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# Pre-calculus/trig

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Note: I believe the answers are the second and third option, I just need validation.

Mar 30, 2019

#1
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Cupcake, put your graph into the desmos program like I did with theother one and then you can see for yourself.

https://www.desmos.com/calculator

I suppose you should work it out first and then use desmos for checking your answer.

Mar 30, 2019
#2
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I did check on desmos, now I think it's the second, third, and fourth option, do you agree?

Cupcake  Mar 30, 2019
#3
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https://www.desmos.com/calculator/kwk7kktciz

The green ones do not cross so they are asymptotes.

The orange ones do cross so they aren't

So do you want to try again?

(next time maybe you could give me a link to your desmost graph   )

Mar 30, 2019
#4
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Okay, I guess I'm viewing something wrong, could you tell me what exactly you think the answer is?

Cupcake  Mar 30, 2019
#5
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OK :)

It is the second and the third one.

The first and the forth BOTH cross the graph, so they are not asymptotes.

Melody  Mar 30, 2019
#6
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Oh okay, so I was correct at first, thank you!

Cupcake  Mar 30, 2019
#9
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Yes you were

Melody  Mar 30, 2019
#7
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Lets look at this anyway. You should not rely on desmos, that is just for checking.

FIRST WAY  (The most formal way)

$$y=3cot(0.5x)-4$$

The 4 only pushed the graph up or down so it has nothing to do with aspymptotes.

The 3 just exaggerates the graph in a up down direction so that has nothing to do with it either.

so that asymptotes will be the same as for

$$y=cot(0.5x)$$

now

$$y=cot(0.5x)=\frac{cos(0.5x)}{sin(0.5x)}\\ \text{You cannot divide by 0 so the asymptotes will be where sin(0.5x)=0}\\ 0.5x\ne n\pi \qquad where \;n\;is\;an\;integer\\ so\\ x\ne 2n\pi$$

so x canot be 0 or  pi or -pi    they are three of the asymptotes.

Mar 30, 2019
#8
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SECOND WAY (probably easier for you)

Use you calculator to get values for     cot(0.5*3pi),    cot(0.5*0),    cot (0.5*2pi),  Cot(0.5*-2pi)   and cot(0.5*pi/2)

If you get a an error then that vaue of x is an asymptote.

If your calc does not have cot then use   1/tan( )

Mar 30, 2019
#10
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Awesome... Thank you so much!!

Cupcake  Mar 30, 2019
#11
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You are very welcome :)

Melody  Mar 30, 2019