Create a rational function, g(x) that has the following properties.

[Disclaimer: The answer is not complete. The problem turns out to be much more diffult than expected. We have tried our best and give our - unfortunately negative - results below.]

A rational function means g(x) = A(x) / B(x) where A, B are polynomials.

i) no V.A. means that when B(x)=0 we must also have A(x)=0 (otherwise we have a V.A. where B(x)=0)

ii) no O.A. means that degree(A) must not be = degree( B ) + 1  − otherwise we have an O.A.

iii) H.A. y=1 means that degree(A) = degree(B) and the leading coefficient of A must be the same than that of B; without loss of generality we can take it = 1, i.e., A and B are both product of terms of the form (x ± ak) and/or (x² + bk x + ck).

iv) Hole at x = −4 means that B(x = −4) = 0 ⇔  B(x) = (x+4) Q(x)  and according to (i) we must also have A(-4) = 0  ⇔   A(x) = (x+4) P(x).

Then at any x ≠ −4 we have g(x) = P(x) / Q(x) =: h(x) and we can use this simplified expression (with factors (x+4) cancelled) to compute values and derivatives for extrema, etc.

Also, to have no other holes we must have Q(x) ≠ 0 for all x, so Q must be a product of irreducible factors (x² + bx + c) with negative discriminant b²-4c.

In particular, we must have even degree  deg(Q) = 2m.

v + vi) : min at  x = −3 and max at  x = 1  means that the derivative of g or h = P/Q must be zero at these points:  h'(−3) = h(1) = 0. 
We'll come back to this after vii & viii.

vii) x-intercept -1 means that $g(−1)  = 0  \iff   P(−1) = 0  \iff$  $P(x) = (x+1)^{2k} R(x)$ 
where  R  must have no real root, in order to have no other x-intercept (i.e., no other zero).
So, as for Q,  R  is the product of quadratic factors with negative discriminant and therefore of even degree (possibly zero: deg(R)=0  <=>  R=1),
and since  deg P = deg Q  from (iii),  (x+1)  must appear to an even power  2k = deg(Q) − deg(R) = 2m − deg(R)  (from  iv).

viii) y-intercept 1/3 means that we must have   1/3 = g(0) = h(0) = P(0)/Q(0)  ⇔  Q(0) = 3 P(0) = 3 R(0).

So we must have Q(0) = 3 R(0), for example Q = x² + bx + 3 R(0)  or  Q = (x² + $b_1$ x + a) (x² + $b_2$ x + 3 R(0)/a)  for some  a ≠ 0. 

Let's now turn back to (v + vi). To have extrema we need  g'(x) = 0, so let's compute

$g'(x) = h'(x) = P'(x) / Q(x) − P(x) Q'(x) / Q(x)² $
$= (2k (x+1)^{2k-1} R(x) + (x+1)^{2k} R'(x))/Q(x) − (x+1)^{2k} R(x) Q'(x) / Q(x)² $.

Dividing this by  $(x+1)^{2k-1}$ we find  g'(x) = 0  ⇔  (2k R(x) + (x+1) R'(x)) Q(x) = (x+1) R(x) Q'(x).

From (v + vi), we need  g'(x) = 0  at  x = 1  and  x = −3:

x = 1: (2k R(1) + 2 R'(1)) Q(1) = 2 R(1) Q'(1)  ⇔   (k R(1) + R'(1)) Q(1) = R(1) Q'(1)  (eq.1)

x = −3: (2k R(1) - 2 R'(1)) Q(−3) = −2 R(−3) Q'(−3)  ⇔  (k R(1) - R'(1))  Q(−3) = − R(−3) Q'(−3) . (eq.2)

If  m = 1, i.e.,  Q(x) = x² + bx + c  ⇒  Q'(x) = 2x+b, we must have  k=1, R=1, R'=0;
we had  c = 3  (from viii)  and so  at x = 1 (eq.1): 1+b+3 = 2+b : impossible.


If  m = 2, Q = (x² + b1 x + a) (x² + b2 x + 3/a)
⇒ Q' = (2x + b1 ) (x² + b2 x + 3/a) + (x² + b1 x + a) (2x + b2)

From (v + vi) we must also have  g(−3) = −1/6  and  g(1) = 1/2 
which gives, with  Q(x) = P(x) / g(x) = (1+x)^{2k} R(x) / g(x):

$Q(-3) = (-2)^{2k} R(-3) (−6) =  (9 − 3 b_1 + a) (9 − 3 b_2  + 3 R(0)/a)$  for  m = 2
and  $Q(1) = 2^{2k} R(1) \times 2 = 2 \cdot 4^k R(1)  =   (1 + b_1 + a) (1 + b_2  + 3 R(0)/a)$   for  m = 2.

