Graph y=2sec(x/2)

Problem

y=2*sec(x/2)

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 or vertical stretch. The coefficient 2 indicates a vertical stretch by a factor of 2 The local minima of the 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 of sec(b*x) is (2*π)/|b| Here, b=1/2

P=(2*π)/(1/2)=4*π

4. Find the vertical asymptotes by setting the argument of the secant function to the locations where the parent secant is undefined. The parent sec(θ) is undefined at θ=π/2+n*π

x/2=π/2+n*π

x=π+2*n*π

For n=0 x=π For n=1 x=3*π For n=−1 x=−π

5. Plot key points within one period [0,4*π]
   At x=0 y=2*sec(0)=2*(1)=2
   At x=2*π y=2*sec(π)=2*(−1)=−2
   At x=4*π y=2*sec(2*π)=2*(1)=2

6. Sketch the graph by drawing the vertical asymptotes at x=…,−π,π,3*π,… and drawing the U-shaped branches between them, alternating between opening upward (starting at y=2 and opening downward (starting at y=−2.

Final Answer

y=2*sec(x/2)

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