Posted by **Hannah** on Sunday, February 14, 2010 at 11:31am.

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.

- Pre-Cal(Please check) -
**Anonymous**, Sunday, February 14, 2010 at 11:57am
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

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