# Homework Help: Horizontal asymtotes

1. Apr 24, 2010

### thereddevils

When the power of the leading coefficient of the numerator of a rational function is a lot greater than the power of the leading coefficient of the denominator , ie

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

The horizontal asymtote is y=x^2 according to the book . Is that true ? Is there any proof for this . I only know when its one power difference , that would be oblique asymtote .

2. Apr 24, 2010

### cragar

I think that is right , look at the highest power on top and the highest power on the bottom and divide them .

3. Apr 24, 2010

### HallsofIvy

I really doubt that your text book says horizontal asymptote. A horizontal asymptote is a straight horizontal line and only occurs in a rational function when the numerator and denominator have the same degree. A "slant" or "skew" asymptote, a non-horizontal line, occurs when the numerator has degree one higher than the denominator.

I've never seen a curve called an "asymptote" before but since
$$\frac{x^3}{x- 1}= x^2+ x+ 1+ \frac{2}{x-1}$$
so for large x, the graph comes close to $x^2+ x+ 1$ which itself will be close to $x^2$. (Though I would say the "asymptote" was $x^2+ x+ 1$, not just $x^2$.)

4. Apr 24, 2010

### thereddevils

thank you . It's not horizontal asymtote , i just didn't know what to call it , is there a name for that kind of asymtote ?

Is it necessary to include this asymtote in my sketching ? Usually , i would just include the vertical asymtote in my sketching if there is no horizontal asym.

5. Apr 24, 2010

### cragar

I think you might be able to call it a slant asymptote.