Please help

find an equation of a rational function given the following properties
Vertical asymptote: x=-2
Slant Asymptote: y=x-5
domain : (-inf, -2)U(-2,-1)U(-1,inf)
x intercept: x=3 Y intercept: y=12
hole:(1,4)

Vertical asymptote: x=-2

y = a/(x+2)

x intercept: x=3
y = a(x-3)/(x+2)

hole:(1,4)
y = a(x-3)(x-1) / (x-1)(x+2)

domain : (-inf, -2)U(-2,-1)U(-1,inf)
you also need a hole at x = -1
y = a(x-3)(x-1)(x+1) / (x+1)(x-1)(x+2)

Slant Asymptote: y=x-5
You need the above to get small as x gets big, but without adding any more holes or asymptotes
y = x-5 + (a(x-3)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))

Y intercept: y=12
solve for a in y(0)=12 gives a = -34/3
y = x-5 + ((-34/3)(x-3)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))
now collect terms to make that a single rational expression
y = ((3x^4-9x^3-27x^2-43x+72)(x-1)(x+1)) / ((x+1)(x-1)(x+2)(x^2+1))

so the denominator going to be (x+1)(x-1)(x+2)(x^2+1)

To find the equation of a rational function with the given properties, we can start by considering the form:

f(x) = (Ax + B) / (Cx + D)

To determine the values of A, B, C, and D, let's analyze the given characteristics of the function one by one.

1. Vertical asymptote: x = -2
A rational function has a vertical asymptote at x = -2 when the denominator of the function becomes zero at that point. Since (Cx + D) represents the denominator, we can set it equal to zero and solve for C and D:

Cx + D = 0
-2C + D = 0
D = 2C

Therefore, the denominator is (Cx + 2C).

2. Slant asymptote: y = x - 5
A slant asymptote of y = x - 5 means that as x approaches infinity or negative infinity, the function gets closer and closer to the line y = x - 5. We can match this by setting the numerator degree higher than the denominator degree. Since (Ax + B) represents the numerator, its degree must be one greater than the denominator, which means the denominator should be linear.

3. Domain: (-inf, -2) U (-2, -1) U (-1, inf)
The given domain specifies the restrictions for x-values. As we can see, x = -2 and x = -1 are excluded from the domain, so these are the values our vertical asymptote and hole. Hence, the denominator should be (x + 2)(x + 1).

4. X-intercept: x = 3, Y-intercept: y = 12
The x-intercept occurs when the function crosses the x-axis, which happens when the numerator is zero. Therefore, we can set Ax + B equal to zero and solve for A and B:

3A + B = 0
B = -3A

To find the y-intercept, we can substitute x = 0 into the equation:

f(0) = A(0) + B / (C(0) + D)
12 = B / D

Since B = -3A and D = 2C (from the vertical asymptote equation above), we can rewrite this as:

12 = -3A / 2C
24C = -3A

Substituting B = -3A and D = 2C back into the equation:

12 = -3A / (2C)
24C = -3A

We can solve these equations simultaneously to find A and C.

Now, we can summarize the information we have gathered:

Vertical asymptote: x = -2 (denominator: (Cx + 2C))
Slant asymptote: y = x - 5 (numerator: Ax + B, denominator: (Cx + 2C))
Domain: (-inf, -2) U (-2, -1) U (-1, inf)
X-intercept: x = 3 (numerator: Ax + B)
Y-intercept: y = 12 (B = -3A, D = 2C)

By solving the equations above, we can find the values of A, B, C, and D and construct the equation of the rational function.

To find the equation of a rational function given these properties, we can start by considering the general form of a rational function:

f(x) = (P(x))/(Q(x))

Where P(x) and Q(x) are polynomials.

1. Vertical asymptote: x = -2
A vertical asymptote occurs when the denominator Q(x) equals zero. Therefore, we know that Q(x) must have a factor of (x + 2) to have a vertical asymptote at x = -2.

2. Slant Asymptote: y = x - 5
A slant asymptote occurs when the degree of the numerator P(x) is greater than the degree of the denominator Q(x) by one. In this case, the degree of P(x) must be one greater than the degree of Q(x).

Since the equation of the slant asymptote is y = x - 5, the leading term of P(x) will be (x - 5).

3. Domain: (-∞, -2) U (-2, -1) U (-1, ∞)
The domain specifies the values of x for which the function is defined. Based on the given domain, we know that there are no restrictions on the values x can take. Therefore, both P(x) and Q(x) can be any polynomials with no restrictions on their factors.

4. x-intercept: x = 3
An x-intercept occurs when the numerator P(x) equals zero. Therefore, we know that P(x) must have a factor of (x - 3) to have an x-intercept at x = 3.

5. y-intercept: y = 12
A y-intercept occurs when x equals zero. So we can plug in x = 0 into the equation f(x) = (P(x))/(Q(x)) and solve for P(0)/Q(0). Given that y-intercept is y = 12, we have:

12 = P(0)/Q(0)

6. Hole: (1, 4)
A hole occurs when both the numerator P(x) and the denominator Q(x) have a common factor that cancels out. In this case, we know that P(x) - 4 = (x - 1), and Q(x) - 1 = (x - 1), so (x - 1) is the common factor that creates the hole at (1, 4). We can remove this common factor from both the numerator and the denominator.

Now, with all this information, we can construct the equation of the rational function:

f(x) = ((x - 5)(x - 3))/(Q(x))

Since the slant asymptote is given as y = x - 5, we know that the degree of Q(x) must be equal to or greater than the degree of P(x). Let's assume the degree of Q(x) is equal to the degree of P(x). Therefore, Q(x) will be a linear polynomial.

We can now find Q(x) by setting Q(x) = x + b, where b is a constant. Using the vertical asymptote, we know that Q(-2) must equal zero, so we have:

-2 + b = 0
b = 2

Therefore, the equation of the rational function given the properties would be:

f(x) = ((x - 5)(x - 3))/(x + 2)