What is the period and asymptote in y= tan(2x-pi)
for y = tan kØ , the period of the tangent curve is π/k
(notice that this differs from the period definition for sine and cosine)
so the period of tan (2x-π) is π/2 radians or 90°
We know that tan (π/2) is undefined (a vertical asymptote)
so 2x - π = π/2
2x = 3π/2
x = 3π/4
So your function will have a vertical asymptote at
x = 3π/4 , and one every π/2 to the right or to the left after that
vertical asymptotes:
in radians : x = 3π/4 + kπ/2 , where k is an integer
in degrees : x = 135° + 90k° , where k is an integer
To find the period and asymptotes of the function y = tan(2x - π), we can use the properties of the tangent function.
1. Period:
The period of the tangent function, denoted as T, is given by the formula T = π/b, where b is the coefficient of x in the argument of the tangent function. In this case, the coefficient of x is 2. Therefore, the period of the function is T = π/2.
2. Asymptotes:
The asymptotes of the tangent function occur when the tangent function is undefined, which happens when the angle in the argument of the tangent function is equal to (2k + 1)π/2, where k is a whole number.
In the given function y = tan(2x - π), the argument of the tangent function is (2x - π).
Setting (2x - π) equal to (2k + 1)π/2, we can find the values of x where the asymptotes occur.
(2x - π) = (2k + 1)π/2
Solving for x, we have:
2x = (2k + 1)π/2 + π
2x = (2k + 3)π/2
x = (2k + 3)π/4
Therefore, the asymptotes occur at x = (2k + 3)π/4 for any whole number k.
In summary:
- The period of y = tan(2x - π) is π/2.
- The asymptotes of y = tan(2x - π) occur at x = (2k + 3)π/4 for any whole number k.