Horizontal Asymptote

[Blitze105]

Hello,
I have this problem:

f(x)= x^2-1 / x

Can some one show me how to find the horizontal asymptote on a problem like this? i got y= 1, and y = -1 but that isn't correct

 

[mmm4444bot]

f(x)= x^2-1 / x

Hello Blitze:

Since you typed spaces around the slash, I'm going to assume that your ambiguous expression is intended as follows.

f(x) = (x^2 - 1)/x

If my assumption is correct, then who told you that the graph of this function has an horizontal asymptote?

It does not. It has a slant asymptote.

When we have rational functions (i.e., both the numerator and denominator are polynomials), we can use the degrees of these polynomials to determine whether or not there is a horizontal or slant asymptote.

Let n be the degree of the polynomial in the numerator.

Let d be the degree of the polynomial in the denominator.

If n < d, then y = 0 is a horizontal asymptote.

If n = d, then y = a/b is a horizontal asymptote, where a is the leading coefficient in the numerator and b is the leading coefficient in the denominator.*

If n > d, then there is no horizontal asymptote.

If n is one greater than d, then there is a slant asymptote.

In your exercise, another way to see the slant asymptote is to rewrite the ratio as a difference of fractions.

f(x) = x^2/x - 1/x

Simplify the right-hand side, and then let x go to both plus infinity and minus infinity to see what happens to the value of f.

Horizontal and slant asymptotes both arise as x becomes very large in absolute value.

When working with rational functions involving more complicated expressions, we sometimes divide numerator and denominator by the highest power of x (as long as we know it's not zero). In other situations, polynomial division (either longhand or synthetic) is a good choice.

Please ask any questions if you are still unsure about this.

Cheers,

~ Mark

* Thanks to wjm11 for the error alert