Asymptotes

Q: Hi Joel, could you run through the basics of asymptotes?

The fundamental idea of asymptotes is that it is a limit. For example, consider the function \displaystyle y=\frac{1}{x}. What does it tend to as x gets really, really large? The answer, of course, is 0, since 1 divided by a reaaaaaaaally huge number gives us a reaaaaaaaally small number, which approaches 0. We can graph this function this way:

asymptotes-1-1

Since y tends toward 0, we call y=0 our horizontal asymptote.

What about the y? How do we make that go to infinity? We do this by considering 1 divided by a reaaaaaaaally small number. At the threshold, we take 1 divided by 0 and get an undefined number. So to get to that extreme, we let the denominator equal 0. In this case,  x=0, which just so happens to be our vertical asymptote.

BUT, what about this function, \displaystyle y=\frac{1}{x-2}, which is the initial function but translated by 2 units in the positive xdirection? We ask the same questions: what happens when x gets really, really large and how can we make y really, really large?

When x gets really, really large, it follows that x-2 gets really, really large and y equals 1 divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches 0. Thus the horizontal asymptote remains as y=0.

How can we make y go to infinity? We do this by letting the denominator equal 0. In this case,  x-2=0 and  x=2. Hence, that is our asymptote.

asymptotes

BUT, what about this function, \displaystyle y=\frac{1}{x-2}+3, which is the second function but translated by 3 units in the positive ydirection? We ask the same questions: what happens when x gets really, really large and how can we make y really, really large?

When x gets really, really large, it follows that x-2 gets really, really large and \displaystyle \frac{1}{x-2} equals 1 divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches 0. Thus, overall\displaystyle y=\frac{1}{x-2}+3 tends towards \displaystyle y=3 since the \displaystyle \frac{1}{x-2} tends toward 0. Therefore \displaystyle y=3 is our horizontal asymptote.

How can we make y go to infinity? We do this by letting the denominator equal 0. In this case,  x-2=0 and  x=2. Hence, that is our asymptote.

asymptotes-1

BUT, what about this function, \displaystyle y=x+3+\frac{1}{x-2}, which is the third function plus x??? We ask the same questions: what happens when x gets really, really large and how can we make y really, really large?

When x gets really, really large, it follows that x-2 gets really, really large and \displaystyle \frac{1}{x-2} equals 1 divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches 0. Thus, overall\displaystyle y=x+3+\frac{1}{x-2} tends towards \displaystyle y=x+3 since the \displaystyle \frac{1}{x-2} tends toward 0. Therefore \displaystyle y=x+3 is our asymptote. It’s not horizontal, though. We call these slanted asymptotes oblique.

How can we make y go to infinity? We do this by letting the denominator equal 0. In this case,  x-2=0 and  x=2. Hence, that is our vertical asymptote.

asymptotes-2

BUT, what about this function, \displaystyle y=ax+b+\frac{p}{qx+r}, which is the generalised third function? Bonus marks if you can describe the sequence of transformations from the third function to this. When x gets really, really large, it follows that qx+r gets really, really large and \displaystyle \frac{p}{qx+r} equals p divided by a reaaaaaaaally huge number, which gives us a reaaaaaaaally small number, which approaches 0. Thus, overall, \displaystyle y=ax+b+\frac{p}{qx+r} tends towards \displaystyle y=ax+b since the \displaystyle \frac{p}{qx+r} tends toward 0. Therefore \displaystyle y=ax+b is our oblique asymptote.

How can we make y go to infinity? We do this by letting the denominator equal 0. In this case,  qx+r=0 and \displaystyle  x=-\frac{r}{q}. Hence, that is our vertical asymptote.

In sum, to find the asymptotes, we ask the two questions

  1. What happens when  x gets really, really large? (to obtain horizontal or oblique asymptotes)
  2. What happens when the denominator equals zero? (to obtain vertical asymptotes)

*Footnote: The last function assumes q \neq 0 as division by zero is undefined. Also, if q = 0, we actually get a linear function \displaystyle y=ax+b+\frac{p}{r}, now assuming that r \neq 0 as division by zero is undefined. Note also that if a = 0 then we get the initial examples without oblique asyptotes.

 

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Author: joelkindiak

Build people up. Point them to Christ.

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