# Where are the asymptotes of f(x) = tan(4x - π) from x = 0 to x = 2 ?

**Solution:**

f(x) = tan(4x - π)

We can write it as

tan(4x - π) = sin(4x - π)/ cos(4x - π)

We get vertical asymptotes for the function f(x) = tan(4x - π)

Asymptotes are the straight lines that approach the curve but do not meet it.

We obtain asymptotes when the graph is undefined when the inside is -π/2, π/2, 3π/2

cos (4x - π) = 0

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 is outside the interval (5π/8 > π/2) so try -π/2

4x - π = -π/2

4x = -π/2 + π

4x = π/2

x = π/8

Therefore, the asymptotes are π/8 and 3π/8.

## Where are the asymptotes of f(x) = tan(4x - π) from x = 0 to x = 2 ?

**Summary:**

The asymptotes of f(x) = tan(4x - π) from x = 0 to x = π/2 are π/8 and 3π/8.

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