Find the amplitude, period, phase shift, & the zeros then sketch a graph of the function.

Question

y=tan(1/2x-pi/2), 0≤x≤4pi

Answer 1

First off, you need to know all about the base function (tan x in this case).

For tan(x), the period is P = pi, the amplitude A is none (there is no max or min value), phase shift and vert shift are 0.

For y=tan(1/2x-pi/2), A is still none

 

P is a little trickier. You have to factor out from the parentheses whatever the coefficient of x is, in this case 1/2. So, y=tan1/2(x-pi). Divide the period (of the base function, i.e. pi) by what you've factored out of the parentheses: pi / 1/2, so P=2pi (dividing by 1/2 is equivalent to multiplying by 2/1).  

 

IF THE COEFFICIENT OF x IS 1, THE PHASE OR HORIZONTAL SHIFT IS VERY EASY TO DETERMINE.  Take whatever is inside the parentheses of the argument of the tangent function (x-pi), set it equal to 0 and solve for x.  x-pi=0, so x=pi and thus the phase shift is pi units to the right.  

x+pi would result in a PS of -pi (pi units to the left).  

 

Vertical shift is easy but it isn't required to answer the problem.  

 

For the zeros, find ONE zero for the base tangent function and add the phase shift (pi to the right, so add +pi) and then add multiples of the period (2pi) as long as you're within the given domain (0≤x≤4pi).  One zero of tan(x) is at x=0 (0,tan(0)) = (0,0), SO 0+pi(the phase shift) will give you a zero at pi, then from there start adding multiples of the period (2pi,4pi,6pi,...) as long as you stay between 0 and 4pi (0≤x≤4pi).  The first zero for your shifted function was at pi, so pi + 2pi = 3pi, pi + 4pi = 5pi <== OUTSIDE OUR DOMAIN.

Zeros are at pi and 3pi.