Answer
$$y = - \frac{3}{2}{\text{ and }}y = \frac{3}{2}{\text{ are both horizontal asymptotes}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{\sqrt {9{x^2} - 2} }}{{2x + 1}} \cr
& {\text{From the graph we have:}} \cr
& \mathop {\lim }\limits_{x \to - \infty } \frac{{3x}}{{\sqrt {{x^2} + 2} }} = - \frac{3}{2}{\text{ and }}\mathop {\lim }\limits_{x \to \infty } \frac{{3x}}{{\sqrt {{x^2} + 2} }} = \frac{3}{2} \cr
& {\text{Then,}} \cr
& y = - \frac{3}{2}{\text{ and }}y = \frac{3}{2}{\text{ are both horizontal asymptotes}} \cr
& \cr
& {\text{Graph}} \cr} $$
