Graph y=2csc(2x)

Problem

y=2*csc(2*x)

Solution

1. Identify the parent function and its properties. The function is y=a*csc(b*x) which is the reciprocal of y=a*sin(b*x) The parent function is csc(x)

2. Determine the amplitude of the related sine function. The coefficient a=2 indicates a vertical stretch. The local minima of the cosecant curves will be at y=2 and the local maxima will be at y=−2

3. Calculate the period of the function. The period P for cosecant is found using the formula P=(2*π)/|b|

P=(2*π)/2

P=π

4. Locate the vertical asymptotes. Asymptotes for csc(2*x) occur where sin(2*x)=0 This happens when 2*x=n*π for any integer n

x=(n*π)/2

For one period starting at x=0 the asymptotes are at x=0 x=π/2 and x=π

5. Find key points between the asymptotes. The midpoints between asymptotes provide the vertices of the curves.

At *x=π/4,y=2*csc(2⋅π/4)=2*csc(π/2)=2*(1)=2

At *x=(3*π)/4,y=2*csc(2⋅(3*π)/4)=2*csc((3*π)/2)=2*(−1)=−2

6. Sketch the graph by drawing the vertical asymptotes and plotting the vertices. Draw U-shaped curves opening upward from (π/4,2) and downward from ((3*π)/4,−2)

Final Answer

y=2*csc(2*x)

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