How do you graph #3sec(3x)#?

1 Answer
Jul 25, 2018

Answer:

See graph and details.

Explanation:

#y = 3 sec ( 3 x ) = 3 / cos ( 3 x ) notin 3 ( - 1, 1 ) = ( - 3, 3 )#

Period = period of #cos ( 3 x ) = ( 2 pi ) / 3#

Asynptotes# x =# zeros of the denominator #cos ( 3 x ) #.

#= ( 2 k + 1 )(pi/6) , k = 0, +-1, +-2, +-3, ...#

See Graph.
graph{(y cos (3x ) - 3)(y^2-9) = 0[-20 20 -10 10]}
Graph for one period #(2pi)/3, x in [ - pi/3, pi/3 ]#, sans asymptotic

#x = +- 1/6pi#:
graph{(y cos (3x ) - 3)(y^2-9)(x-1.04+0.001y) (x+1.04+0.001y)(x-0.52+0.001y) (x+0.52+0.001y)= 0[-3 3 -6 6 ]}