# For what values of x, if any, does f(x) = tan((5pi)/4-x)  have vertical asymptotes?

Feb 27, 2018

$f \left(x\right)$ has vertical asymptotes at $\ldots . \frac{3 \pi}{4} , - \frac{\pi}{4} , - \frac{5 \pi}{4} , \ldots$

We can say that $f \left(x\right)$ is undefined for any applicable $\frac{3 \pi}{4} + n \pi$, where $n \in \mathbb{Z}$.

#### Explanation:

$\tan \left(x\right)$ is undefined at certain points:

$\frac{\pi}{2} , \frac{3 \pi}{2} , \frac{5 \pi}{2}$, and so on.

Graphing $\tan \left(x\right) ,$ we find that, where $h$ is a point where $\tan \left(x\right)$ is undefined:

${\lim}_{x \rightarrow {h}^{+}} \tan \left(x\right) = \infty$, and ${\lim}_{x \rightarrow {h}^{-}} \tan \left(x\right) = - \infty$

We'll find the first three vertical asymptotes.

So here, $f \left(x\right) \text{ DNE}$ when $\left(\frac{5 \pi}{4} - x\right) \rightarrow h$.

Taking $h = \frac{\pi}{2}$, we have:

$\left(\frac{5 \pi}{4} - x\right) \rightarrow \frac{\pi}{2}$, which we can write as:

$\frac{5 \pi}{4} - x = \frac{\pi}{2}$

$- x = \frac{\pi}{2} - \frac{5 \pi}{4}$

$- x = - \frac{3 \pi}{4}$

$x = \frac{3 \pi}{4}$

Taking $h = \frac{3 \pi}{2}$, we have:

$\frac{5 \pi}{4} - x = \frac{3 \pi}{2}$

$- x = \frac{3 \pi}{2} - \frac{5 \pi}{4}$

$- x = \frac{\pi}{4}$

$x = - \frac{\pi}{4}$

For $h = \frac{5 \pi}{2}$, we have:

$\frac{5 \pi}{4} - x = \frac{5 \pi}{2}$

$- x = \frac{5 \pi}{2} - \frac{5 \pi}{4}$

$- x = \frac{5 \pi}{4}$

$x = - \frac{5 \pi}{4}$

This series continues to the left and right.