Question

Graph y=3cos(2x-(pi/2))

Answer 1

How do u graph y=3cos(2(x+pi/4))+2??
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y=3cos(2(x+pi/4))+2
Equation for graphing sin function: y=Acos(Bx-C), with A=amplitude, Period=2π/B,
Phase shift=C/B.
For given sin function: y=3cos(2(x+pi/4))+2=3cos(2x+pi/2))+2
Amplitude=3
B=2
Period=2π/B=2π/2=π
1/4 period=π/4
Phase shift=C/B=(π/2)/2=π/4 (shift to the left)
Curve is shifted vertically 2 units
y-intercept
set x=0
y=3cos(2x+pi/2))+2=3cos(π/2)+2=0+2=2
..
Graphing for one period:
Without any phase shift or vertical shift, you can plot the given cos curve (with amplitude=3) on an (x,y) coordinate system with the following coordinates:
(0,3), (π/4,0), (π/2,-3), (3π/4,0), (π,3)
With a phase shift of π/4 to the left, the x-coordinates change to read as follows:
(-π/4,3), (0,0), (π/4,-3), (π/2,0), (3π/4,3)
Adding a vertical shift of 2 units cause the y-coordinates to change, resulting in the final configuration of the graph as follows:
(-π/4,5), (0,2), (π/4,-1), (π/2,2), (3π/4,5)

Source: algebra.com

Answer 2

A plot is attached.

Answer 3

Ah I see, I know what I have to do now. Thanks!