude and the phase shift, the answers by @MathLower1 are incorrect.


Edwin correctly found the amplitude as positive 3, but regarding the phase shift, he dodged a direct answer.