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Horizontal Asymptote of Rational Function

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nycmathdad

Member
Mar 21, 2021
75
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.
 

jonah

Member
Feb 21, 2015
85
Beer soaked ramblings follow.
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
 
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nycmathdad

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Mar 21, 2021
75
The horizontal asymptotes are given to be y = sqrt{2}/2 & y = -sqrt{2}/2. Which is correct?
 

jonah

Member
Feb 21, 2015
85
Beer soaked ramblings follow.
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.
 
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nycmathdad

Member
Mar 21, 2021
75
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.
 

jonah

Member
Feb 21, 2015
85
Beer soaked query follows.
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?
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
699
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 \(\displaystyle \frac{\sqrt{2}}{2}\).
 
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nycmathdad

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Mar 21, 2021
75
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.
 

jonah

Member
Feb 21, 2015
85
Brandy induced realization follows.
...
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)
... 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 $y=-\frac{\sqrt{2}}{2}$
 

nycmathdad

Member
Mar 21, 2021
75
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 $y=-\frac{\sqrt{2}}{2}$
I'm not too familiar with desmos.
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
699
Is the "heavy lifting" lifting the glass of brandy?
 
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jonah

Member
Feb 21, 2015
85
Beer soaked recommendation follows.
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.
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.
 

Country Boy

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MHB Math Helper
Jan 30, 2018
699
I think the brandy would be the limit of my exercises!
 

jonah

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Feb 21, 2015
85

Country Boy

Well-known member
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Jan 30, 2018
699
Lifting a glass of brandy repeatedly tends to leave me in no condition to do exercises!
 

jonah

Member
Feb 21, 2015
85
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.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,429
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?
 

jonah

Member
Feb 21, 2015
85
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.
 

nycmathdad

Member
Mar 21, 2021
75
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?
 

jonah

Member
Feb 21, 2015
85
Beer soaked ramblings follow.
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.
 

jonah

Member
Feb 21, 2015
85
I just noticed that nycmathdad just got banned again.