Question

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

Answer 1

D 3pi/8 , 5pi/8

Explanation:

Answer 2

`tan(4x- \pi )= (sin(4x- \pi ))/(cos(4x- \pi ))`

The asymptotes are where the graph is undefined. Since: tan(x) =sin(x)/cos(x)
It is where cos(4x-π) = 0

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

4x - π = π/2
4x = π/2 + π
4x = 3π/2
x = 3π/8

4x - π = 3π/2
4x = 3π/2 + π
4x = 5π/2
x = 5π/8
This ones outside the interval (5π/8 > π/2) , try -π/2

4x - π = -π/2
4x = -π/2 + π
4x = π/2
x = π/8

Asymptotes are π/8 and 3π/8