Graph y=2sec(1/2x)

Problem

y=2*sec(1/2*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 (vertical stretch). The coefficient 2 indicates a vertical stretch by a factor of 2 The local minima of the secant 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 is found using the formula P=(2*π)/|b| where b=1/2

P=(2*π)/(1/2)=4*π

4. Find the vertical asymptotes by setting the argument of the secant function equal to the locations of the parent function's asymptotes, π/2+n*π

1/2*x=π/2+n*π

x=π+2*n*π

For n=0 and n=−1 the asymptotes are at x=π and x=−π

5. Plot key points within one period. Since the period is 4*π we evaluate the function at x=0 x=2*π and x=4*π

y(0)=2*sec(0)=2

y*(2*π)=2*sec(π)=−2

y*(4*π)=2*sec(2*π)=2

6. Sketch the curves between the asymptotes. The graph consists of U-shaped branches opening upward from (0,2) and downward from (2*π,−2)

Final Answer

y=2*sec(1/2*x)

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