Answer
$$\eqalign{
& {\text{Vertical asymptote: }}x = 4 \cr
& {\text{Horizontal asymptote: }}y = 2 \cr} $$
Work Step by Step
$$\eqalign{
& h\left( x \right) = \frac{{2x + 3}}{{x - 4}} \cr
& {\text{The function is not continuous at }}x = 4,{\text{ so there is a vertical}} \cr
& {\text{asymptote at }}x = 4 \cr
& {\text{Calculate the horizontal asymptotes, evaluating }} \cr
& \mathop {\lim }\limits_{x \to - \infty } h\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to \infty } h\left( x \right) \cr
& \mathop {\lim }\limits_{x \to - \infty } h\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{2x + 3}}{{x - 4}} = \frac{{\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{{2x}}{x} + \frac{3}{x}} \right)}}{{\mathop {\lim }\limits_{x \to - \infty } \left( {\frac{x}{x} - \frac{4}{x}} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to - \infty } \left( {2 + \frac{3}{x}} \right)}}{{\mathop {\lim }\limits_{x \to - \infty } \left( {1 - \frac{4}{x}} \right)}} = \frac{2}{1} = 2 \cr
& \mathop {\lim }\limits_{x \to \infty } h\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{2x + 3}}{{x - 4}} = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{2x}}{x} + \frac{3}{x}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {\frac{x}{x} - \frac{4}{x}} \right)}} \cr
& = \frac{{\mathop {\lim }\limits_{x \to \infty } \left( {2 + \frac{3}{x}} \right)}}{{\mathop {\lim }\limits_{x \to \infty } \left( {1 - \frac{4}{x}} \right)}} = \frac{2}{1} = 2 \cr
& {\text{Therefore, }} \cr
& {\text{Vertical asymptote: }}x = 4 \cr
& {\text{Horizontal asymptote: }}y = 2 \cr} $$
