How do you graph $y = sec (3 x - \frac{2 π}{3}) - 1$ $y=sec(3x-(2pi)/3)-1$ ?

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Answer 1 · 2018-07-27T14:52:50

See graph and explanation.

A sec value $\notin [- 1, 1)$ $notin [ - 1, 1 )$

$y = sec (3 x - \frac{2}{3} π) - 1 \notin {- 20)$ $y = sec ( 3 x - 2/3pi ) - 1 notin { -2 0 )$

As sec ( .. ) = 1/ cos ( .., )

the asymptotes are given by the zeros of $cos (3 x - \frac{2}{3} π)$ $cos ( 3x - 2/3pi )$

$\Rightarrow$ $rArr$ the asymptotes are

$3x - 2/3 pi = ( 2 k + 1 ) pi/2, k = 0, +-1, +-2, +-3, ...$ , giving

$x = 1/3 ( ( 2 k + 1 ) pi/2 + 2/3 pi ) = (( 3 ( 2 k + 1 ) + 4 ) pi/18$

$= 1/18 (6 k + 7) pi = ....- 5/18pi, 1/18pi, 7/18pi, ...$

See graph, with two asymptotes, near O.

graph{((y+1)cos(3x-2/3pi)-1)(y)(y+2)(x+5/18 pi+0.001y)(x-1/18pi+0y)=0}

How do you graph y=sec(3x-(2pi)/3)-1y=sec(3x−2π3)−1y=sec(3x-(2pi)/3)-1?