Condition for $f(x)=ax^5+bx^4+cx^2+d$ is Symmetric about line $x=k$ Which of the following is the Condition  for $f(x)=ax^5+bx^4+cx^2+d$ is Symmetric about line $x=k$
$1.$ $k=a$
$2.$ $k=b$
$3.$ $k=c$
$4.$ $k=d$
My Try:
$f(x)$ is Symmetric about $x=k$ when
$$f(2k-x)=f(x)$$
$$a(2k-x)^5+b(2k-x)^4+c(2k-x)^2+d=ax^5+bx^4+cx^2+d$$
Now comparing coefficient of $x^5$ we get $a=0$ which implies
$f(x)=bx^4+cx^2+d$ which becomes even function and hence symmetric about $x=0$, i.e., $k=0$
hence $k=a$
Is this right appraoch?
 A: Something is not right in this exercise, maybe you should check the original and type the exact question.
The way it is expressed here the phrase

the condition for "$f(x)=ax^5+\cdots$ is symmetric about line $x=k$" is $k=a$

means something like

for every $k$ (say, in $\mathbb R$) and for every $f$ such that $f(x)=ax^5+\cdots$, if $k=a$ then $f(x)$ is symmetric about line $x=k$.

And this not true, since (as you showed) that's only true if $k=a=0$ (and so is not of degree 5).
Actually is intuitive that the graph of a polynomial function of odd degree has no axial simmetry and also (maybe a little less intuitive) that the graph of a polynomial of positive even degree cannot have more that one vertical line as symmetry axis. So if necessarily $k=0$, there is not another possible value of $k$.
A possible correct version would be:

If $f(x)=ax^5+\cdots$ is symmetric about line $x=k$, then...

and yes, $k=a$ would be a right answer since it's a logical consequence of the hypothesis (of course $k=0$ and $a=0$ will also be logical consequences, but they're not between the options).

NOTE: It's an important detail that is not $f(x)$ —the function itself— which is symmetric around this or that line, but it's graph. That is: the subset of $\mathbb R^2$ given by all the points $(x,y)$ such that $f(x)=y$ (the graph of $f$) is symmetric with respect to the line $\{(x,y)\in\mathbb R^2\colon x=k\}$.
