# For what values of x, if any, does f(x) = tan((-7pi)/4-2x)  have vertical asymptotes?

Aug 2, 2018

$x = \left(2 k + 1\right) \frac{\pi}{4} - \frac{7}{8} \pi = \left(4 k - 5\right) \frac{\pi}{8} ,$
$k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$
$x = \ldots , - \frac{5}{8} \pi , - \frac{1}{8} \pi , \frac{3}{8} \pi , \frac{7}{8} \pi , \ldots$

#### Explanation:

y = tan ( -7/4 pi - 2 x) = - tan ( 2 ( x + 7/8 pi )

The period = $\frac{\pi}{2}$.
The vertical asymptotes of $y = a \tan \left(b x + c\right) + d$ are

$b x + c = \left(2 k + 1\right) \frac{\pi}{2} , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

Here, they are given by

$2 \left(x + \frac{7}{8} \pi\right) = \left(2 k + 1\right) \frac{\pi}{2}$

$\Rightarrow x = \left(2 k + 1\right) \frac{\pi}{4} - \frac{7}{8} \pi = \left(4 k - 5\right) \frac{\pi}{8} ,$

$k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

$= \ldots , , - \frac{5}{8} \pi , - \frac{1}{8} \pi , \frac{3}{8} \pi \ldots$

See graph:
graph{(y+tan(2(x+7/8pi)))=0[-pi pi -pi/2 pi/2]}