Asymptote of Tangent Function

[harpazo]

Find the asymptote of
y = (3/2) tan(2x + pi/2).

Within parentheses, factor out 2.

y = (3/2) tan(2(x + pi/4))

The asymptote of tan x is x = pi/2 + pi•n, where n is any integer.

Set 2(x + pi/4) to the asymptote of tan x and solve for x.

2(x + pi/4) = pi/2 + pi•n

2x + pi/2 = pi/2 + pi•n

Subtracting pi/2 from both sides leads to

2x = pi•n

x = pi•n/2

1. Am I right?

2. What does the answer mean?

How is this done for csc x, cot x and sec x?

 

[raquelc]

You'll have the same graph than tan, but instead of having the asymptotes separated on intervals of pi, you have them every pi/2 and instead of starting at pi/2, you start at 0

 

[harpazo]

You'll have the same graph than tan, but instead of having the asymptotes separated on intervals of pi, you have them every pi/2 and instead of starting at pi/2, you start at 0

Can you provide help by finding the asymptote of y = (3/2) cot(x - pi/2) using algebra in place of graphing?

 

[raquelc]

Cot is cos/sin. If for tan you looked at when the cos was 0, this time you'll look at when the sin is 0

 

[harpazo]

Cot is cos/sin. If for tan you looked at when the cos was 0, this time you'll look at when the sin is 0

To observe when sine is 0, I must convert the function to y = (3/2) [cos(x - pi/2)]/[sin(x - pi/2)]. Right?

 

[MarkFL]

To observe when sine is 0, I must convert the function to y = (3/2) [cos(x - pi/2)]/[sin(x - pi/2)]. Right?

No, you just observe that:

[MATH]\sin(\theta)=0[/MATH]
implies:

[MATH]\theta=\pi k[/MATH] where \(k\in\mathbb{Z}\).

 

[raquelc]

To observe when sine is 0, I must convert the function to y = (3/2) [cos(x - pi/2)]/[sin(x - pi/2)]. Right?

Yes,
So x-pi/2=pi*k, and solve for x

 

[harpazo]

Yes,
So x-pi/2=pi*k, and solve for x

MarkFL said no. Thus, it is sufficient for me that he has spoken.

 

[harpazo]

No, you just observe that:

[MATH]\sin(\theta)=0[/MATH]
implies:

[MATH]\theta=\pi k[/MATH] where \(k\in\mathbb{Z}\).

No need to convert. I got it.

 

[MarkFL]

MarkFL said no. Thus, it is sufficient for me that he has spoken.

It could be that I misunderstood, or was just wrong. I went back and found your question was:

Can you provide help by finding the asymptote of y = (3/2) cot(x - pi/2) using algebra in place of graphing?

I mistakenly thought you were asking about the cotangent function in general.

I would rewrite the given function as:

[MATH]y=-\frac{3}{2}\tan(x)[/MATH]
And then we see that raquelc is correct when stating the asymptotes are at:

[MATH]x=\frac{\pi}{2}(2k+1)[/MATH]

 

[harpazo]

It could be that I misunderstood, or was just wrong. I went back and found your question was:



I mistakenly thought you were asking about the cotangent function in general.

I would rewrite the given function as:

[MATH]y=-\frac{3}{2}\tan(x)[/MATH]
And then we see that raquelc is correct when stating the asymptotes are at:

[MATH]x=\frac{\pi}{2}(2k+1)[/MATH]

Cool.

 

[harpazo]

Yes,
So x-pi/2=pi*k, and solve for x

I appreciate your input.