Give the cubic polynomial $ y = ax^3 + bx^2 + cx + d$ which is symmetric about the line $x=k$ then derive a relation between $a,b, c, d$ $and$ $k$.
I have attempted this question and used the fact that at the point of symmetry either the first derivative must be zero or not defined. However I am not convinced by my method and feel I have left something out and not considered it. Any help would be greatly appreciated
Hint: If the polynomial is $p(x)$, symmetry about the line $x=k$ implies $p(k+t)=p(k-t)$ for all $t\in\mathbb{R}$.
EDIT: I got $6\,a{k}^{2}t+2\,a{t}^{3}+4\,bkt+2\,ct=t\cdot(6\,a{k}^{2}+2\,a{t}^{2}+4\,bk+2\,c)=0$. Since this must hold for all $t\in\mathbb{R}$, it follows that $6\,a{k}^{2}+2\,a{t}^{2}+4\,bk+2\,c=0$. Once again, that this holds for all $t\in\mathbb{R}$ implies $a=0$. Hence, the relations we derive are
$$a=2bk+c=0.$$
Notice that we could have derived $a=0$ a priori from the fact that $p$ has odd degree. Indeed, if that were not the case, then $\lim_{x\to\infty}p(x)$ and $\lim_{x\to-\infty}p(x)$ would have opposite signs, so the graph of $p$ couldn't be symmetric about a vertical line.
Then, the quesiton would reduce to finding a relation on a quadratic polynomial. It is well known that these polynomials have parabolas for graphs, and that their only vertical line of symmetry is the one that crosses the point of minimum/maximum, at $x=\frac{-c}{2b}$. If it sounds unfamiliar, notice that these polynomials are usually described as $q(x)=ax^2+bx+c$, and this point is then usually written as $x=\frac{-b}{2a}$.