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

Option A) x=3π/4, x=5π/4

The asymptotes of the function tan (z) are the values of z that are the odd multiples of π/2:
z=(2n+1)π/2, witn n= ..., -3, -2, -1, 0, 1, 2, 3, ...

In this case z=2x − π, then:
2x − π=(2n+1)π/2

Solving for x: Adding π both sides of the equation:
2x − π+ π=(2n+1)π/2+ π

Adding the terms on the right side of the equation:
2x=[(2n+1)π+2π]/2

Common factor π:
2x=[(2n+1)+2]π/2
2x=(2n+1+2)π/2
2x=(2n+3)π/2

Multiplying both sides of the equation by 1/2:
(1/2)(2x)=(1/2)[(2n+3)π/2]
(2/2)x=(2n+3)π/[(2)(2)]
x=(2n+3)π/4

For n=-1:
x=[2(-1)+3]π/4
x=(-2+3)π/4
x=π/4
x=π/4<π/2  No

For n=0:
x=[2(0)+3]π/4
x=(0+3)π/4
x=3π/4
π/2<x=3π/4<3π/2  Ok

For n=1:
x=[2(1)+3]π/4
x=(2+3)π/4
x=5π/4
π/2<x=5π/4<3π/2  Ok

For n=2:
x=[2(2)+3]π/4
x=(4+3)π/4
x=7π/4
x=7π/4>3π/2  No

x=3π/4, x=5π/4

The second step on the test was the move he was supposed to