# Asymptote

• Jun 10th 2011, 05:41 PM
Resilient
Asymptote
Is it correct to say that $f(x) = \frac{4x^2+4x+1}{x^3-x}$ has an oblique asymptote $f(x) = \frac{4}{x}$ and horizontal asymptotes $x = 0$, $x = 1$ and $x = -1$?
• Jun 10th 2011, 05:48 PM
Ackbeet
The degree of the numerator must be one greater than the degree of the denominator in order to have a slant asymptote. In addition, 4/x is not a straight line. The horizontal asymptotes are all y = 0, because the degree of the denominator is greater than the degree of the numerator.
• Jun 10th 2011, 05:53 PM
Resilient
Quote:

Originally Posted by Ackbeet
The degree of the numerator must be one greater than the degree of the denominator in order to have a slant asymptote.

Can you explain this condition bit more, please?
• Jun 10th 2011, 05:59 PM
Ackbeet
As an example: the function

$\frac{x^{2}-1}{x}=x-\frac{1}{x}$

has a slant asymptote of x, because as x gets very large, the 1/x is very small in comparison to the x, and hence becomes negligible. See here for a more full explanation.
• Jun 10th 2011, 06:35 PM
Resilient
Quote:

Originally Posted by Ackbeet
As an example: the function

$\frac{x^{2}-1}{x}=x-\frac{1}{x}$

has a slant asymptote of x, because as x gets very large, the 1/x is very small in comparison to the x, and hence becomes negligible. See here for a more full explanation.

okay, thank you. just one more question, so that I get this.

what can we say about the asymptotes of $f(x) = \frac{x^4+1}{x^3-1}$? I see that it has vertical asymptote at x = 1, x = -1 and x = 0, but I'm not sure what else
• Jun 10th 2011, 10:58 PM
pickslides
Why does it have a vertical asymptote at x = -1 and x = 0?
• Jun 10th 2011, 11:47 PM
Resilient
Quote:

Originally Posted by pickslides
Why does it have a vertical asymptote at x = -1 and x = 0?

Sorry, I meant

$f(x) = \frac{x^4+1}{x^3-x}$
• Jun 11th 2011, 12:08 AM
earboth
Quote:

Originally Posted by Resilient
...

what can we say about the asymptotes of $f(x) = \frac{x^4+1}{x^3-x}$? I see that it has vertical asymptote at x = 1, x = -1 and x = 0, but I'm not sure what else

Quote:

Originally Posted by Resilient
Sorry, I meant

$f(x) = \frac{x^4+1}{x^3-x}$

1. As you stated correctly the vertical asymptotes correspond to the zeros of the denominator.

2. To get the horizontal or slanted asymptotes transform the term of the function into a complete polynomial and a fraction:
Code:

                                      x^2 + 1  (x^4        +  1) ÷ (x^3 - x) = x + ----------- -(x^4 - x^2)                          x^3 - x ------------         x^2 + 1
The none-fractional part of the term is the term of the slanted asymptote. That means in your case the slanted asymptote has the equation:

$y = x$