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

1 Answer
Aug 30, 2016

Answer:

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

Explanation:

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!