# Horizontal and vertical asymptotes

• September 17th 2009, 07:08 PM
Asuhuman18
Horizontal and vertical asymptotes
Find the vertical and horizonal asymtotes of the function:

(x-9)/sqrt(3x^2+4x+3)
• September 17th 2009, 07:31 PM
redsoxfan325
Quote:

Originally Posted by Asuhuman18
Find the vertical and horizonal asymtotes of the function:

$\frac{x-9}{\sqrt{3x^2+4x+3}}$

Vertical asymptotes occur when the denominator is zero. Horizontal asymptotes are found by taking the limits as $f(x)$ approaches positive and negative infinity.

Spoiler:
$3x^2+4x+3=0$ has no real solutions, so there are no vertical asymptotes.

Before I take the infinite limits, I'll do a little work on $f(x)$.

$\frac{x-9}{\sqrt{3x^2+4x+3}} = \frac{x-9}{\sqrt{3x^2(1+\frac{4}{3x}+\frac{1}{x^2})}}$ $= \frac{x(1-\frac{9}{x})}{\sqrt{3}\cdot|x|\cdot\sqrt{1+\frac{4 }{3x}+\frac{1}{x^2}}}=\frac{sign(x)}{\sqrt{3}}\cdo t\frac{1-\frac{9}{x}}{\sqrt{1+\frac{4}{3x}+\frac{1}{x^2}}}$, where $sign(x)$ is $1$ if $x$ is positive and $-1$ if $x$ is negative.

$\lim_{x\to\infty}\frac{sign(x)}{\sqrt{3}}\cdot\fra c{1-\frac{9}{x}}{\sqrt{1+\frac{4}{3x}+\frac{1}{x^2}}}= \frac{1}{\sqrt{3}}\cdot\frac{1-0}{\sqrt{1+0+0}} = \frac{1}{\sqrt{3}}$

$\lim_{x\to-\infty}\frac{sign(x)}{\sqrt{3}}\cdot\frac{1-\frac{9}{x}}{\sqrt{1+\frac{4}{3x}+\frac{1}{x^2}}}= \frac{-1}{\sqrt{3}}\cdot\frac{1-0}{\sqrt{1+0+0}} = \frac{-1}{\sqrt{3}}$

So horizontal asymptotes occur at $\boxed{y=\pm\frac{1}{\sqrt{3}}}$