[SOLVED] asymptotes of a tan function

**brentwoodbc** · April 25th 2009, 05:03 PM

what are the equations of the asymptotes for the function y=tan((2pi)/4)x where 0<x<4

I have trouble with this style of question whether it be find the x intercepts etc.

I got the normal asymptotes (pi/2, 3pi/2, etc.)
Edit-b=pi/2
but if i multiply pi/2 by pi/2, 3pi/2 It gives me the wrong answer.
HOw do I solve these kind of questions?

Thanks.

**mr fantastic** · April 25th 2009, 05:46 PM

Originally Posted by brentwoodbc

what are the equations of the asymptotes for the function y=tan((2pi)/4)x where 0<x<4

I have trouble with this style of question whether it be find the x intercepts etc.

I got the normal asymptotes (pi/2, 3pi/2, etc.)
Edit-b=pi/2
but if i multiply pi/2 by pi/2, 3pi/2 It gives me the wrong answer.
HOw do I solve these kind of questions?

Thanks.

Since $\tan \left( \frac{2 \pi}{4} x \right) = \frac{\sin \left( \frac{2 \pi}{4} x\right)}{\cos \left( \frac{2 \pi}{4} x\right)}$ , you just need to solve $\cos \left( \frac{2 \pi}{4} x\right) = 0$ .

Note that you should first simplify $\frac{2 \pi}{4} x$ .

**brentwoodbc** · April 25th 2009, 06:16 PM

so that would give you "one" value for x when y=0 correct, what's that have to do with the asymptotes?
I still dont understand why you cant multiply the original asymptotes by the b value.

**mr fantastic** · April 25th 2009, 06:19 PM

Originally Posted by brentwoodbc

so that would give you "one" value for x when y=0 correct, what's that have to do with the asymptotes?
I still dont understand why you cant multiply the original asymptotes by the b value.

Do you understand the definition of a vertical asymptote? Do you see why I said to solve the equation I gave you? Did you try solving that equation?

Re-read my first reply.

**brentwoodbc** · April 25th 2009, 06:26 PM

Originally Posted by mr fantastic

Do you understand the definition of a vertical asymptote? Do you see why I said to solve the equation I gave you? Did you try solving that equation?

Re-read my first reply.

I do understand what an asymptote is
I "kind of" understand why you told me to solve it. maybe to find the first 0, then multiply that by the b value?
I got x=1?

**mr fantastic** · April 25th 2009, 06:34 PM

Originally Posted by brentwoodbc

I do understand what an asymptote is
I dont understand why you told me to solve it. maybe to find the 0's?
I got x=1?

A vertical asymptote is a vertical line that passes through the value of x for which the function is undefined. When you're function has the form of a fraction, it will obviously be undefined when the denominator is equal to zero. tan can be written in the form of a fraction. I told you to solve when the denominator of this fraction is equal to zero.

x = 1 is one solution to the equation I told you to solve. There are others lying in the domain [0, 4].

**brentwoodbc** · April 25th 2009, 06:42 PM

Originally Posted by mr fantastic

A vertical asymptote is a vertical line that passes through the value of x for which the function is undefined. When you're function has the form of a fraction, it will obviously be undefined when the denominator is equal to zero. tan can be written in the form of a fraction. I told you to solve when the denominator of this fraction is equal to zero.

x = 1 is one solution to the equation I told you to solve. There are others lying in the domain [0, 4].

wow ok I think I understand, so to find the asymptotes do I multiply 1 by the b value? ie
pi/2, pi, 3pi/2, 2pi?

I have graphed it and the 1'st asymptote appears to be 1, then every 2, like 1,3,5...

so I guess since the period is 2, it's the 1 "first asymptote" plus the period (2) = 1,3,5 but since x is 0<x<4 its 1,3.

So when you solve for x like you got me to do it will always give the first asymptote/zero?

**mr fantastic** · April 25th 2009, 06:57 PM

Originally Posted by brentwoodbc

wow ok I think I understand, so to find the asymptotes do I multiply 1 by the b value? ie
pi/2, pi, 3pi/2, 2pi?

I have graphed it and the 1'st asymptote appears to be 1, then every 2, like 1,3,5...

so I guess since the period is 4, it's the 1 "first asymptote" times 4/2, 2 times 4/2, = 1,3,5 but since x is 0<x<4 its 1,3.

Forget about recipes. Understand and apply the basic principles. Solve $\cos \left( \frac{\pi}{2} x \right) = 0$ :

$\frac{\pi}{2} x = \frac{\pi}{2}$ , or $\frac{\pi}{2} x = \frac{3\pi}{2}$

**brentwoodbc** · April 25th 2009, 07:01 PM

Originally Posted by mr fantastic

Forget about recipes. Understand and apply the basic principles. Solve $\cos \left( \frac{\pi}{2} x \right) = 0$ :

$\frac{\pi}{2} x = \frac{\pi}{2}$ , or $\frac{\pi}{2} x = \frac{3\pi}{2}$

I edited that post.
I forget, when you solve for x when y = 0 what x is given if there are multiple x's, the first one, or the one closest to 0?

**mr fantastic** · April 25th 2009, 07:08 PM

Originally Posted by brentwoodbc

I edited that post.
I forget, when you solve for x when y = 0 what x is given if there are multiple x's, the first one, or the one closest to 0?

I don't understand what you're asking. I've given you the equation to solve and I've explained why you have to solve it. And I've all but solved that equation in my previous post.

The vertical asymptotes are x = 1 and x = 3. Try to see why.

**brentwoodbc** · April 25th 2009, 07:14 PM

Originally Posted by mr fantastic

I don't understand what you're asking. I've given you the equation to solve and I've explained why you have to solve it. And I've all but solved that equation in my previous post.

The vertical asymptotes are x = 1 and x = 3. Try to see why.

I 100% understand how to solve this now, (thanks very much.)
I am just asking if when you solve an equation for an x value like this equation "or any other equation" that has multiple x values, which x value will you get, the first, the closest to y=0 etc.

this is just purley out of curiosity.

an example could be (x-5)^2-4=0
x=8
why does solving for x give you 8 and not 4?

**The Second Solution** · April 25th 2009, 07:35 PM

Originally Posted by brentwoodbc

....
an example could be (x-5)^2-4=0
x=8
why does solving for x give you 8 and not 4?

Because you made a mistake.

x = 7, 3.

$(x - 5)^2 = 4 \Rightarrow x - 5 = \pm 2$ and so on.

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