# 009A Sample Final A, Problem 6

6. Find the vertical and horizontal asymptotes of the function  ${\displaystyle f(x)={\frac {\sqrt {4x^{2}+3}}{10x-20}}.}$

Foundations:
Vertical asymptotes occur whenever the denominator of a rational function goes to zero, and  it doesn't cancel from the numerator.
On the other hand, horizontal asymptotes represent the limit as ${\displaystyle x}$ goes to either positive or negative infinity.

Solution:

Vertical Asymptotes:
Setting the denominator to zero, we have
${\displaystyle 0=10x-20=10(x-2),}$
which has a root at ${\displaystyle x=2.}$ This is our vertical asymptote.
Horizontal Asymptotes:
More work is required here. Since we need to find the limits at ${\displaystyle \pm \infty }$, we can multiply our ${\displaystyle f(x)}$ by

${\displaystyle {\frac {\sqrt {\frac {1}{x^{2}}}}{\,\,\,{\frac {1}{x}}}}.}$

This expression is equal to ${\displaystyle 1}$ for positive values of ${\displaystyle x}$, and is equal to ${\displaystyle -1}$ for negative values of ${\displaystyle x}$. Since multiplying ${\displaystyle f(x)}$ by an expression equal to ${\displaystyle 1}$ doesn't change the limit, we will add a negative sign to our fraction when considering the limit as ${\displaystyle x}$ goes to ${\displaystyle -\infty }$. Thus,

${\displaystyle \lim _{x\rightarrow \pm \infty }{\frac {\sqrt {4x^{2}+3}}{10x-20}}\,\,\cdot \,\,\pm {\frac {\sqrt {\frac {1}{x^{2}}}}{\,\,\,{\frac {1}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {{\frac {4x^{2}}{x^{2}}}+{\frac {3}{x^{2}}}}}{{\frac {10x}{x}}-{\frac {20}{x}}}}=\lim _{x\rightarrow \pm \infty }\pm {\frac {\sqrt {4+{\frac {3}{x^{2}}}}}{10-{\frac {20}{x}}}}=\pm {\frac {2}{10}}=\pm {\frac {1}{5}}}$

Thus, we have a horizontal asymptote at ${\displaystyle y=-1/5}$ on the left (as ${\displaystyle x}$ goes to ${\displaystyle -\infty }$), and a horizontal asymptote at ${\displaystyle y=1/5}$ on the right (as ${\displaystyle x}$ goes to ${\displaystyle +\infty }$).