Answer
$f(x)=\frac{2x^2+1}{x^2+x+2}$
Work Step by Step
We are looking for a rational function with denominator which has no real zeros and the same degree of the numerator and denominator, so that the quotient of the leading coefficients is $2$.
Take a rational function:
$$f(x)=\frac{2x^2+1}{x^2+x+2}$$
Finding the vertical asymptote by equating the denominator to $0$:
$$x^2+x+2=0$$
$$x=\frac{-1\pm\sqrt{1^2-4(1)(2)}}{2(1)}=\frac{-1\pm\sqrt{-7}}{2}$$
Since the $x$-values are not real numbers, there is no vertical asymptote.
Since the degrees of the numerator and the denominator are the same, take the quotient of leading coefficients of the numerator and denominator to find the horizontal asymptote:
$$y=\frac{2}{1}$$ $$y=2$$
