Question

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

Answer 1

There are vertical asymptotes at `x=pi/2+pik,k in ZZ`

Explanation:

To look at this problem I will use the identity:
`sec(x)=1/cos(x)`

From this we see that there will be vertical asymptotes whenever `cos(x)=0`. Two values for when this occurs spring to mind, `x=pi/2` and `x=(3pi)/2`. Since the cosine function is periodic, these solutions will repeat every `2pi`.

Since `pi/2` and `(3pi)/2` only differ by `pi`, we can write all these solutions like this:
`x=pi/2+pik`, where `k` is any integer, `k in ZZ`.

The function has no holes, since holes would require both the numerator and the denominator to equal `0`, and the numerator is always `1`.