Graph y=3csc(2x)

Problem

y=3*csc(2*x)

Solution

1. Identify the parent function and its properties. The function is a transformation of y=csc(x) which has vertical asymptotes where sin(x)=0

2. Determine the amplitude of the related sine function. The coefficient 3 indicates a vertical stretch, meaning the local minima of the cosecant curves will be at y=3 and the local maxima will be at y=−3

3. Calculate the period of the function. The period of csc(B*x) is (2*π)/B Here, B=2 so the period is (2*π)/2=π

4. Locate the vertical asymptotes by finding where the argument of the cosecant function equals multiples of π Set 2*x=n*π which gives x=(n*π)/2 for any integer n

5. Identify key points for one cycle. Between the asymptotes at x=0 and x=π/2 there is a local minimum at the midpoint x=π/4 with y=3*csc(2⋅π/4)=3*(1)=3 Between x=π/2 and x=π there is a local maximum at x=(3*π)/4 with y=3*csc(2⋅(3*π)/4)=3*(−1)=−3

6. Sketch the graph by drawing the vertical asymptotes at x=0,±π/2,±π and plotting the U-shaped curves opening upward from (π/4,3) and downward from ((3*π)/4,−3)

Final Answer

y=3*csc(2*x)

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