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
 A: 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 $y = -x$.
A: Let $p$ be a real-valued function defined on some nonempty subset $D$ of $\mathbb{R}$.

If $G$ is the graph of the equation $p(x)+p(y)=0$, then:
\begin{align*}
&\text{The point}\;(a,b)\;\text{is on}\;G\\[4pt]
\iff\;&p(a)+p(b)=0\\[4pt]
\iff\;&p(b)+p(a)=0\\[4pt]
\iff\;&\text{The point}\;(b,a)\;\text{is on}\;G\\[4pt]
\end{align*}
hence $G$ is symmetric about the line $y=x$.

Similarly, if $H$ is the graph of the equation $p(x)-p(y)=0$, then:
\begin{align*}
&\text{The point}\;(a,b)\;\text{is on}\;H\\[4pt]
\iff\;&p(a)-p(b)=0\\[4pt]
\iff\;&p(b)-p(a)=0\\[4pt]
\iff\;&\text{The point}\;(b,a)\;\text{is on}\;H\\[4pt]
\end{align*}
hence $H$ is also symmetric about the line $y=x$.

Moreover, for all $d\in D$, the point $(d,d)$ is on $H$.

However, neither of the graphs $G,H$ is necessarily symmetric about the line $y=-x$.

For example:


*

*If $p(t)=t-1$, the graph $G$ of the equation $p(x)+p(y)=0$ is the graph of the equation $x+y=2$, and $G$ is not symmetric about the line $y=-x$.$\\[4pt]$

*If $p(t)=t^2-2t$, the graph $H$ of the equation $p(x)-p(y)=0$ is the union of the graphs of the equations $x+y=2$ and $x=y$, and $H$ is not symmetric about the line $y=-x$.

A: $C$ 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 $a>0$ 
