Here is the answer I just want to know how people find it st

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

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)=2-3^x+1

Here the constant is the horizontal asymptote which is 2.Hence the horizontal asymptote y=2