How do you graph $y = sec x + 2$ $y=secx+2$ ?

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How do you graph $y = sec x + 2$ $y=secx+2$ ?

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Answer 1 · 2018-06-14T06:33:34

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Explanation:

Given: $y = sec x + 2$ $y = sec x + 2$

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

Since $sec x = \frac{1}{cos x}$ $sec x = 1/(cos x)$ , sketch a dashed cosine function

$y = cos x + 2 \Rightarrow amplitude = 1 and period = 2 π$ $y = cos x + 2 => " amplitude" = 1 " and period " = 2 pi$

Remember that a cosine with a period of $2 π$ $2 pi$ needs to be divided into 4 sections: $0, \frac{π}{2}, π, \frac{3 π}{2}, 2 π$ $0, pi/2, pi, (3pi)/2, 2 pi$ .

Wherever the cosine function crosses the $y = 2$ $y = 2$ 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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How do you graph $y = sec x + 2$ $y=secx+2$ ?

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