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how do i find the two asymptotes and period for y=tan(x/3)?
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y=tan(x/3)=sin(x/3)/cos(x/3) Asymptotes occur when cosx/3=0. x=(2n+1)*(pi/2) put n=0,1,2 and generally period of tan(kx)=pi/k
why am i putting n=0,1 & 2 if i only need 2 asymptotes?
it should be 0,1
so for 0 i got pi/2
and for 1 i got 3pi/2. would those be the two asymptotes?
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or would one of the asymptotes be 3?
it should be x/3=(2n+1)*(pi/2)
okay so then 3pi/2 and 9pi/2 is what i got. is that correct?
and the period would be pi/3?
here k=1/3
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is k=1/3 the period of part of the formula to find the asymptotes?
i think asymptotes are all right and for period p=pi/k=pi/(1/3)=3(pi)
oh okay, thank you for your helP! :)
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