First let see which zeroes have the numerator and the denominator.
For the numerator the zeros are:
`x_1 =0` and `x^2+x-2 =0 rArr x_2 =1 and x_3 =-2`
For the denominator the zeros are:
`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`