Graph y=5sec(x)

Problem

y=5*sec(x)

Solution

1. Identify the parent function and its properties. The function is y=a*sec(b*x−c)+d where a=5 b=1 c=0 and d=0 The parent function is y=sec(x) which is the reciprocal of y=cos(x)

2. Determine the period of the function. The period of sec(x) is 2*π Since b=1 the period remains 2*π

3. Locate the vertical asymptotes by finding where the reciprocal function cos(x) is equal to zero. This occurs at x=π/2+n*π for any integer n

4. Determine the range and vertical stretch. The coefficient a=5 indicates a vertical stretch. While sec(x) has a range of (−∞,−1]∪[1,∞) the function y=5*sec(x) has a range of (−∞,−5]∪[5,∞)

5. Identify key points for one period. At x=0 y=5*sec(0)=5*(1)=5 At x=π y=5*sec(π)=5*(−1)=−5 At x=2*π y=5*sec(2*π)=5

6. Sketch the graph by drawing the vertical asymptotes at x=−π/2,π/2,(3*π)/2 plotting the local minimums at (0,5) and (2*π,5) and plotting the local maximum at (π,−5) Draw the U-shaped curves approaching the asymptotes.

Final Answer

y=5*sec(x)

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