Understanding Trig Function Behavior: Period, Amplitude, and Shifts

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The discussion focuses on analyzing the trigonometric function y=-3cos(2x-π/6)+1. The user correctly identifies the period as π, the horizontal phase shift as 15 degrees to the left, and the vertical displacement as 1 unit up. They determine the maximum value to be 4, the minimum value to be -2, the amplitude as 3, and the range as {y: -2 ≤ y ≤ 4}. Other participants confirm that the user's calculations for the period, domain, and shifts are accurate.
aisha
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Hi just need a little help with the behaviour of this trig function

y=-3\cos (2x-\frac {\pi} {6}) +1
I converted the pi over 6 to degrees and got -30

need to state the period, amplitude, max/min values, range, domain, horizontal phase shift, and vertical dispacement.

So far after rearranging the the equation i have

period= pi
range= ?
amplitude=?
domain= all values XER
horizontal phase shift= 15 degrees left
vertical displacement= up 1
max\min ? :redface: not sure how to figure this out from the equation i tried graphing using the graphing calculator but it won't give me the values for max and min. Therefore I cannot solve for amplitude yet and same with range.

CAN SOMEONE PLEASE HELP ME JUST NEED HELP WITH THE MAX AND MIN THANKS :blushing:
 
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I got
max = 4, min = -2
amplitude = 3
range = {y:-2 <= y <= 4}

is this correct or have i done something wrong..please check and reply soon! :blushing:
 
what are the max and min values of the cos function?
Use that to figure out the max and min values of y.
 
aisha said:
I got
max = 4, min = -2
amplitude = 3
range = {y:-2 <= y <= 4}

is this correct or have i done something wrong..please check and reply soon! :blushing:
yes, those are correct.
 
is everything else correct too the period domain and shifts?
 
Yes - everything else looks ok.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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