Trigonometric Functions Graphing

[majinknight]

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:
verticle 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.

 

[Morgan]

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.

 

[majinknight]

Ok well i figured out what i did wrong so the help isn't needed but thank you anyways