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

Jan 2, 2018

There are vertical asymptotes at $x = \frac{\pi}{2} + \pi k , k \in \mathbb{Z}$

#### Explanation:

To look at this problem I will use the identity:
$\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

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

Since $\frac{\pi}{2}$ and $\frac{3 \pi}{2}$ only differ by $\pi$, we can write all these solutions like this:
$x = \frac{\pi}{2} + \pi k$, where $k$ is any integer, $k \in \mathbb{Z}$.

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