SOLUTION: state the amplitude, period, phase shift, and vertical shift for the function. Then graph the function. y=4cos[3theta+(3/2)pi]+2

Question 750083: state the amplitude, period, phase shift, and vertical shift for the function. Then graph the function. y=4cos[3theta+(3/2)pi]+2
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
state the amplitude, period, phase shift, and vertical shift for the function. Then graph the function. y=4cos[3theta+(3/2)pi]+2
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Equation for cos function: y=Acos(Bx-C), A=amplitude, period=2π/B, phase shift=C/B
For given equation: y=4cos(3x+3π/2)+2
A=4
B=3
period=2π/B=2π/3
1/4 period=2π/12=π/6
C=3π/2
phase shift=C/B=(3π/2)/3=π/2 (shift to the left)=3π/6
curve is bumped up 2 units
..
Graphing function:
On an x-y coordinate system with the x-axis scaled in radians, start with coordinates of the basic cos functon with period=2π/3,amplitude =1, phase shift=0, vertical shift=0, for one period.
coordinates of basic cos function:
(0,1), (π/6,0), (π/3,-1), (π/2,0), (2π/3,1), (5π/6,0)
..
shift curve π/2 to the left:
(-π/2,1), (-π/3,0), (-π/6,-1), (0,0), (π/6,1), (π/3,0)
..
change amplitude to 4:
(-π/2,4), (-π/3,0), (-π/6,-4), (0,0), (π/6,4), (π/3,0)
..
vertical shift up 2 units (final configuration)
(-π/2,6), (-π/3,2), (-π/6,-2), (0,2), (π/6,6), (π/3,2)
set x=0
y-intercept: 4cos(3x+3π/2)+2=4cos(3π/2)+2=-4+2=2

..
It is easy to make a mistake here. Please check my calculations.
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