1. ## Slant Asymptotes

Can someone explain slant asymptotes as simple as you can manage, perhaps with an example?

All I got out of my text for solving is to use long division on the functions which confuses me because they didn't show any of the middle steps and I've never done that before.

2. for an irreducible, rational polynomial function, if the degree of the numerator is one greater than the degree of the denominator, then the graph of the rational function will have a slant asymptote.

for example ...

$\displaystyle y = \frac{x^3 - x^2 + 1}{x^2 + 1}$

... note the polynomial is irreducible and the degree of the numerator is one greater than the degree of the denominator.

perform the long division.

you should get the quotient $x - 1$ with a remainder.

the slant asymptote is $y = x-1$

3. Originally Posted by skeeter
for an irreducible, rational polynomial function, if the degree of the numerator is one greater than the degree of the denominator, then the graph of the rational function will have a slant asymptote.

for example ...

$\displaystyle y = \frac{x^3 - x^2 + 1}{x^2 + 1}$

... note the polynomial is irreducible and the degree of the numerator is one greater than the degree of the denominator.

perform the long division.

you should get the quotient $x - 1$ with a remainder.

the slant asymptote is $y = x-1$
Specifically, you get $x- 1$ with remainder $-x+ 2$ which means that
$\frac{x^3- x^2+ 1}{x^2+ 1}= x- 1+ \frac{-x+ 2}{x^2+ 1}$

Since the denominator of that last fraction has higher degree than the numerator, as x goes to infinity, the fraction goes to 0 so the graph of $\frac{x^3- x^2+ 1}{x^2+ 1}$ gets closer and closer to the graph of y= x- 1, a "slant" line.

4. Originally Posted by skeeter
for an irreducible, rational polynomial function, if the degree of the numerator is one greater than the degree of the denominator, then the graph of the rational function will have a slant asymptote.

for example ...

$\displaystyle y = \frac{x^3 - x^2 + 1}{x^2 + 1}$

... note the polynomial is irreducible and the degree of the numerator is one greater than the degree of the denominator.

perform the long division.

you should get the quotient $x - 1$ with a remainder.

the slant asymptote is $y = x-1$
I understood that from the text. My real problem is that I don't know how to do long division with functions. The textbook skipped the middle steps and my prof rushed through it in class.

5. Originally Posted by ImaCowOK
I understood that from the text. My real problem is that I don't know how to do long division with functions. The textbook skipped the middle steps and my prof rushed through it in class.
most profs assume their calculus students know how to perform long division.

Polynomial Long Division

if the above link fails to provide a satisfactory explanation, then I recommend you google the subject polynomial long division ... you will use it again in your study of the calculus.

next time, please be specific about where you are having a problem.

6. Originally Posted by skeeter
most profs assume their calculus students know how to perform long division.