Question

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

Answer 1

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!