f(x) = x^3 + 6x^2 - 15x - 40/x^3 + 7x^2 - 8x (1.how many holes are present in this rational function. (2. what are the holes in this rational function

Answer 1

There are no holes. The denominator factors as x(x+8)(x-1), so there are zeroes at -8, 0, and 1.

Substituting these three values into the numerator, you get -48, -40, and -48, respectively. Thus, we have vertical asymptotes at -8, 0, and 1.