A Question About A Polynomial Problem I'm trying to understand the solution to this problem:https://imgur.com/E2Jm63r. The solution states that the answer is $C$ and $D$, because if $-p(x)=p(y)$, then the equation of the values that satisfy the $-p(x)=p(y)$ must by symmetric with respect to$ y=-x$. I have 2 questions about this.

$1$. Why does the fact that $-p(x)=p(y)$ mean that the equation must be symmetrical over y=-x? Is this only true for polynomials, or is it true for any function?

$2$. Does the fact that $p(x)=p(y)$ mean that the equation must be symmetrical over y=x? Is this only true for polynomials, or is it true for any function?

Thanks in advance
Let $S$ be the set of points $(x, y)$ such that $p(x) + p(y) = 0$.

Suppose $(x, y) \in S$. Then $p(x) + p(y) = 0$. But then also $p(y) + p(x) = 0$, and so $(y, x) \in S$. This means that $S$ is symmetric with respect to the line $y = x$. Thus, the answer (A) is wrong.

In general, $S$ might not be symmetric with respect to the line $y = -x$. Indeed, if $p(x) = x + 1$, then the equation $p(x) + p(y) = 0$ amounts to $x + y + 2 = 0$, which represents a line parallel to the line $y = -x$.
$C$ is true because you can define $p(x)=x^2-\frac{1}{2}$ and you have that

$p(x)+p(y)=x^2+y^2-1=0$

That is the equation of the circumference;

$D$ is true because you can define 

$p(x)=x$ 

and you have that 

$p(x)+p(y)=x+y=0$

that is the equation of the second and fourth quadrant;

$A$ is not true because it is the locus of $y=ax^2$ where $a&gt;0$