Answer
Vertical asymptote: $x=0$
Horizontal asymptote: $y=3$
Work Step by Step
$r(x)=\dfrac{6x^{3}-2}{2x^{3}+5x^{2}+6x}$
Vertical asymptote
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:
$2x^{3}+5x^{2}+6x=0$
$x(2x^{2}+5x+6)=0$
$x=0$
$2x^{2}+5x+6=0$
Use the quadratic formula, which is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ to solve this second equation. Here, $a=2$, $b=5$ and $c=6$:
$x=\dfrac{-5\pm\sqrt{5^{2}-4(2)(6)}}{2(2)}=\dfrac{-5\pm\sqrt{25-48}}{4}$
The operation inside the square root will yield a negative number, so this equation has complex roots. The only vertical asymptote this function has is $x=0$
Horizontal asymptote
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:
$y=\dfrac{6}{2}=3$