Answer
vertical asymptotes: $x=1$ and $x=-1.5$.
horizontal asymptote: $y=2$
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:
$4x^2+2x-6=0\\2(x-1)(2x+3)=0$
$x=1$ and $x=-1.5$.
Thus the vertical asymptotes are $x=1$ and $x=-1.5$.
Here, the degrees of the numerator and the denominator are the same. Thus, the horizontal asymptote is the quotient of the leading coefficients:
$y=\frac{8}{4}=2$
