# Thread: Horizontal asymptotes

1. ## Horizontal asymptotes

So Im having a hard time remembering the rules for them...

In these equations what would be the horizontal asymptotes?

F(x)= x^4+6/x^2-4

F(x)= 8+3x/1-x^2

Since the equation #1 has higher exponent in the numerator is the horizontal asymptote equal to 0? what about the 2nd one? thanks

oh and one more thing. the vertical asymptotes are 2, -2 for the first equation, and -1, 1 for the 2nd one, right ?

2. You need to be more careful with parentheses: x^4+(6/x^2)-4 is very different from x^4+6/(x^2-4), which is very different from (x^4+6)/(x^2-4). Technically, you wrote the first. However, since that function does not have a horizontal asymptote, I conclude you didn't mean that.

3. Ok im going to type it like id type it in my calculator sorry about that:

y= (x^4+6)/(x^2-4)

y= (8+3x)/(1-x^2)

4. Ah, thanks. That makes more sense.

Equation 1 has, as you say, a higher power in the numerator. That usually means there is no horizontal asymptote. If you think about it in a back-of-the-envelope sort of way, the numerator is going to "beat" the denominator and get big quicker. In fact, it's going to get bigger like $x^{2}$, since the power of the numerator is to the fourth, as opposed to the denominator's second power. Therefore, there are no horizontal asymptotes.

Equation 2 gets bigger (in magnitude) in the denominator than in the numerator. Therefore, it's going to have a horizontal asymptote. How do you find it? Well, you want to take the limits as $x\to\pm\infty.$ Do you remember how to do those?

5. Yes I remember. thanks for the help!

6. You're very welcome. Have a good one!

7. Another way to think about it, just a tiny bit less "back of the envelope" than Ackbeet's, is to divide both numerator and denominator by the highest power of x.

$\frac{x^4+ 6}{x^2- 4}= \frac{1+ \frac{6}{x^4}}{\frac{1}{x^2}- \frac{4}{x^2}}$
where I have divided both numerator and denominator by $x^4$. Now, as x goes to either plus or minus infinity, which is the whole point of a "horizontal asymptote", the fractions with an "x" in the denominator go to 0. Here, the entire denominator becomes 0 so there is result- there is NO horizontal asymptote.

$\frac{8+ 3x}{1- x^2}= \frac{\frac{8}{x^2}+ \frac{3}{x}}{\frac{1}{x^2}- 1}$ where I have divided both numerator and denominator by $x^2$. Again, each of the fractions with x in the denominator goes to 0 but now it is the numerator of the entire fraction that goes to 0 while the denominator does not- the horizontal aymptote is 0.

You can see from that that: If the numerator has higher degree than the denominator, there is no horizontal asymptote, if the denominator has higher degree than the numerator, the horizontal asymptote, and if they numerator and denominator have the same degree, the horizontal asymptote is the ratio of the coefficients of those higher powers.

8. Incidentally, if the power of the numerator is one greater than the power of the denominator, you won't get horizontal asymptotes. But you can get what are called slant or oblique asymptotes, where the function approaches the equation of a straight line.