Asymptotes

[KennyD94]

Hi, i'm looking to get a function that expresses these characteristics and i'm not sure what approach to take
When x = 90 , y = 95
As x -> infinity , y -> 100 but never gets to 100, however y should be 99.9 when x = 100

What function could be used that has similar characteristics or what approach could I take to find this function? Thank you!

 

[JeffM]

One class of functions that you probably have worked with before and may have those chacteristics is a subset of rational functions.

With that as a hint, what can you do?

If you cannot get to a final answer, please tell us what you tried or thought about.

 

[Dr.Peterson]

Another possibility would be an exponential function.

 

[HallsofIvy]

Specifically, \(\displaystyle y= e^{-x}\) has y= 0 as asymptote. So both \(\displaystyle y= 100+ Ae^{-x}\) and \(\displaystyle y= 100- Ae^{-x}\) have y= 100 as asymptote for any number A. Find A so that y(90)= 99.

Also \(\displaystyle y= \frac{1}{x}\) also has y= 0 as asymptote and you can do the same thing.

 

[KennyD94]

One class of functions that you probably have worked with before and may have those chacteristics is a subset of rational functions.

With that as a hint, what can you do?

If you cannot get to a final answer, please tell us what you tried or thought about.

Hi Jeff, so for rational functions what I have looked at so far is: the coefficient of the highest term in the numerator should be 100 times that of the coefficient of the highest term in the denominator and for it to be a horizontal asymtote, the polynomials should be of the same degree eg. y = 100(x)^2/x^2 this will give me a horizontal asymptote at y = 100, however i'm unsure of how to get what the additional polynomials should be in the equation to achieve x = 90 when y = 95 and when x = 100 y = 99.9

 

[KennyD94]

Specifically, \(\displaystyle y= e^{-x}\) has y= 0 as asymptote. So both \(\displaystyle y= 100+ Ae^{-x}\) and \(\displaystyle y= 100- Ae^{-x}\) have y= 100 as asymptote for any number A. Find A so that y(90)= 99.

Also \(\displaystyle y= \frac{1}{x}\) also has y= 0 as asymptote and you can do the same thing.

Ah, think I have it, so:
\(\displaystyle y= 100- Ae^{-x}\)
\(\displaystyle 95= 100- Ae^{-90}\)
\(\displaystyle 5= Ae^{-x}\)
\(\displaystyle ln5= lnA+ lne^{-90}\)
\(\displaystyle lnA = 91.6094...\)
\(\displaystyle A = e^{91.6094}\)
\(\displaystyle A = 6.101e+39\)
\(\displaystyle y(90) = 95\)
\(\displaystyle y(100) = 99.99999\)

Skipped a few steps for the calculations, but looks like it worked out

 

[KennyD94]

Ah, think I have it, so:
\(\displaystyle y= 100- Ae^{-x}\)
\(\displaystyle 95= 100- Ae^{-90}\)
\(\displaystyle 5= Ae^{-x}\)
\(\displaystyle ln5= lnA+ lne^{-90}\)
\(\displaystyle lnA = 91.6094...\)
\(\displaystyle A = e^{91.6094}\)
\(\displaystyle A = 6.101e+39\)
\(\displaystyle y(90) = 95\)
\(\displaystyle y(100) = 99.99999\)

Skipped a few steps for the calculations, but looks like it worked out

So, I have a follow on question from this, if my formula is
\(\displaystyle y= 100- (6.101E+39) e^{-x}\)
\(\displaystyle y(90) = 95.00\)
\(\displaystyle y(91) = 98.16\)
\(\displaystyle y(92) = 99.34\)
\(\displaystyle y(93) = 99.91\)
\(\displaystyle y(94) = 99.99\)
.
.
.
\(\displaystyle y(100) = 99.99\)

how do I change the formula to get a slower ramp towards 99.99 as x -> 100
example
\(\displaystyle y(90) = 95.00\)
\(\displaystyle y(95) = 97.50\)
\(\displaystyle y(100) = 99.99\)