Graph y=sec(x-pi/4)

Problem

y=sec(x−π/4)

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 specifically at x=π/2+n*π

2. Determine the phase shift by looking at the argument of the function. The term (x−π/4) indicates a horizontal shift to the right by π/4 units.

3. Find the new vertical asymptotes by setting the argument equal to the original asymptote locations.

(x−π/4)=π/2+n*π

x=(3*π)/4+n*π

4. Identify key points by shifting the local minima and maxima of the parent function. The parent function has a local minimum at (0,1) and a local maximum at (π,−1) Shifting these right by π/4 gives:

Local Minimum: *(π/4,1)

Local Maximum: *((5*π)/4,−1)

5. Sketch the graph by drawing the vertical asymptotes at x=−π/4 x=(3*π)/4 and x=(7*π)/4 then plotting the key points and drawing the U-shaped curves that approach the asymptotes.

Final Answer

y=sec(x−π/4)

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