Graph sec(x/2)

Problem

sec(x/2)

Solution

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

2. Determine the period of the function. The period of sec(b*x) is (2*π)/|b| Here, b=1/2

Period=(2*π)/(1/2)=4*π

3. Find the vertical asymptotes by setting the argument of the secant function equal to the locations where the cosine function is zero. For sec(θ) asymptotes occur at θ=π/2+n*π

x/2=π/2+n*π

x=π+2*n*π

The first two positive asymptotes are at x=π and x=3*π

4. Identify key points for one period. Evaluate the function at values between the asymptotes.

At *x=0,sec(0)=1

At *x=2*π,sec(π)=−1

At *x=4*π,sec(2*π)=1

5. Sketch the graph by drawing the vertical asymptotes at x=…,−π,π,3*π,… and plotting the local minima at (0,1),(4*π,1) and local maxima at (2*π,−1),(−2*π,−1) Draw the characteristic "U" shapes approaching the asymptotes.

Final Answer

sec(x/2)* has period *4*π* and asymptotes at *x=π+2*n*π

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