$$y= -\tan(x+\pi/4).$$How do I find the asymptote of this?
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$\begingroup$We need to find asymptotes of $$y = -\tan \left (x + \frac{\pi}{4} \right).$$
It is well-known that the $\tan$ function is undefined whenever $\cos(x) = 0$, which happens at $x = \frac{\pi}{2} + \pi k$ for integer $k$.
Hence, we must have: $$x + \frac{ \pi}{4} = \frac{\pi}{2} + \pi k.$$
Simplifying, we get: $$x = \pi k + \frac{ \pi}{4}.$$
Thus, the asymptotes are $$x = \pi k + \frac{\pi}{4}$$ for integer $k$.
$\endgroup$ $\begingroup$Normally, I wouldn't answer, because the OP didn't show work. But since another answer has already been given ...
Alternative approach
$$y= -\tan(x+\pi/4).$$How do I find the asymptote of this?
Use that
$\tan(a+b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)},$
$\tan(\pi/4) = 1$
$\tan(x) = 1 \iff x \in \{\pi/4, 5\pi/4\}$, within a modulus of $2\pi$.
$$y = - \tan(x + \pi/4) = - \frac{\tan(x) + 1}{1 - \tan(x)}.$$
For an asymptote, it is sufficient that the denominator $= 0$ and the numerator $\neq 0$. Sufficiency refers to the fact that when the numerator and denominator are both $(0)$, then the situation is unclear. In this problem, no such complication is present.
Therefore, it is immediate that (within a modulus of $2\pi$), the asymptotes occur at $\{\pi/4, 5\pi/4\}.$
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