Answer
There is no vertical asymptote.
$y=\displaystyle\frac{1}{2}$ is horizontal asymptote.
Work Step by Step
\[y=\frac{x-9}{\sqrt{4x^2+3x+2}}\]
For vertical asymptotes denominator should be zero.
\[\sqrt{4x^2+3x+2}=0\]
\[\Rightarrow 4x^2+3x+2=0\]
Here $a=4, b=3$ and $c=2$
\[\Rightarrow D=b^2-4ac=9-4(4)(2)=-23<0\]
Therefore there is no real root so there is no vertical asymptotes.
Let \[L=\lim_{x\rightarrow\infty}\displaystyle\frac{x-9}{\sqrt{4x^2+3x+2}}\]
\[L=\lim_{x\rightarrow\infty}\frac{\displaystyle\frac{x-9}{x}}{\displaystyle\frac{\sqrt{4x^2+3x+2}}{x}}\]
\[L=\lim_{x\rightarrow\infty}\displaystyle\frac{1-\displaystyle\frac{9}{x}}{\displaystyle\sqrt{4+\displaystyle\frac{3}{x}+\displaystyle\frac{2}{x^2}}}\]
\[L=\displaystyle\frac{1-0}{\displaystyle\sqrt{4+0+0}}=\frac{1}{2}\]
$\Rightarrow y=\displaystyle\frac{1}{2}$ is horizontal asymptote.
