1) Find the Asymptotes:

Question

1) Find the Asymptotes:
 
    y = -2 sec 2pi x

pi/2b and 3pi/2b 
  
   pi/2(2pi) = pi/4   and 3pi/4

Is this correct?

2) Find the period and asymptotes:
    
   y = 1/2 sec pi x / 2

b = pi/2    I do not know how to find the period of asymptotes for this one.  My original answer was 1 for the period and pi and 3pi for the asymptotes.

Anonymous · Answer

No.
y = -2 sec 2pi x

when the argument is PI/2, or 3PI/2 of the secant function, it is inf

2PI x =PI/2
x= 1/4 or x= 3/4 is the x value of the asymptote.

2. PIx/2=PI/2 or 3PI/2
x=1 or x=3 is asympotes

period? here is a neat way to memorize finding frequency and period.

You should know this could also be written as sin w*x, where w(omega) is the angular frequency (2PI f) or 2PI/period

So where we have sec PIx/2
convert it PI*x/2=2PIx/2 * 1/period
period= 1/f= 1