Find Asymptote of Cosine Functions

harpazo

harpazo

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I searched online and You Tube for a lesson on how to find asymptotes of trig functions algebraically but only found one on the tangent function. My textbook does not cover this topic. How is it done?

Find the asymptote of the functions given below algebraically. Please, show the steps.

A. y = cos(3x + pi/3)

B. y = -3 cos(2pi•x - pi/4)
 
Cosines don't have asymptotes!

Do you not see that from a glance at the graph?
 
Please answer Dr P's question. Don't you see that from the graph? Do you know what an asymptote looks likes? There are three types of asymptotes, can you name them and draw a diagram of each one?
 
Not true in general, sorry. Look at the continuous graph of y = e^-x. It is continuous and has a horizontal asymptote.
 
harpazo said:
The graph is continuous and thus there are not asymptotes for cosine.
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I believe what you mean is that the cosine is defined (and, more important, bounded) for all real values of x, so it doesn't have a vertical asymptote. Continuity is somewhat related, but to a mathematician is quite different.

At a vertical asymptote, a function would "become infinite", so that it would not be defined there, and there would be no maximum value (bound). This is what makes the tangent, for example, quite different. And a horizontal asymptote is another beast altogether.
 
Dr.Peterson said:
I believe what you mean is that the cosine is defined (and, more important, bounded) for all real values of x, so it doesn't have a vertical asymptote. Continuity is somewhat related, but to a mathematician is quite different.

At a vertical asymptote, a function would "become infinite", so that it would not be defined there, and there would be no maximum value (bound). This is what makes the tangent, for example, quite different. And a horizontal asymptote is another beast altogether.
Click to expand...

What makes the horizontal asymptote any different? A beast in what way? Can you give an example?
 
harpazo said:
What makes the horizontal asymptote any different? A beast in what way? Can you give an example?
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The function can be continuous and still have a horizontal asymptote. I already gave you an example, did you not look at it?

Any function of the form y=a^x, where o<a<1 OR of the form y=a^x, where a>1 (so for example y= .5^x and y=3^x) will be continuous and have horizontal asymptotes.

Then there are oblique asymptotes! They are yet another beast!
 
And what makes them different? Horizontal asymptotes are horizontal (and can be continuous w/o violating the definition of a function) while vertical asymptote are vertical and MUST be discontinuous.
 
What about my work for this one?

Find asymptote of y = - 3 sec(pi•x/2 + pi/4) - 1.

Use x = kx + pi/2, where k is any integer.

Set x = argument of secant.
I found x to be

kx + pi/2 = pi•x/2 + pi/4

x = 2k + (1/2), where k is any integer is the general equation for the asymptote of the given function.

Yes?
 
harpazo said:
What about my work for this one?

Find asymptote of y = - 3 sec(pi•x/2 + pi/4) - 1.

Use x = kx + pi/2, where k is any integer.

Set x = argument of secant.
I found x to be

kx + pi/2 = pi•x/2 + pi/4

x = 2k + (1/2), where k is any integer is the general equation for the asymptote of the given function.

Yes?
Click to expand...
Your answer is correct, though the work looks like nonsense. I'm not sure how much was just typos; maybe you did the right work on paper but copied it wrong.

First, you used x to mean two things, which is dangerous. You don't really mean that x = kx + pi/2! Evidently you mean that the argument of the secant is kx + pi/2. But I'm not sure what that even means.

What you should have done is to say that the argument, at an asymptote, is an odd multiple of pi/2, that is, (2k+1)pi/2 for any integer k. This can be written as pi(k + 1/2), or k pi + pi/2. That's probably what you meant, but you replaced a pi with x when you wrote it.

So what you want is pi*x/2 + pi/4 = k pi + pi/2. Dividing by pi, this is x/2 + 1/4 = k + 1/2, and doubling, x + 1/2 = 2k + 1, so we end up with x = 2k + 1/2 (for any integer k). That's what you got, so I guess you must really have done something like the right work.
 
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Dr.Peterson said:
Your answer is correct, though the work looks like nonsense. I'm not sure how much was just typos; maybe you did the right work on paper but copied it wrong.

First, you used x to mean two things, which is dangerous. You don't really mean that x = kx + pi/2! Evidently you mean that the argument of the secant is kx + pi/2. But I'm not sure what that even means.

What you should have done is to say that the argument, at an asymptote, is an odd multiple of pi/2, that is, (2k+1)pi/2 for any integer k. This can be written as pi(k + 1/2), or k pi + pi/2. That's probably what you meant, but you replaced a pi with x when you wrote it.

So what you want is pi*x/2 + pi/4 = k pi + pi/2. Dividing by pi, this is x/2 + 1/4 = k + 1/2, and doubling, x + 1/2 = 2k + 1, so we end up with x = 2k + 1/2 (for any integer k). That's what you got, so I guess you must really have done something like the right work.
Click to expand...

It is a typo at my end.
 
Jomo said:
… Horizontal asymptotes are … continuous … while [a] vertical asymptote [is] … discontinuous.
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It seems like you're confusing 'asymptote' with 'asymptotic behavior', Jomo.

All asymptotes are continuous, straight lines; they go forever in each direction.

When a function's outputs begin to exhibit such a linear trend (that is, when some part of a function's graph constantly approaches an asymptote), then we say the function has asymptotic behavior (over the corresponding interval).

It is functions that may be continuous or discontinuous; not the asymptotes.

?
 
You need to revisit lessons on asymptotes, harpazo. (Cohen's trigonometry text appears extensive; it's hard for me to believe that he didn't cover the topic at all.)

Previously, you told me that this reference is a "nice link about graphs of functions".

Don't just flip through the pictures; step through reading all of the information, and try to answer their questions as you go.

?
 
Otis said:
You need to revisit lessons on asymptotes, harpazo. (Cohen's trigonometry text appears extensive; it's hard for me to believe that he didn't cover the topic at all.)

Previously, you told me that this reference is a "nice link about graphs of functions".

Don't just flip through the pictures; step through reading all of the information, and try to answer their questions as you go.

?
Click to expand...

Ok. Will do.
 
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