## You know that tan(Ï€/2) is undefined so there is a vertical asymptote at x=Ï€/2

then x-Ï€/3 = Ï€/2

6x - 2Ï€ = 3Ï€

x = 5Ï€/6

there are vertical asymptotes at

x = 5Ï€/6 + kÏ€, where k is an integer.

Reiny
answered

13 years ago

13 years ago

then x-Ï€/3 = Ï€/2

6x - 2Ï€ = 3Ï€

x = 5Ï€/6

there are vertical asymptotes at

x = 5Ï€/6 + kÏ€, where k is an integer.

Step-by-Step Bot
answered

1 year ago

1 year ago

The tangent function has vertical asymptotes where the cosine function, which is the denominator of tangent, equals zero. In other words, the vertical asymptotes occur when cos(x - Ï€/3) = 0.

To solve this equation, we set the argument of the cosine function equal to Ï€/2, since the cosine function has a zero at Ï€/2:

x - Ï€/3 = Ï€/2

Next, we solve for x:

x = Ï€/2 + Ï€/3

Simplifying further:

x = (3Ï€ + 2Ï€) / 6

x = 5Ï€ / 6

Therefore, the vertical asymptote of y = tan(x - Ï€/3) is x = 5Ï€ / 6.

Explain Bot
answered

11 months ago

11 months ago

The tangent function, denoted as tan(x), has vertical asymptotes at x = (Ï€/2) + Ï€k and x = -(Ï€/2) + Ï€k, where k is an integer.

In this case, the function y = tan(x - Ï€/3) has a horizontal shift of Ï€/3 units to the right compared to the standard tangent function.

Therefore, to find the vertical asymptote of y = tan(x - Ï€/3), we need to find the x-values that satisfy the vertical asymptote condition for the standard tangent function.

For the standard tangent function, the vertical asymptotes occur at x = (Ï€/2) + Ï€k and x = -(Ï€/2) + Ï€k, where k is an integer.

However, due to the horizontal shift of Ï€/3 units to the right in the given function y = tan(x - Ï€/3), we need to replace "x" in the equation of the vertical asymptote with "(x - Ï€/3)".

So, the vertical asymptotes of y = tan(x - Ï€/3) occur at:

x = (Ï€/2) + Ï€k - Ï€/3, and

x = -(Ï€/2) + Ï€k - Ï€/3, where k is an integer.

Simplifying these expressions, we get the vertical asymptotes of the given function as:

x = Ï€/6 + Ï€k, and

x = -5Ï€/6 + Ï€k, where k is an integer.