Graph y=sec(2x)

Problem

y=sec(2*x)

Solution

1. Identify the parent function and its properties. The function y=sec(2*x) is the reciprocal of y=cos(2*x)

2. Determine the period of the function. The period of sec(b*x) is (2*π)/|b|

Period=(2*π)/2=π

3. Find the vertical asymptotes by identifying where the cosine function is zero. For cos(2*x) zeros occur when 2*x=π/2+n*π

x=π/4+(n*π)/2

The first two positive asymptotes are x=π/4 and x=(3*π)/4

4. Identify key points by finding the local extrema. These occur where cos(2*x) is 1 or −1

At *x=0,y=sec(0)=1

At *x=π/2,y=sec(π)=−1

At *x=π,y=sec(2*π)=1

5. Sketch the graph by drawing the vertical asymptotes and plotting the local minima and maxima. Draw the U-shaped curves approaching the asymptotes from these points.

Final Answer

y=sec(2*x)

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