Horizontal Asymptote of Rational Function

In summary, the conversation discusses finding the horizontal asymptote for the function f(x) = [sqrt{2x^2 - x + 10}]/(2x - 3). The participants mention that the top degree does not equal the bottom degree and if the top degree is greater than the bottom degree, the horizontal asymptote does not exist. They also mention rewriting the radical using a fractional degree, but it does not lead to a solution. One participant suggests using Desmos to find the horizontal asymptotes, while the other is not familiar with it. They eventually realize that there is a typo in the textbook's answer section and there are indeed two horizontal asymptotes: y = sqrt{2}/2 and y = -sqrt
  • #1
nycmathdad
74
0
Given f(x) = [sqrt{2x^2 - x + 10}]/(2x - 3), find the horizontal asymptote.

Top degree does not = bottom degree.

Top degree is not less than bottom degree.

If top degree > bottom degree, the horizontal asymptote DNE.

The problem for me is that 2x^2 lies within the radical. I can rewrite the radical using a fractional degree (2x^2 - x + 10)^(1/2) but leads no where, I think.
 
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  • #2
Beer soaked ramblings follow.
nycmathdad said:
Given f(x) = [sqrt{2x^2 - x + 10}]/(2x - 3), find the horizontal asymptote.

Top degree does not = bottom degree.

Top degree is not less than bottom degree.

If top degree > bottom degree, the horizontal asymptote DNE.

The problem for me is that 2x^2 lies within the radical. I can rewrite the radical using a fractional degree (2x^2 - x + 10)^(1/2) but leads no where, I think.
I see one vertical asymptote and one horizontal asymptote but your textbook's answer section says there are two horizontal asymptote. Probably a typo. (1.5.65)
Screenshot_20210402-070255_Adobe Acrobat.jpg

https://www.desmos.com/calculator/ggerdmqli0
 
Last edited:
  • #3
The horizontal asymptotes are given to be y = sqrt{2}/2 & y = -sqrt{2}/2. Which is correct?
 
  • #4
Beer soaked ramblings follow.
nycmathdad said:
The horizontal asymptotes are given to be y = sqrt{2}/2 & y = -sqrt{2}/2. Which is correct?
Click on the desmos link and find out for yourself.
 
  • #5
I cannot tell from the graph if the textbook has a typo or not.

sqrt{2}/2 = 0.7071067812

-sqrt{2}/2 = -0.7071067812

I don't see this on the graph.
 
  • #6
Beer soaked query follows.
nycmathdad said:
I cannot tell from the graph if the textbook has a typo or not.

sqrt{2}/2 = 0.7071067812

-sqrt{2}/2 = -0.7071067812

I don't see this on the graph.
Have you typed those alleged horizontal asymptotes on desmos?
 
  • #7
nycmathdad said:
Given f(x) = [sqrt{2x^2 - x + 10}]/(2x - 3), find the horizontal asymptote.

Top degree does not = bottom degree.

Top degree is not less than bottom degree.

If top degree > bottom degree, the horizontal asymptote DNE.

The problem for me is that 2x^2 lies within the radical. I can rewrite the radical using a fractional degree (2x^2 - x + 10)^(1/2) but leads no where, I think.
For very, very large x, "-x+ 10" negligible compared to "\(\displaystyle 2x^2\)". For very, very large x,\(\displaystyle \frac{\sqrt{2x^2- x+ 10}}{2x-3}\) is indistinguishable from \(\displaystyle \frac{\sqrt{2x^2}}{2x- 3}= \frac{\sqrt{2}x}{2x- 3}\). In the limit, it IS true that "top degree= bottom degree".

Another way to see this: divide both numerator and denominator of \(\displaystyle \frac{\sqrt{2x^2- x+ 10}}{2x- 3}\) by x. In the numerator the x goes into the square root as \(\displaystyle x^2\) so it becomes \(\displaystyle \frac{\sqrt{2- \frac{1}{x}+ \frac{10}{x^2}}}{2- \frac{3}{x}}\).

As x goes to infinity, the fractions with x or \(\displaystyle x^2\) in the denominator go to 0 so that, again, has limit \(\displaystyle \frac{\sqrt{2}}{2}\).
 
  • #8
Country Boy said:
For very, very large x, "-x+ 10" negligible compared to "\(\displaystyle 2x^2\)". For very, very large x,\(\displaystyle \frac{\sqrt{2x^2- x+ 10}}{2x-3}\) is indistinguishable from \(\displaystyle \frac{\sqrt{2x^2}}{2x- 3}= \frac{\sqrt{2}x}{2x- 3}\). In the limit, it IS true that "top degree= bottom degree".

Another way to see this: divide both numerator and denominator of \(\displaystyle \frac{\sqrt{2x^2- x+ 10}}{2x- 3}\) by x. In the numerator the x goes into the square root as \(\displaystyle x^2\) so it becomes \(\displaystyle \frac{\sqrt{2- \frac{1}{x}+ \frac{10}{x^2}}}{2- \frac{3}{x}}\).

