Graph y=sec(x+pi)

Problem

y=sec(x+π)

Solution

1. Identify the parent function and its properties. The function y=sec(x+π) is a transformation of the parent function y=sec(x) The parent function has vertical asymptotes where cos(x)=0 which occurs at x=π/2+n*π for any integer n

2. Determine the phase shift by looking at the argument of the secant function. The expression x+π indicates a horizontal shift to the left by π units.

3. Find the new vertical asymptotes by setting the argument equal to the locations where the cosine function is zero.

cos(x+π)=0

x+π=π/2+n*π

x=−π/2+n*π

4. Identify key points for one period. In the parent function y=sec(x) a local minimum occurs at (0,1) and a local maximum occurs at (π,−1) Applying the shift of π to the left:

Point 1: *(0−π,1)⇒(−π,1)

Point 2: *(π−π,−1)⇒(0,−1)

5. Sketch the graph by drawing the vertical asymptotes at x=…,−(3*π)/2,−π/2,π/2,(3*π)/2,… and plotting the U-shaped curves (parabolic-like branches) that approach these asymptotes from the key points. Note that sec(x+π)=−sec(x) so the graph is a reflection of the parent secant graph across the x-axis.

Final Answer

y=sec(x+π)

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