Answer
$$\eqalign{
& \left( {\text{a}} \right)\infty ,\infty \cr
& \left( {\text{b}} \right){\text{There is no slant asymptote}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{x{{\left( {x + 2} \right)}^3}}}{{3{x^2} - 4x}} \cr
& f\left( x \right) = \frac{{{{\left( {x + 2} \right)}^3}}}{{3x - 4}} \cr
& {\text{Expand the numerator}} \cr
& f\left( x \right) = \frac{{{x^3} + 6{x^2} + 12x + 8}}{{3x - 4}} \cr
& \left( {\text{a}} \right){\text{ Calculate }}\mathop {\lim }\limits_{x \to \infty } f\left( x \right){\text{ and }}\mathop {\lim }\limits_{x \to - \infty } f\left( x \right) \cr
& \mathop {\lim }\limits_{x \to \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^3} + 6{x^2} + 12x + 8}}{{3x - 4}} \cr
& = \frac{{\overbrace {\mathop {\lim }\limits_{x \to \infty } \left( {{x^3} + 6{x^2} + 12x + 8} \right)}^{{\text{arbitrary large to positive infinity }}}}}{{\underbrace {\mathop {\lim }\limits_{x \to \infty } \left( {3x - 4} \right)}_{{\text{arbitrary large to positive infinity }}}}} = \infty \cr
& and \cr
& \mathop {\lim }\limits_{x \to - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^3} + 6{x^2} + 12x + 8}}{{3x - 4}} \cr
& = \frac{{\overbrace {\mathop {\lim }\limits_{x \to - \infty } \left( {{x^3} + 6{x^2} + 12x + 8} \right)}^{{\text{arbitrary large to negative infinity }}}}}{{\underbrace {\mathop {\lim }\limits_{x \to - \infty } \left( {3x - 4} \right)}_{{\text{arbitrare large to negative infinite }}}}} = \infty \cr
& \cr
& \left( {\text{b}} \right){\text{The slant asymptote occurs with rational functions only }} \cr
& {\text{when the degree of the polynomial in the numerator exceeds }} \cr
& {\text{the degree of the polynomial in the denominator by exactly 1.}} \cr
& {\text{For this exercise, the degree of the numerator is equal to the}} \cr
& {\text{degree of the denominator.}}{\text{ So there is no slant asymptote}}{\text{.}} \cr} $$
