Yes. If $f$ has an oblique asymptote (call it $y=ax+b$), you will have: $$a=\lim_{x\to\pm\infty}\frac{f(x)}{x}$$
$$b=\lim_{x\to\pm\infty} f(x)-ax$$
In your example, $\displaystyle\lim_{x\to+\infty}\frac{\sqrt{4x^2+x+6}}{x}=2$ and $\displaystyle\lim_{x\to+\infty}\sqrt{4x^2+x+6}-2x=\frac{1}{4}$
The asymptote as $x\to+\infty$ is therefore $y=2x+\dfrac{1}{4}$