How to determine at which point a polynomial function with even and odd x-exponents is symmetrical, algebraically? Say you have a polynomial function like $f(x) = 2x^3 - 6x^2 + 8$. This function won't be point symmetrical over $(0, 0)$, nor will it be symmetrical over the y-axis, but after plotting it, you can see that it's point symmetrical over the point $(1, 4)$.
For a polynomial function like $g(x) = (x+1)^3$, you can tell that it was shifted by one unit to the left on the x-axis, $h(x) = x^3$ is symmetrical over the point $(0, 0)$, so $g(x)$ will be symmetrical over the point $(-1, 0)$.
I don't really see how you could use this approach for $f(x)$ yet though.
One approach I found that works for $f(x)$ is to take the second derivative and set it to $0$.
$$f'(x) = 6x^2 - 12x$$ $$f''(x) = 12x - 12$$ $$f''(x) = 0$$ $$12x = 12$$ $$\fbox{x = 1}$$
Now to find the corresponding y value, plug in $x = 1$ into $f(x)$.
This second derivative approach doesn't seem to work for all functions though, there could be a point where the graph changes concavity but still isn't point symmetrical over.
Now my question is, is there a way to determine over which point any function like this will be symmetrical over without plotting it?

In the comments, it was stated that for a point of symmetry to exist, all the derivatives of the function must be (anti-)symmetric with respect to a line x = a. How can you check if a function is (anti-)symmetric to this line algebraically?
 A: If $f(x) = a_n x^n + a_{n-1} x^{n-1} +...$, the only possible abscissa for a symmetry point is the roots mean, i.e. $-\frac {a_{n-1}} {n \, a_n}$.
First proof
As was said in the comments, if $f$ is symmetric or anti-symmetric, all its derivatives are symmetric or anti-symmetric around the same abscissa. This alternates: even degree polynomials are symmetric (this is because of the behavior on $\infty$), odd degree polynomials are anti-symmetric.
So we can find this abscissa $a$ with the $n-1$th derivative: $f^{(n-1)}(x) = a_n \, n! \, x + (n-1)! \, a_{n-1} = 0$
$x=-\frac {a_{n-1}} {n \, a_n}$.
Tentative of a second proof
Another possible proof would proceed like this.
If there is a symmetry or anti-symmetry point $(a,b)$, we can vertically translate $f$ by subtracting $b$: $g(x)=f(x)-b$.
Now $g$ is (anti-)symmetric around $(a,0)$. So its real roots are symmetric around $a$.
If $g$ has all its $n$ roots real, because of the symmetry, the roots mean will be $a$, so $a=-\frac {a_{n-1}} {n \, a_n}$.
Now how to deal with the case where $g$ has some non-real complex roots?
One possible way would be to show that we can build a polynomial $h$ such that $h-g$ is a polynomial with degree at most $n-2$, and $h$ is (anti-)symmetric around a point with abscissa $a$ such as $g$. We would then be able to apply $a=-\frac {a_{n-1}} {n \, a_n}$: as we have only added an $n-2$ degree polynomial, $h$ has same first two coefficients $a_n$ and $a_{n-1}$ as $f$ and $g$.
Unfortunately I don't succeed in proving that.
Third proof, which gives the complete list of conditions
The method here is to directly expand the (anti-)symmetry equation.
If $n$ is odd, $f$ is anti-symmetric, so
$f(a+x)+f(a-x)=2b$
$\sum_{k=0}^n a_k(a+x)^k + \sum_{k=0}^n a_k(a-x)^k=2b$
$\sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} x^j a^{k-j} + \sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} (-1)^j x^j a^{k-j}=2b$
This can be split into separate equations, one for each power of $x$, i.e. one for each $j$. Equations for $j$ odd give $0=0$. Equations for $j$ even and $\ge 2$ give the following:
$\forall j$ even and $\ge 2$, $\sum_{k=j}^n a_k {k \choose j} a^{k-j} + \sum_{k=j}^n a_k {k \choose j} (-1)^j a^{k-j} = 0$
$\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$
For $j=n-1$, we get
$\sum_{k=n-1}^n a_k {k \choose {n-1}} a^{k-(n-1)} = 0$
$a_{n-1} a^0 + a_n \, n \, a^1 = 0$
$a = - \frac {a_{n-1}} {n \, a_n}$
The other conditions are obtained with the other even values for $j$. The value for $b$ can be computed from $j=0$ (but that's not much different from computing $b=f(a)$). Unfortunately, the conditions grow ugly when $j$ decreases; but it is the necessary and sufficient list and there does not seem to be a simpler way.
As example, the next condition, for $j=n-3$, expands into:
$3n^2a_n^2a_{n-3} - 3n(n-2)a_na_{n-1}a_{n-2}+(n-1)(n-2)a_{n-1}^3=0$
If $n$ is even, $f$ is symmetric, so
$f(a+x)-f(a-x)=0$
$\sum_{k=0}^n a_k(a+x)^k - \sum_{k=0}^n a_k(a-x)^k=0$
$\sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} x^j a^{k-j} - \sum_{k=0}^n a_k \sum_{j=0}^k {k \choose j} (-1)^j x^j a^{k-j}=0$
As above, this splits into separate equations, one for each $j$. Equations for $j$ even give $0=0$. Equations for $j$ odd give:
$\forall j$ odd and $\ge 1$, $\sum_{k=j}^n a_k {k \choose j} a^{k-j} - \sum_{k=j}^n a_k {k \choose j} (-1)^j a^{k-j} = 0$
so $\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$ as above.
For $j=n-1$, we get $a = - \frac {a_{n-1}} {n \, a_n}$ as before.
The two cases (n even; n odd) can be summarized in the following conditions:
$f(x) = a_n x^n + a_{n-1} x^{n-1} +...$ is symmetric (if $n$ even) or anti-symmetric (if $n$ odd) around $a=-\frac {a_{n-1}} {n \, a_n}$ (and $b=f(a)$, for anti-symmetry) iff:
$\fbox {$\forall j=n-1, n-3, ... \text{ 2 or 1, }\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$}$
Example
Using $f(x)=(x-4)(x-2)(2x+1)(x+3)(x+5)-209$
which means we have an anti-symmetry around $(a=-\frac 1 2,b=-209)$
$f(x)=2x^5+5x^4-48x^3-77x^2+214x-89$
(The reason why I don't use the example in the question post, is because I would like to have no null coefficient, and $a_n \ne 1$).
$a=-\frac {a_{n-1}} {n\,a_n}=-\frac {5} {5 \times 2}=-\frac 1 2$
$3 n^2 a_n^2 a_{n-3} - 3n(n-2)a_n a_{n-1} a_{n-2}+(n-1)(n-2)a_{n-1}^3 =$
$3 \times 5^2 \times 2^2 \times (-77) - 3 \times 5 \times 3 \times 2 \times 5 \times (-48) + 4 \times 3 \times 5^3=-23100+21600+1500=0$
There is no other condition (the one above is $\sum_{k=j}^n a_k {k \choose j} a^{k-j}=0$ for $j=n-3=2$), so the polynomial is anti-symmetric around $a=-\frac 1 2$ and $b=f(a)=-209$.
