Graph y=1/2sec((pix)/2)

Problem

y=1/2*sec((π*x)/2)

Solution

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

2. Determine the amplitude (vertical stretch). The coefficient 1/2 indicates a vertical compression. The local minima of the secant curves will be at y=1/2 and the local maxima will be at y=−1/2

3. Calculate the period of the function. The period P is found using the formula P=(2*π)/B where B=π/2

P=(2*π)/π/2=4

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

(π*x)/2=π/2+n*π

x=1+2*n

The asymptotes occur at odd integers: x=…,−3,−1,1,3,…

5. Identify key points for one period. Since there is no horizontal or vertical shift, the center of a "U-shaped" branch occurs at x=0

y(0)=1/2*sec(0)=1/2

The center of an "n-shaped" branch occurs at x=2

y(2)=1/2*sec(π)=−1/2

6. Sketch the graph by drawing the vertical asymptotes at x=1 and x=3 then plotting the vertex (0,1/2) opening upward and the vertex (2,−1/2) opening downward.

Final Answer

y=1/2*sec((π*x)/2)

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