Is it correct to say that $\displaystyle f(x) = \frac{4x^2+4x+1}{x^3-x}$ has an oblique asymptote $\displaystyle f(x) = \frac{4}{x}$ and horizontal asymptotes $\displaystyle x = 0$, $\displaystyle x = 1$ and $\displaystyle x = -1$?

Printable View

- Jun 10th 2011, 05:41 PMResilientAsymptote
Is it correct to say that $\displaystyle f(x) = \frac{4x^2+4x+1}{x^3-x}$ has an oblique asymptote $\displaystyle f(x) = \frac{4}{x}$ and horizontal asymptotes $\displaystyle x = 0$, $\displaystyle x = 1$ and $\displaystyle x = -1$?

- Jun 10th 2011, 05:48 PMAckbeet
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 PMResilient
- Jun 10th 2011, 05:59 PMAckbeet
As an example: the function

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

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 PMResilient
- Jun 10th 2011, 10:58 PMpickslides
Why does it have a vertical asymptote at x = -1 and x = 0?

- Jun 10th 2011, 11:47 PMResilient
- Jun 11th 2011, 12:08 AMearboth
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

$\displaystyle y = x$