# Thread: Stuck on certain Oblique asymptote

1. ## Stuck on certain Oblique asymptote

Hello there,

I am currently stuck on a problem to find the oblique asymptote with long division. The equation is f(x) = 4x/ (x^2 - 4x-5) . Now I using long division you cannot get an answer because the degree of the divisor is greater than the degree of the dividend. Would that mean that there is just a remainder of 4x OR 4x/(x^2-4x-5). That would basically mean that the oblique asymptote of this equation is 4x, right?

Or am I going in the wrong direction? Thanks

Sean

2. Hi

The long division of 4x by (x²-4x-5) gives 0, which means that the asymptote is the x-axis

3. Hello,

This function f(x) = 4x/ (x^2 - 4x-5) doesn't have an oblique asymptote .

In order for a function like f(x)=P(x)/Q(x) to have a oblique asymptote is a must to have the following relation:

degree of P(x)= degree of Q(x) + 1;

Have a nice day !

4. Thank you Hush_Hush and running-gag
I had that in mind but was debating whehter that was true for oblique asymptotes or not.

Thanks,

Sean

5. A rational function's graph can have an horizontal asymptote (such as the x-axis) or an oblique ("slant") asymptote, but it will not have both! (And, if the degree of the numerator is larger than the degree of the denominator by 2 or more, it won't have a slant asymptote, either.)

To learn how this works, try here.