Consider a function of the type $y=f(x)$. This means the function is surjective: every element of its co-domain is an image of one or more objects of its domain (please correct me if I'm wrong).
So I've been told that a graph of such function can be described as this set: $$\left\{\left(x,y\right)\in \mathbb{R}^2:\left(y+1\right)^2=y^2\right\}$$
Now I can't get my head around this. Let me tell you what I can get from this. The set of all points belongs to the $\mathbb{R}^2$ plane such that they can be found in the line $y=-\frac{1}{2}$. Why does this make sense?
Also, why is this set not applicable: $$\left\{\left(x,y\right)\in \mathbb{R}^2:\left(x-1\right)^2=x^2\right\}$$?
A set $G\subset X\cdot Y$ is a graph of a function $f:X\rightarrow Y$ when and only when $\forall x\in X$ there is one and only one $y\in Y$ such that $\left(x,y\right)\in G$.So $\left\{\left(x,y\right)\in \mathbb{R}^2:\left(x-1\right)^2=x^2\right\}$ can never represent $G$, since it yields a line parallel to the $y-axis$: meaning, there are multiple images for one lonely $x$.
The set $\left\{\left(x,y\right)\in \mathbb{R}^2:\left(y+1\right)^2=y^2\right\}$ describes $G$ because it yields a line parallel to the $x-axis$: meaning, each value $x$ has one and only one image, although that image is the same for every $x$. That is $-\frac{1}{2}$.