SOLUTION: oblique asymptote: If a given rational function of higher degree (say 6) does not have a horizontal asymptote, to check for the oblique asymptote... ...Do you just keep doing

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Question 1163607: oblique asymptote:
If a given rational function of higher degree (say 6) does not have a horizontal asymptote, to check for the oblique asymptote...
...Do you just keep doing polymomial division repeatedly till you come down to the quotient that is of form mx+b ?
Thank you
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

A rational function is a ratio of two polynomials


    f(x) = p%28x%29%2Fq%28x%29.


Oblique asymptote is a straight line of the form  y = ax + b.


Oblique asymptote arises when the degree of the polynomial  p(x)  in the numerator is 1 unit greater 
than the degree of the polynomial  q(x)  in the denominator.


To get the equation of the oblique asymptote under appropriate condition, you divide the polynomial in the numerator
by the polynomial in the denominator (long division).


Then you get a linear binomial function as a quotient, and it is the desired equation of the oblique asymptote.


Example:


    f(x) = %28x%5E6+-+2%29%2F%28x%5E5+%2B+x%5E4+%2B+x%5E3+%2B+x%5E2+%2B+x+%2B+1%29 


The numerator is  p(x) = x%5E6+-+2%29.


The denominator is  q(x) = %28x%5E5+%2B+x%5E4+%2B+x%5E3+%2B+x%5E2+%2B+x+%2B+1%29.


p%28x%29%2Fq%28x%29 = x - 1 - 1%2F%28x%5E5+%2B+x%5E4+%2B+x%5E3+%2B+x%5E2+%2B+x+%2B+1%29.


The quotient is  x-1,  and  the equation of the oblique asymptote is  y = x-1  in this case.


See the plot below.


    


    Plot  y = %28x%5E6+-+2%29%2F%28x%5E5+%2B+x%5E4+%2B+x%5E3+%2B+x%5E2+%2B+x+%2B+1%29 (the given rational function, red) and y = x-1 (oblique asymptote, green)



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Comment from student: What if the p(x) is degree 8 & q( x) is of degree 3
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My response :


        Then there is  NO  oblique asymptote.

        There is a   "curvilinear asymptote",  instead.


See this Wikipedia article

https://en.wikipedia.org/wiki/Asymptote#Oblique_asymptotes


/\/\/\/\/\/\/

It is good that you ask questions.   I like it . . .


Do not forget to post your  "THANKS"  for my teaching.


/ / / / / / / / / / / /

I got next your "comment" post with several other questions.

The answers to all these questions are in that Wikipedia article I referred to you.

You will find answers to all your questions in that article.

I don't want to re-tell it here, in my post.

Wikipedia makes it better and in more authoritative way . . .


Happy learning (!)



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There is an oblique (linear) asymptote only if the degree of the numerator is 1 more than the degree of the denominator.

So there is never any need to "keep doing polynomial division" to find an asymptote. If there is an oblique asymptote, a single division will find it.

Example 1: %28x%5E2%2B2x%29%2F%28x%2B1%29+=+%28x%2B1%29-1%2F%28x%2B1%29

A graph of the rational function and the oblique asymptote, y=x+1:

graph%28400%2C400%2C-5%2C5%2C-10%2C10%2C%28x%5E2%2B2x%29%2F%28x%2B1%29%2Cx%2B1%29

Example 2: %28x%5E3%2B3x%5E2%2B3x%29%2F%28x%5E2%2B2x%2B1%29+=+%28x%2B1%29-1%2F%28x%5E2%2B2x%2B1%29

A graph of the rational function and the oblique asymptote, y=x+1:



If the degree of the numerator is more than 1 greater than the degree of the denominator, then you get an asymptote which is not oblique (linear).

For example, if the degree of the numerator is 2 more than the degree of the denominator, the asymptote is degree 2 (quadratic -- a parabola).

Example 3: %28x%5E3%2B3x%5E2%2B3x%29%2F%28x%2B1%29+=+%28x%5E2%2B2x%2B1%29-1%2F%28x%2B1%29

A graph of the rational function and the quadratic asymptote:



In general, if the degree of the numerator is n more than the degree of the denominator, the asymptote will be of degree n.

Example 4: numerator degree 4, denominator degree 1; asymptote degree 3:

A graph....