Trigonometric Functions Graphing

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SUMMARY

The discussion focuses on graphing the trigonometric function y=3cos(2(x-120°))-3, with specific parameters including a vertical translation of 3 units down, a phase shift of 120° to the right, an amplitude of 3, and a period of 180°. The user clarifies their understanding of the horizontal scale set at 15° and seeks assistance in determining the correct intervals for graphing the function. Ultimately, the user resolves their confusion regarding the graphing process and confirms their approach using an x/y chart to plot key points.

PREREQUISITES
  • Understanding of trigonometric functions and their properties
  • Knowledge of graph transformations including amplitude, phase shift, and vertical translation
  • Familiarity with cosine function behavior and periodicity
  • Ability to create and interpret x/y charts for function plotting
NEXT STEPS
  • Study the properties of cosine functions and their transformations
  • Learn how to calculate and interpret the period of trigonometric functions
  • Practice graphing various trigonometric functions with different transformations
  • Explore the use of graphing tools or software for visualizing trigonometric functions
USEFUL FOR

Students learning trigonometry, educators teaching trigonometric functions, and anyone interested in mastering graphing techniques for periodic functions.

majinknight
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Ok I have hit a little bumb in the road on my trigonometric function learning. Ok i need to graph the function y=3cos2(x-120degrees)-3, for x is less than or equal to 360degrees but is greater than or equal to -360degrees. Ok so i know the:
vertical translation is 3 down
the phase shift is 120 to the right
amplitude is 3
period is 180 degrees
and so i have the horizontal scale at 15degrees.
so far so good?

Ok now here is the one problem, i do not know if i am to go and as sin grpahs go it starts at highest then to middle then lowest then middle then highest and is finished one wave. Well since my hs=15 each line on graph is 15. now i start on 120 as i have to and herei s my question. When i go down to the middle line which would be -3 would it be on the interval of 150degrees or 135 degrees? Or am i doing it wrong as i thought period meant one full wavelength is 180 degrees so it is making me think i am doing it wrong. Please help if possible.
 
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Alright from this equation y=3cos(2(x-120degrees))-3

I have the rule (.5x+120, 3y-3) is that what you have?
then with a cosine funtion you start on the y-axis at point (0,1) and then you move to (90, 0) then to (180, -1) then to (270, 0) then to (360, 1) correct? so what you are correct with your beginning point being (120, 0) {thats what you have right} what i normally do when approaching this type of problem is very very basic.. because i am a square who doesn't like to very risky.. i used my rule with an x/y chart. soo...

original points
x | y
0 | 1
90 | 0
180 | -1
270 | 0
360 | 1

New rule added ones
x | y
120 | 0
165 | -3
210 | -6
255 | -3

so you are basically adding 45 to the x every time and using 0, -3, and -6 over and over again.. does this answer your question and help you at all. I am slightly rusty on all the information on this because i havnt worked with this in about 3months.
 
Ok well i figured out what i did wrong so the help isn't needed but thank you anyways
 

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