Where are the asymptotes of f(x) = tan(4x − π) from x = 0 to x = pi over 2?
x = pi over 4, x = 3 pi over 4
x = 0, x = pi over 4
x = pi over 2, x = 3 pi over 2
x = 3 pi over 8, x = 5 pi over 8

Thus the asymptotes are π/8 and 3π/8  

Step-by-step explanation:

An asymptote is a line that a graph approaches without touching. The asymptotes are where the graph is undefined.

tan(x)= sin(x)/cos(x), where cos(4x-π) = 0

cos(4x-π) = 0 when inside is - π/2 , π/2 , 3π/2

4x-π = π/2

Add π at both sides:

4x-π+π = π/2 +π

4x= π+2π/2

4x= 3π/2

x= 3π/8

4x-π =  3π/2

4x = 3π/2 + π

4x = 5π/2  

x = 5π/8

This one is outside the interval.

4x-π = -π/2

4x = -π/2 +π

4x = -π+2π/2

4x=π/2

x = π/8

Thus the asymptotes are π/8 and 3π/8

ok

Step-by-step explanation: