So here is the question, Write an equation for the rational function with vertical asymptotes x=4 and x= -3/2. The function has horizontal asymptote y= 1/2 and vertical intercept (0, -1/6).
Help please!!

You're looking for the expression of a rational function F = P/Q, where P and Q are polynomial functions.

x=4 and x=-3/2 as vertical asymptotes give you the roots of the denominator Q.
Hence, Q(X) = ( X -4 )^a . ( X + 3/2 )^b . R
with R a polynomial function with no real roots, and a, b are natural numbers.

The simpliest Q is : Q(X) = (X-4)(X+3/2). Let's take it.

y=1/2 as horizontal asymptote shows that degree(P) = degree(Q) and that the leading term of P is 1/2 the leading term of Q.

Let's take : P(X) = 1/2 X² + c.X + d, with c, d real numbers.

Vertical intercept gives you the value of F(0).

F(0) = -1/6
and
F(0) = P(0) / Q(0) = d / ((- 4)(3/2)) = - d / 6

so : -d/6 = -1/6, d=1.
You can now choose c=0 :

F(x) = ( 1/2 X² + 1 ) / ((X-4)(X+3/2))

(for example)