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

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Answer 1 · 2018-07-25T09:39:44

See graph and details.

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

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 ]}

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