Answer
vertical asymptotes: $x=0.5$ and $x=-1$
horizontal asymptote: $y=3$
Work Step by Step
Vertical asymptotes occur where the denominator is zero. Thus, we set the denominator to zero and solve for the $x$ values to find the vertical asymptotes:
$2x^2+x-1=0\\(2x-1)(x+1)=0$
$x=1/2$ and $x=-1$
Thus the vertical asymptotes are $x=0.5$ and $x=-1$.
Here, the degrees of the numerator and the denominator are the same. Thus, the horizontal asymptote is the quotient of the lead coefficients:
$y=\frac{6}{2}=3$
