`f(x) =(2x^3+2x^2-4x)/(16x-x^3)`

First let see which zeroes have the numerator and the denominator.

For the numerator the zeros are:

`2x^3+2x^2-4x =0`

`2x(x^2+x-2) =0`

`x_1 =0` and `x^2+x-2 =0 rArr x_2 =1 and x_3 =-2`

For the denominator the zeros are:

`16x-x^3 =0`

`x(16-x^2) =0`

`x(4-x)(4+x)=0`

`x_1 =0 , x_2=4 and x_3 =-4`

Since `x=0` is a common root for both numerator and denominator, it is a hole.

The x intercepts are the values of `x` that make `f(x) =0` . These are the other two roots of the numerator. So the x intercepts are `x=1` and `x=-2`

The domain of the function are all values of x for which the `f(x)` expression makes sense. Since `f(x)` does not make sense in the points x where the denominator becomes zero (and is not a hole), we need to exclude this values from the domain. Except these values the function is well defined on all real axis. So that the domain is:

`D = (-oo,-4) uu (-4,+4) uu (+4,+oo)`

The hole `x=0` has not been excluded from the definition domain because the limit of the function in this point is well defined and finite (see l'Hopital rule).

`lim_(x->0)f(x) =lim_(x->0)(6x^2+4x-4)/(16-3x^2) =-4/16 =-1/4`

To find the horizontal asymptotes of the function we compute its limits when `x -> +oo and -oo`

`lim_(x->oo) f(x) =lim_(x->oo) (2x^3)/((-x^3)) =2/(-1) =-2`

`lim_(x->-oo) f(x) =lim_(x->oo) (2x^3)/((-x^3)) =(-2)/(+1) =-2`

Therefore there is only one horizontal asymptote `y=-2` .

The graph of the function is below attached.

**Answer: The hole of the function is 0, the x intercepts are 1 and -2, the domain is `(-oo,-4)uu(-4,4)uu(4,+oo)` and the horizontal asymptote is `y=-2` **