SOLUTION: I'm having trouble with the following problem: Find the domain, range, amplitude, period, phase shift, and verticle shift of y=3cos1/2(x-2pi) +1. I know the equation is of the ge

Question 403781: I'm having trouble with the following problem: Find the domain, range, amplitude, period, phase shift, and verticle shift of y=3cos1/2(x-2pi) +1.
I know the equation is of the general form y=a*cos(bx+c) + d, but the 1/2 behind the cos has me stumped. What effect does the 1/2 have?
Thanks,
DS
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
I'm having trouble with the following problem: Find the domain, range, amplitude, period, phase shift, and verticle shift of y=3cos1/2(x-2pi) +1.
I know the equation is of the general form y=a*cos(bx+c) + d, but the 1/2 behind the cos has me stumped. What effect does the 1/2 have?
..
The standard form of the cos function is y=Acos(Bx-C)
For the given function,
y=3cos1/2(x-2pi) +1 =3cos(x/2-pi)+1
The 1/2 after the Cos is B factored out of the expression,(x/2-pi)
For example,(Bx-C)=B(x-C/B) As you can see, B and the phase shift,C/B, can be readily identified with this change.
phase shift =2pi
period=2pi/B=2pi/(1/2)=4pi
(1/4) period = (4pi/4)=pi
without vertical or horizontal shifting, (x,y)coordinates would be as follows:
(0,3),(1/4pi,0),(1/2pi,-3),(3/4pi,0),(4pi,3)
with a horizontal shift of 2pi,(x,y)coordinates would be as follows:
(2pi,3),(9/4pi,0),(5/2pi,-3),(11/4pi,0),(6pi,3)
with a vertical shift shift of +1,(x,y)coordinates would finally be as follows:
(2pi,4),(9/4pi,1),(5/2pi,-2),(11/4pi,1),(6pi,4)
..
Domain [2pi,6pi] (1-cycle)
Range [4,-2]
Amplitude 4max,2min
Period 4pi
Hor. Phase Shift 2pi (to the right)
Vert Shift 1 up
..
See the graph below for the final translated configuration
..

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