Answer
Vertical asymptotes: $x=\dfrac{1}{2}$ and $x=-1$
Horizontal asymptote: $y=3$
Work Step by Step
$s(x)=\dfrac{6x^{2}+1}{2x^{2}+x-1}$
Vertical asymptotes
A rational function has vertical asymptotes where the function is undefined, that is, where the denominator is zero.
Set the denominator equal to $0$ and solve for $x$ to find the vertical asymptotes of this function:
$2x^{2}+x-1=0$
$(x+1)(2x-1)=0$
$x+1=0$
$x=-1$
$2x-1=0$
$2x=1$
$x=\dfrac{1}{2}$
Horizontal asymptotes
Since the degrees of the numerator and the denominator are the same, divide the leading coefficient of the numerator by the leading coefficient of the denominator to obtain the horizontal asymptote:
$y=\dfrac{6}{2}=3$