Thread: Help trying to find horizontal Asymptotes.

[itex]y=\frac{x}{x^2+5x+4x}[/itex]

I am stuck on finding the horizontal asymptotes. I know for sure that there is one at y=0. I have graphed this on my calculator and it seems that there is another between 0 and 1.
Am i on the right track?
I did take the inverse of this function and solved it and it came up with 3 answers.

Originally Posted by johnsy123

$\displaystyle y=\frac{x}{x^2+5x+4x}$

I am stuck on finding the horizontal asymptotes. I know for sure that there is one at y=0. I have graphed this on my calculator and it seems that there is another between 0 and 1.
Am i on the right track?
I did take the inverse of this function and solved it and it came up with 3 answers.

Is that x after the 4 in the denominator correct? Is this actually

$y=\dfrac{x}{x^2+9x}$ or do you mean $y=\dfrac{x}{x^2+5x+4}$

Originally Posted by romsek

Is that x after the 4 in the denominator correct? Is this actually

$y=\dfrac{x}{x^2+9x}$ or do you mean $y=\dfrac{x}{x^2+5x+4}$

Hey, yes the one on the right is correct......x/(x^2+5x+4).
Apologies for the poorly inputed text.

Originally Posted by johnsy123

Hey, yes the one on the right is correct......x/(x^2+5x+4).
Apologies for the poorly inputed text.

You find horizontal asymptotes by letting $x \to \pm \infty$. In this case 0 is the only horizontal asymptote.

This is seen by noting that as $x \to \pm \infty$ the $x^2$ term in the denominator overwhelms the others and you are left with $\dfrac{x}{x^2}=\dfrac{1}{x}$

Clearly $\dfrac{1}{x} \to 0$ as $x \to \pm \infty$

Do you also need to find the vertical asymptotes?