From (iv) we have  h(−4) = −3/19  = P(−4) / Q(−4)  
⇔  Q(−4) = −(19/3) P(−4) = −(19/3) $(-3)^{2k}R(-4) = 19 \,(-3)^{2k-1}R(-4) = (16 − 4 b_1 + a) (16 − 4 b_2  + 3 R(0)/a)$  for  m = 2.

Let us show that it is impossible to find a solution with k = m = 2, R = 1:

For m = 2, the above 3 equations allow us to compute the 3 parameters b1, b2 and a:

(Q(-3) = −6 ∙ 16) / 36  −  (Q(1) = 2 ∙ 16) / 4  ⇔   $(1 + 1/a) b_1  +  (1+ a/3) b2  =  12$ (eq.3)
(Q(-4) = −27 ∙ 19) / 80  −  (Q(1) = 2 ∙ 16) / 5   ⇔  $(1 + 3/4a) b1  +  (1 + a/4) b2  =  61/4$ (eq.4)

Then:  4 (eq.4) − 3 (eq.3)  ⇔  $b_1 + b_2  = 25$  and we can substitute this back to get :

b1 = (25*a + 39)*a/(a^2 - 3)
b2 = −(39 a + 75) / (a^2 - 3)

But if we substitute this in  Q'(1) = 2,  then we find  a ( a² - 3 ) = 0, however, neither a=0 nor a² = 3 is possible in view of the definition of  Q (with 3/a)  and the b-values we got.
So it is impossible to have a solution with  m = k = 2, R = 1.

So we must try $m = 2 > k = 1, ~ R(x) = x² + a_1 x + a_0 \ne 1$, or else $m \ge 3$.

With  R = 1, we get a system for the 6 unknowns (a0, a1, b1, b2, c1 = a, c2 = 3 R(0)/a = 3 a0 / a) which can be simplified to:

[1] c2*c1 = 3*a0

[2] ((a1 - a0)*b2 + (2*a1*c2 + (2*a1 - 2*a0)))*b1 + (2*a1*c1 + (-2*a0 - 12))*b2 + ((3*a1 + a0)*c2 + 4*a1)*c1 + (a1 - a0)*c2 + 26/3*a1 + 16/3*a0 = 20/3

[3] ((-6*a1 + 2*a0)*b2 + (-1/3*a0*c2 + (-12*a1 + 4*a0)))*b1 + (-1/3*a0*c1 + 18*a1 -3*a0 + 27)*b2 + ((-8/3*a1 - 2/3*a0)*c2 - 2/3*a0)*c1 + (-6*a1 + 2*a0)*c2 + 61*a1 + -19*a0 = 94

[4] (3*b2 + 6)*b1 + 3*c2 + -8*a1 + 5*a0 + 8 = 0

[5] c2*b1 + (c1 + 3)*b2 + 2*c1 = 0

[6] 3*b2 - a1 + a0 + 4 = 0

But then, when we substitute c1*c2 from the first and b2 from the last into the other equations, it seems that there is no possible solution. (We get two quadratic equations for a2 that are incompatible.)

So, it appears we must consider $m = 3$, i.e., Q = (x² + b1 x + c1) (x² + b2 x + c2) (x² + b3 x + c3)
with again Q(0) = c1 c2 c3 = 3  (from the y-intercept), and similar (but much more complicated) equations for the values 1/2, -1/6 and -3/19  at  x = 1, -3 and -4
(which we will call eq.5, eq.3 and eq.4)

and the equations for the derivative, g'(x)=1  <=>  3 Q(1) = Q'(1),  3 Q(-3) = -Q'(-3)
[which we will call again eq.1 & eq.2]

This makes up 6 equations that should allow us to find the 6 unknowns (b1, b2, b3, c1, c2, c3).
We can again take linear combinations that make some monomials disappear.
For example 9(eq.1)+(eq.2) does not have b1 b2 b3 any more.

Proceeding this way, adding multiples of one equation to the other in order to simplify by eliminating one of the terms, we obtain the following equivalent system of equations :

c1*c2*c3 = 3,

b1 + b2 + b3 = 31,

c3*c2*b1 + c3*c1*b2 + b3*c2*c1 = 159,

(b2 + b3)*b1 + b3*b2 + c1 + c2 + c3 = 101,

(b3*b2 + (c2 + c3))*b1 + (c1 + c3)*b2 + b3*c1 + b3*c2 = -6

(c3*b2 + b3*c2)*b1 + b3*c1*b2 + (c2 + c3)*c1 + c3*c2 = -161 

But it seems quite difficult to solve these equations...
(...)