# For what values of x, if any, does f(x) = sec((-15pi)/8+2x)  have vertical asymptotes?

Oct 29, 2016

$x = \frac{\pi \left(19 + 8 k\right)}{16} , k \in \mathbb{Z}$

#### Explanation:

$f \left(x\right) = \sec \left(\frac{- 15 \pi}{8} + 2 x\right)$

Rewriting using the definition of secant:

$f \left(x\right) = \frac{1}{\cos} \left(\frac{- 15 \pi}{8} + 2 x\right)$

This function will have vertical asymptotes when its denominator equals zero, or when:

$\cos \left(\frac{- 15 \pi}{8} + 2 x\right) = 0$

Note that the function $\cos \left(\theta\right) = 0$ when $\theta = - \frac{\pi}{2} , \frac{\pi}{2} , \frac{3 \pi}{2}$, which can be generalized as saying that $\cos \left(\theta\right) = 0$ for $\theta = \frac{\pi}{2} + k \pi , k \in \mathbb{Z}$. Note that $k \in \mathbb{Z}$ is the mathematical way of writing that $k$ is an integer.

Thus, we see that there is a vertical asymptote in $f \left(x\right)$ when:

$\frac{- 15 \pi}{8} + 2 x = \frac{\pi}{2} + k \pi \text{ "" "" "" } , k \in \mathbb{Z}$

Adding $\frac{15 \pi}{8}$ to both sides:

$2 x = \frac{\pi}{2} + \frac{15 \pi}{8} + k \pi \text{ "" "" "" } , k \in \mathbb{Z}$

$2 x = \frac{19 \pi}{8} + k \pi \text{ "" "" "" } , k \in \mathbb{Z}$

$x = \frac{19 \pi}{16} + \frac{k \pi}{2} \text{ "" "" "" } , k \in \mathbb{Z}$

$x = \frac{\pi \left(19 + 8 k\right)}{16} \text{ "" "" "" } , k \in \mathbb{Z}$

So, there are an infinite number of times when there are vertical asymptotes. We can identify those close to the origin by trying $k = - 1$, which gives an asymptote at $x = \frac{11 \pi}{16}$, or any other value of $k$ for a different asymptote.