<s>[INST] How to find the vertical asymptotes of this equation, $f(x)=2\tan(4x-32)$? Could someone show how to use the vertical asymptote formula?  I am having a hard time getting it into the right form. I thought maybe I had to put $(4x-32)$ equal to the vertical asymptote equation.  I can't get the math to work out.
Asymptote formula $$x= \frac CB + \frac{\pi}{ 2 \lvert B \rvert} k $$
where $k$ is an integer. Here is my attempt but clearly it is wrong.  $$ 4x-32 = \frac{32}{4} + \frac {\pi}{2 \lvert 4 \rvert}k$$  When I solve the equation, I get: $$ x= 2+\frac{\pi}{32}k +8$$ When I put a $0$ in for $k$, the equation $= 10$.  It should $= 8$.
What am I doing wrong?
 [/INST] The tangent function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \tan x$ has vertical asymptotes at $x = \dfrac{\pi}{2} + k\pi, k \in \mathbb{Z}$.
We can solve for the vertical asymptotes of $g(x) = A\tan(Bx - C)$, where $B \neq 0$, by setting
$$Bx - C = \frac{\pi}{2} + k\pi, k \in \mathbb{Z}$$
then solving for $x$.  Doing so yields
\begin{align*}
Bx & = C + \frac{\pi}{2} + k\pi, k \in \mathbb{Z}\\[2 mm]
x & = \frac{C}{B} + \frac{\pi}{2B} + \frac{k\pi}{B}, k \in \mathbb{Z}
\end{align*}
which does not agree with the formula you stated.
If we substitute $4$ for $B$ and $32$ for $C$ in the above formula, we obtain the vertical asymptotes
\begin{align*}
x & = \frac{32}{4} + \frac{\pi}{2 \cdot 4} + \frac{k\pi}{4}, k \in \mathbb{Z}\\
  & = 8 + \frac{\pi}{8} + \frac{k\pi}{4}, k \in \mathbb{Z}
\end{align*}
If $k = 0$, then we obtain
$$x = 8 + \frac{\pi}{8} \approx 8.392699082$$
which is a vertical asymptote of the graph of $f(x) = 2\tan(4x - 32)$, as you can check with a graphing calculator.
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