Graph y=4sec(2x)

Problem

y=4*sec(2*x)

Solution

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

2. Determine the amplitude of the related cosine function. The coefficient 4 indicates a vertical stretch, meaning the local minima of the secant curves will be at y=4 and the local maxima will be at y=−4

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

4. Find the vertical asymptotes by setting the argument of the secant function equal to the locations where the cosine function is zero.

cos(2*x)=0

2*x=π/2+n*π

x=π/4+(n*π)/2

5. Identify key points for one cycle. At x=0 y=4*sec(0)=4 At x=π/2 y=4*sec(π)=−4

6. Sketch the graph by drawing the vertical asymptotes at x=−π/4,π/4,(3*π)/4 and plotting the U-shaped branches between them starting from the local extrema.

Final Answer

y=4*sec(2*x)

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