Question 543649: how do you graph y=1/2 cot(2x- pi)?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! how do you graph y=1/2 cot(2x- pi)
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Equation for graphing cot function: y=Acot(Bx-C), A=multiplier, Period=π/B, phase shift=C/B
For given cot function: y=1/2 cot(2x-π)
A=1/2
B=2
Period=π/B=π/2
1/4 period=π/8
Phase shift=π/2 (shift to the right)
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Graphing for one period on (x,y) coordinate system with x-axis scaled in radians
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Coordinates for basic cot function: (asymptotes at x=0, x=π/2), and x=π
(0,∞), (π/8,1), (π/4,0), (3π/8,-1), (π/2,-∞)
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multiply y-coordinate by 1/2:
(0,∞), (π/8,1/2), (π/4,0), (3π/8,-1/2), (π/2,-∞)
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shift x-coordinates π/2 to the right for the final configuration:
(π/2,∞), (5π/8,1/2), (3π/4,0), (7π/8,-1/2), (π,-∞)
note: you are shifting entire curve one period to the right.
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You now have the coordinates and asymptotes with which you can draw the graph of the given cot function.
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