Question 530012: sketch one period of the given graph .
a. y=2csc(x-π/2)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! sketch one period of the given graph .
a. y=2csc(x-π/2)
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The equation used to graph given function: Acsc(Bx-C), A=multiplier, Period=2π/B, Phase-shift=C/B
For given function:
A=2
B=1
Period=2π/B=2π
Phase Shift=C/B=(π/2)/1=π/2 (to the right)
..
Graphing:
I don't have the means to graph it for you, but I can show you how.
When graphing It would be helpful to relate to the more familiar sin function since csc is its reciprocal.
On the x-axis make tick marks at π/2, π, 3π/2 and 2π representing four 1/4 periods or one period.
For sin(x), coordinates are: (0,0), (π/2,1), (π,0), (3π/2,-1), and(2π,0)
For csc(x), y-coordinates are reciprocals: (0,∞), (π/2,1), (π,∞), (3π/2,-1), and(2π,∞)
For csc(x-π/2), move the x-coordinates π/2 units to the right:
(π/2,∞), (π,1), (3π/2,∞), (2π,-1), and(5π/2,∞)
For the final configuration, 2csc(x-π/2): multiply y-coordinates by 2:
(π/2,∞), (π,2), (3π/2,∞), (2π,-2), and(5π/2,∞)
You now have these 5 points with which you can graph the given function. Note that where you see the y-coordinate, ∞, there is a vertical asymptote at that point. Graphically, the csc curve sits atop and below the max and min of the sin curve.
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