Requirements:
VA: x=4; HA: y=-3; hole=(5,-2)
VA occurs when x-4 is in the denominator.
HA occurs when the numerator and denominator have the same degree and the coefficient of the highest degree of x is -3.
The hole occurs when the x-5 is a common factor in the numerator and denominator, and, if this factor is removed (the hole is removed), the expression evaluates to -2.
Let y=Ax/(x-4)+B(x-5)/(x-5), where A and B are constants to be found.
To satisfy the HA, y=-3 when x is very large, which allows us to ignore the -4 constant in the denominator. Therefore y=A+B=-3.
When x=5 in Ax/(x-4), y=5A+B=-2.
Substituting for B=-3-A, 5A-3-A=-2, 4A=1, A=¼.
Therefore B=-3¼=-13/4.
y=x/(4(x-4))-13(x-5)/(4(x-5)),
y=(x(x-5)-13(x-5)(x-4))/((4(x-4)(x-5)),
y=(x²-5x-13(x²-9x+20))/((4(x-4)(x-5)),
y=(x²-5x-13x²+117x-260)/((4(x-4)(x-5)),
y=(-12x²+112x-260)/((4(x-4)(x-5)),
y=(-3x²+28x-65)/(x²-9x+20).
-3x²+28x-65 can be factorised: (13-3x)(x-5) as was the denominator.
Domain is x≠4, x≠5. Range is y≠-3, y≠-2. Otherwise x and y are unbounded.