How do you graph #y=secx+2#?

1 Answer
Jun 14, 2018

Answer:

See answer below

Explanation:

Given: #y = sec x + 2#

First draw a dashed vertical shift line at #y = 2#

Since #sec x = 1/(cos x)#, sketch a dashed cosine function

#y = cos x + 2 => " amplitude" = 1 " and period " = 2 pi#

Remember that a cosine with a period of #2 pi# needs to be divided into 4 sections: #0, pi/2, pi, (3pi)/2, 2 pi#.

Wherever the cosine function crosses the #y = 2# line there will be a vertical asymptote. At each peak and trough, there will be a point on the secant function that arcs up to the adjacent vertical asymptotes:

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