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 ...