Answer
No vertical asymptote.
Horizontal asymptote: $y=-4$.
Work Step by Step
Finding the vertical asymptote, equate the denominator to $0$:
$$x^2+x+3=0$$
$$x=\frac{-2\pm\sqrt{1^2-4(1)(3)}}{2(1)}=\frac{-2\pm\sqrt{-11}}{2}$$
Since the $x$ values are not real numbers, there is no vertical asymptote.
Since the degree of the numerator and the degree of the denominator are equal, taking the quotient of the leading coefficient of the numerator and the leading coefficient of the denominator, the horizontal asymptote is:
$$y=\frac{-4}{1}$$ $$y=-4$$
