Here is the answer, I just want to know how people find it step by step (6) Find all asymptotes (if any) f (r) In 2a 3) g(a) h (ar) 2-3 Solution f(x)=ln(2x-3) To find the vertical asymptote we have to set 2x-3to zero 2x-3=0 X=3/2 g(x)= (x3-9x)/(x4+3x3+2x2) = (x)(x+3)(x-3)/(x2)(x+1)(x+2) =(x+3)(x-3)/x(x+2)(x+1) Vertical asymtote x(x+2)(x+1)=0 x=-2,-1,0 Horizontal asymtote Since the degree of numerator is less than the degree of denominator. Therefore horizontal asymptote is y=0 h(x)=2-3x+1 Here the constant is the horizontal asymptote which is 2.Hence the horizontal asymptote y=2.