What is an asymptote and why doesn't parabola have one?

[codec9]

The slope of x^2 is 2x. Therefore, as x goes to infinity so does the slope.

 

[MaWM]

MaWM, you are on to something...even though you explaining it mathematically, you are getting the point...my question.

You are saying how it doesn't approach infinity "fast enough"...but why does speed come into play? I mean as long as something is approaching the slope undefined...how can it ever approach infinity? Wouldn't the slope of the parabola become undefined at the point infinity?

Am I being clear?

Yes, clear enough. We generally don't think about asymptotes at infinity. Asymptotes are generally reserved for functions that reach infinity at a *finite* value of x. I suppose you can think of a parabola as having something like an asymptote at infinity, but that description pretty much violates the requirements of the definition of an asymptote. And.. in order to go to infinity at a finite x, you have to be increasing "fast enough". (As I am sure you can appreciate, reaching infinity over finite x, is much more difficult than acheiving it in infinite x)

 

[HallsofIvy]

No, we often talk about "horizontal asymptotes". y= 1/x has y=0 as horizontal asymptote.
y= \frac{x^3- 1}{x+ 1} has the line y= x as an asymptote at infinity because for very large x, x³- 1 is approximately x³ and x+ 1 is approximately x.

 

[Skhandelwal]

Again, I still don't get "fast enough" I mean I know what you mean by it...but mathematically, what is the rate at which something decreases which makes it have an asymptote? Personally, I feel an acceleration will result in it failing the vertical line test and thus qualifying it as nonfunction. So how do we describe that rate? Is that some derivative of acceleration? If yes then which? This is all very confusing.

 

[codec9]

There is no certain rate. Read the thread, or at least the post above yours!

 

[MaWM]

Again, I still don't get "fast enough" I mean I know what you mean by it...but mathematically, what is the rate at which something decreases which makes it have an asymptote? Personally, I feel an acceleration will result in it failing the vertical line test and thus qualifying it as nonfunction. So how do we describe that rate? Is that some derivative of acceleration? If yes then which? This is all very confusing.

Nope, an increasing slope is not enough. It will still take infinitely many xs before the slope is vertical. Nor is having the slope increasing at an accelerting rate enough. In fact, even if *every* derivative of the function is increasing, that is still not fast enough. (functions like 2^x having increasing derivatives of all orders). Its hard to describe how fast a function like 1/x increases before it has an asymptote. Not only does the function itself become undefined at 0, but all of its derivatives do too. That is to say, not only does the magnitude of 1/x grow infinity fast near zero, but the magnitudes of all of its derivatives also grow infinitly fast. So, the kind growth required to have an asymptote is fast beyond the capability of being describable by any sort of derivative.

 

[Skhandelwal]

hmm...codec 9, For some reason, even though I understand the Math of Halls of Ivy, I am having a hard time understanding his point.(Damn that dumb head of mine!)

MaWM, I love your simplicity. However, you are missing a point. It can not really be described in terms of acceleration b/c if the x values were decreasing at a certain rate, they would start going back, and thus start repeating the x values they have already been in. So, if it is not an acceleration, and is beyond all its derivate like you said...how can this be described?(lets not bring summation formulas into this)

But is there another way?

 

[Hurkyl]

So, if it is not an acceleration, and is beyond all its derivate like you said...how can this be described?

As having an asymptote.