# For what values of x, if any, does f(x) = sec((-15pi)/8+9x)  have vertical asymptotes?

Apr 23, 2016

$x = - \frac{11 \pi}{72} + \frac{\pi}{8} k$ $\text{ }$ for $k$ an integer. which may also be written as $x = \frac{5 \pi}{72} + \frac{\pi}{8} k$ $\text{ }$ for #k an integer.

#### Explanation:

Recall $\sec \theta = \frac{1}{\cos} \theta$

The secant has a vertical asymptote where the cosine of the argument is $0$.

The cosine is $0$ where the argument is an odd multiple of $\frac{\pi}{2}$

The odd multiples of $\frac{\pi}{2}$ have form $\left(2 k + 1\right) \frac{\pi}{2}$, which may also be written $\frac{\pi}{2} + \pi k$ for $k$ an integer.

So this function has vertical asymptotes at the solutions to

$\frac{- 15 \pi}{8} + 9 x = \frac{\pi}{2} + \pi k$ $\text{ }$ for $k$ an integer.

$9 x = \frac{4 \pi}{8} - \frac{15 \pi}{8} + \pi k$ $\text{ }$ for $k$ an integer.

$9 x = - \frac{11 \pi}{8} + \pi k$ $\text{ }$ for $k$ an integer.

$x = - \frac{11 \pi}{72} + \frac{\pi}{8} k$ $\text{ }$ for $k$ an integer.

Alternatives

We need not take $- \frac{11 \pi}{72}$ as our base case. We can use any $x$ on the list of solutions (any $x$ that has the proper form).

We got $9 x = - \frac{11 \pi}{8} + \pi k$ $\text{ }$ for $k$ an integer.

Taking $k = 2$ we see that $9 x = \frac{5 \pi}{8}$ is a solution.
So we could replace the solution with

$9 x = \frac{5 \pi}{8} + \pi k$ $\text{ }$ for $k$ an integer.

So, the solutions are:

$x = \frac{5 \pi}{72} + \frac{\pi}{8} k$ $\text{ }$ for $k$ an integer.