finding a horizontal asymptote

[dalasTR]

Ok so I know a few rules, like if the degree in the numerator is one higher than the denominator there is a slant, if the denom is bigger than the numerator it is at y = 0 and if the numerator is bigger than the denom by more than one, there isn't one. But how do I do this one?

f(x) = 4x / sq. rt. (x^2 + 9)

Note: sorry my equation editor is not functioning properly and therefore i couldnt use the proper notation.

any help is greatly appreciated

 

[tkhunny]

It has very similar properties as those with which you are familiar already. Under that radical is a quadratic expression. The square root of a quadratic behaves somewhat like a linear - certainly it an asymptotic sense, In this case, off the 4 in the positive direction and to -4 in the negative direction. Just keep your eyes on different Domain restrictions. For example \(\displaystyle \sqrt{x^{2} - 9}\) would give quite a different result from \(\displaystyle \sqrt{x^{2} + 9}\).

 

[dalasTR]

I'm not quite sure I understand :/

I mean I kind of do but I am not sure how to apply that towards finding the H.A.

 

[dalasTR]

I am thinking it may be y = pos AND neg 4, but I don't know if that is right, or even if it is, why? Haha. I just know that I was wrog when I said y = 0 so +- 4 was my next choice...

 

[tkhunny]

This is the entire theoretical basis for my comments.

\(\displaystyle \sqrt{x^2} = |x|\)

Having said that, out in the nether regions, far away from the Origin, does \(\displaystyle \sqrt{x^{2}+9}\) behave more like \(\displaystyle x^{2}\) or more like \(\displaystyle x\)?

Think on it.