Question

Where are the asymptotes of f(x) = tan(4x − π) from x = 0 to x = pi over 2?

Answer 1

Asymptotes of tan(x) are at (k+1/2)π.  (k=any integer)
So the asymptotes of f(x) are
f(x)=tan(4x-π)=tan((k+1/2)π)
=>
4x-π=(k+1/2)π
=(k+1/2)π+π
=(k+1/2)π
=>
x=(k+1/2)π/4
=(k/4+1/8)π
For
k=-1, x=-π/8  [outside of (0,π/2)]
k=0, x=π/8
k=1, x=3π/8
k=2, x=5π/8  [outside of (0,π/2)]
So the answer is x={π/8, 3π/8}