What are the asymptote(s) and hole(s), if any, of $f(x) = 1/cosx$ ?

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Answer 1 · 2016-08-30T17:50:31

There will be vertical asymptotes at $x = pi/2 + pin$ , n and integer.

There will be asymptotes.

Whenever the denominator equals $0$ , vertical asymptotes occur.

Let's set the denominator to $0$ and solve.

$cosx = 0$

$x = pi/2, (3pi)/2$

Since the function $y = 1/cosx$ is periodic, there will be infinite vertical asymptotes, all following the pattern $x = pi/2 + pin$ , n an integer.

Finally, note that the function $y = 1/cosx$ is equivalent to $y = secx$ .

Hopefully this helps!

What are the asymptote(s) and hole(s), if any, of f(x) = 1/cosx f(x) = 1/cosx?