# Thread: Horizontal and vertical asymptotes

1. ## Horizontal and vertical asymptotes

Find the vertical and horizonal asymtotes of the function:

(x-9)/sqrt(3x^2+4x+3)

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

$\displaystyle \frac{x-9}{\sqrt{3x^2+4x+3}}$
Vertical asymptotes occur when the denominator is zero. Horizontal asymptotes are found by taking the limits as $\displaystyle f(x)$ approaches positive and negative infinity.

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

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

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

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

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

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