As x goes to infinity, the fractions with x or \(\displaystyle x^2\) in the denominator go to 0 so that, again, has limit \(\displaystyle \frac{\sqrt{2}}{2}\).

I thought about dividing the top and bottom by x but wasn't sure if I could do the same on top considering the radical. I see that there's a typo in the textbook answer section, which is very common in math and other science books.
 
  • #9
Brandy induced realization follows.
jonah said:
...
I see one vertical asymptote and one horizontal asymptote but your textbook's answer section says there are two horizontal asymptote. Probably a typo. (1.5.65)
nycmathdad said:
... I see that there's a typo in the textbook answer section, ...
There's nothing like heavy lifting and brandy to make one see one's folly. There was no typo. There are indeed two horizontal asymptotes: $y=\frac{\sqrt{2}}{2}$ and $y=-\frac{\sqrt{2}}{2}$
https://www.desmos.com/calculator/0ewse3snch
 
  • #10
jonah said:
Brandy induced realization follows.There's nothing like heavy lifting and brandy to make one see one's folly. There was no typo. There are indeed two horizontal asymptotes: $y=\frac{\sqrt{2}}{2}$ and $y=-\frac{\sqrt{2}}{2}$
https://www.desmos.com/calculator/0ewse3snch

I'm not too familiar with desmos.
 
  • #11
Is the "heavy lifting" lifting the glass of brandy?
 
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  • #12
Beer soaked recommendation follows.
nycmathdad said:
I'm not too familiar with desmos.
I strongly suggest that you get yourself familiar with it and download the app on your phone. You'll thank me later.
Country Boy said:
Is the "heavy lifting" lifting the glass of brandy?
It does along with some regular barbell and dumbbell exercises, machine exercises, and some negative chins. Repetitive activities can often induce eureka moments.
 
  • #13
I think the brandy would be the limit of my exercises!
 
  • #14
Country Boy said:
I think the brandy would be the limit of my exercises!
Why is that?
I sure hope that you're not injured.
 
  • #15
Lifting a glass of brandy repeatedly tends to leave me in no condition to do exercises!
 
  • #16
Man, you had me worried.
Brandy is just something I occasionally do for those times when my brain needs an extra boost. Otherwise, beer is good enough.
By the way, I don't see you anymore at https://mathforums.com/math/
V8Archie is also MIA lately.
 
  • #17
jonah said:
Man, you had me worried.
Brandy is just something I occasionally do for those times when my brain needs an extra boost. Otherwise, beer is good enough.
By the way, I don't see you anymore at https://mathforums.com/math/
V8Archie is also MIA lately.

I haven't been on MHF for many years, got sick of the constant spam. Has it improved at all?
 
  • #18
Prove It said:
I haven't been on MHF for many years, got sick of the constant spam. Has it improved at all?
A little bit.
Just saw your post there.
 
  • #19
Prove It said:
I haven't been on MHF for many years, got sick of the constant spam. Has it improved at all?

It has improved. However, mo matter what username I select, as soon as they know it is me, the blocking game begins. What about FMH?
 
  • #20
Beer soaked ramblings follow.
nycmathdad said:
It has improved. However, mo matter what username I select, as soon as they know it is me, the blocking game begins. What about FMH?
Maybe it's because you deserved to be blocked because you keep posting political and religious stuff despite warnings not to do so. You know something is hot but you keep touching it anyway. It's like you never learn anything from experience at all. You even irritated Dan (Topsquark, the moderator) recently.
 
  • #21
I just noticed that nycmathdad just got banned again.
 

FAQ: Horizontal Asymptote of Rational Function

What is a horizontal asymptote?

A horizontal asymptote is a straight line that a graph approaches but never touches as the input values of a function increase or decrease without bound.

How do you find the horizontal asymptote of a rational function?

To find the horizontal asymptote of a rational function, you need to compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

Can a rational function have more than one horizontal asymptote?

Yes, a rational function can have up to two horizontal asymptotes. This occurs when the degree of the numerator is exactly one more than the degree of the denominator. In this case, the horizontal asymptotes are the two lines with equations y = a and y = b, where a and b are the ratios of the leading coefficients of the numerator and denominator, respectively.

How does the behavior of a rational function change when the horizontal asymptote is at y = 0?

If the horizontal asymptote is at y = 0, the function will approach the x-axis as the input values increase or decrease without bound. This means that the function will have a horizontal stretch or compression depending on the value of the leading coefficient of the denominator.

What is the significance of the horizontal asymptote in a rational function?

The horizontal asymptote can help us understand the end behavior of a rational function. It can also help us determine the domain and range of the function, as well as any horizontal stretches or compressions that may occur. Additionally, the horizontal asymptote can be used to graph the function and identify any discontinuities.

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