Wednesday

October 22, 2014

October 22, 2014

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

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 11:57amNo.

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

**Answer this Question**

**Related Questions**

trig - Solve cos x-1 = sin^2 x Find all solutions on the interval [0,2pi) a. x=...

Pre-Calculus-check answers - State the period and phase shift of the function y...

Pre-Cal(Please check) - Approximate the equation's solutions in the interval (0,...

pre cal - cosA (-7/25) pi < A < 3pi/2 sin B= -3/5 3pi/2 <B<2pi

calculus - given the graph of f(x) = x sinx, 0<=x<=2pi assuming that a ...

Math - In the interval 0 </ x < 2pi, determine the equations of all ...

trig - I need to state the period and 2 consecutive asymptotes on the graph for ...

TRIG - I need to state the period and 2 consecutive asymptotes on the graph for ...

Trigonometry - 4. Write the equation of the cosine function with an amplitude ...

Calculus - 1) The period of a trig. function y=sin kx is 2pi/k. Then period of y...