Simplify. -14x^3/ x^3-5x^4 a) -14/5x-1; where x is not equal to 1/5,0 b) -14x/1-5x; where x is not equal to 1/5 c) 1-5x/-14x; where x is not equal to 0 d) -14/1-5x; where x is not equal to 1/5, 0

It's D.  Factor x^3 out of the denominator leaving  1 - 5x, then cancel out the x^3 in the numerator with the one you factored out of the denominator.  That leaves you with 1 - 5x cannot equal 0.  Solve that for x and you get 1/5.  However, since we are talking vertical asymptotes and the x^3 is a removable discontinuity, we have to count it as a "problem".  Therefore, the other value of x that is not allowed is x = 0.