Finding oblique asymptotes

  • Thread starter benji
  • Start date
48
0
I'm having trouble figuring the oblique asymptote for this problem...

I haven't ever done one of these before so I really don't know where to start.

Here's the problem:

y=(2x^2+5x+11)/(x+1)
 

Pyrrhus

Homework Helper
2,160
1
http://home.att.net/~srschmitt/precalc/precalc-12-03-00.html [Broken]
 
Last edited by a moderator:
48
0
Hmm... I'm still not quite getting it.

I need to factor the numerator then, correct? But it comes out as an imaginary number?
 

Pyrrhus

Homework Helper
2,160
1
You only need to do the polynomial division, so you can write it in the form of Quotient + Remainder/Divisor, because of Euclides Algorithm Numerator = Quotient*Denominator + Remainder.

The Oblique Asymptote will be the quotient.
 

HallsofIvy

Science Advisor
Homework Helper
41,682
864
The whole point of oblique (or horizontal) asymptotes is "what happens as x goes to infinity or negative infinity?"

With something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] the simplest thing to do is to divide both numerator and denominator by the largest power of x in the denominator- here just x. [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]. Since those fractions with x in the denominator will go to 0 as x goes to either positive or negative infinity, it's easy to see that y approaches 2x+ 5.
 

Galileo

Science Advisor
Homework Helper
1,989
6
I had an oblique asymptote question on my Calc I exam and I totally forgot how to do that. What I eventually did was to get an equation for the tangent at the point (a,f(a)) and then let a approach infinity, figuring that the tangent will coincide with the asymptote in that limit. Although I felt pretty good after coming up with that idea at the time and the answer was correct, I still received a big red cross through my solution :(
 
Just pointing out something...I know it's a really old thread !


Someone pointed out earlier that for something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] you can just divide by "x" and get that the asymptote is: [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]

However, that is not the case I believe. The asymptote is actually 2x+3, and it is found by:

= Lim x -> infinity, f(x) - (mx+b) = 0

So I would do long division, get : 2x + 3 + 8/(x+1) and then use the formula mentioned above:

= lim x-> infinity of 2x + 3 + 8/(x+1) - (2x+3)

that's equal = 0, hence the mx+b part is an oblique asymptote. Which is, 2x+3 = oblique asymptote in this case.

Hope this helps!
 
Hi, dale 123....you can check your answer by graphing it. Use a program if you're having trouble with the graph. Once it's graphed you'll see that your answer is not right. The correct answer is still 2x+5.
 
The whole point of oblique (or horizontal) asymptotes is "what happens as x goes to infinity or negative infinity?"

With something like [tex]y= \frac{2x^2+5x+11}{x+1}[/tex] the simplest thing to do is to divide both numerator and denominator by the largest power of x in the denominator- here just x. [tex]y= \frac{2x+ 5+ \frac{11}{x}}{1+ \frac{1}{x}}[/tex]. Since those fractions with x in the denominator will go to 0 as x goes to either positive or negative infinity, it's easy to see that y approaches 2x+ 5.
Hi, dale 123....you can check your answer by graphing it. Use a program if you're having trouble with the graph. Once it's graphed you'll see that your answer is not right. The correct answer is still 2x+5.
You're both wrong, dale 123 is correct. To find the oblique asymptote, you must use polynomial long division, and then analyze the function as it approaches infinity.

Taking the limit first, like HallsofIvy did, is wrong because 11/x and 1/x approach infinity at different rates, and therefore add to the numerator and denominator in slightly different ways.

If you are not finding the limit at infinity of the entire function, as in this problem, then you MUST write it in the form "term + term + ... + term", and analyze each individual term as x approaches infinity. This is because fractions can do unexpected things when it comes to limits.
 

HallsofIvy

Science Advisor
Homework Helper
41,682
864
Oh, I am so embarassed! When I read my post I thought "wait, that's not right, you divide the polynomial to find the asymptote!". Don't know what I was thinking 5 and a half years ago. Thanks, dale123 and erh5060, for fixing that.
 
867
0
So dividing by the largest power of x works for horizontal asymptotes but not generally for slant asymptotes. That's good to know.


Forgive me; you mean years ago. :wink:
 

HallsofIvy

Science Advisor
Homework Helper
41,682
864
Ahh, yes, of course! I didn't make a mistake- what I wrote clearly must have been true 4½ years ago but not now! :rofl:
 
There is also a nice way to find oblique assymptotes for any function in general, not only rational functions, since the method described here works only for rational functions.

Let f(x) be sucha function and assume that L=ax+b is the oblique assymptote. Let P be any point in the graph of f and Q be a point in the line L, such that d(P,Q) is the shortest distance. Also let denote by K the point on the line L when we draw a perpendicular segment from point P to the x-axis. Now, after a little bit of geometry we find that L is an oblique assymptote iff

[tex]\lim_{x\rightarrow \infty}(f(x)-L)=lim_{x\rightarrow \infty}(f(x)-ax-b)=0[/tex]

If we divide the whole thing by x, and take the limit we find that:

[tex]a=lim_{x\rightarrow\infty}\frac{f(x)}{x}[/tex]

and

[tex]b=\lim_{x\rightarrow\infty}(f(x)-ax)[/tex]
 

Related Threads for: Finding oblique asymptotes

  • Posted
Replies
2
Views
3K
  • Posted
Replies
3
Views
1K
  • Posted
Replies
10
Views
3K
Replies
1
Views
3K
Replies
3
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top