# Find the amplitude, period, phase shift, & zeros then sketch a graph of the function: y=tan(1/2x-pi/2), 0<x<4pi

gsenviro | College Teacher | (Level 1) Educator Emeritus

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For a function written in standard form: y = a tan k(x-b)

amplitude = undefined

period = pi/k and

phase shift = b

In the present case

y = tan (1/2 x - pi/2) = tan 1/2 [x-pi]

The concept of amplitude does not really apply to "tangent or tan" function, since the value is "undefined" whenever the denominator (cosine function) is 0.

Amplitude: undefined

Period:  tangent functions have a period of "pi" with an asymptote separating the periods. However in this case the period = pi/(1/2) = 2 pi.

Phase shift: As mentioned earlier, phase shift is given by the term 'b' in standard equation form and here b= pi

Zeros in the interval, 0<x<4 pi: The zeros of a tangent function occurs at n*pi, i.e. 0, pi, 2pi,3pi,.....The zeros of (x-pi)/2 would be at

(x-pi)/2 = n*pi

or, x = 2n*pi - pi

and in the given interval 0<x<4pi,

the possible zeros are: 0,pi corresponding to x = pi and 3pi

Sources:

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

As for the plot of the given function, it would be a standard tangent graph, with 'undefined' values at x =0, 2pi and 4pi; 0 value at pi and 3pi and as already stated the period is 2 pi.

hope this helps.