How to perform vertical line test algebraically? I understand how a function should have one value mapped to a single value in the range. Thus, when graphing the function I understand how a vertical line can determine if a point or multiple points intersect the vertical line. This is visual. However, how can we perform the same test algebraically across a domain of the function?
I am learning calculus from the start again since college, so in case my question is naive or obvious I apologize.
 A: A vertical line is the set of all points $(x_0,y)$ where $x_0$ is a fixed constant and $y$ is allowed to vary freely. Given an equation in the form $f(x,y)=0$, we say it passes the vertical line test if, for every fixed choice of $x_0$, the resulting equation $f(x_0,y)=0$ has either $0$ or $1$ solutions in terms of $y$.
For example, if we were checking the equation
$$ x = y^2 $$
we would define the function $f(x,y) = x - y^2$. If we choose $x_0=4$, notice that the equation
$$f(4,y) = 4 - y^2 = 0$$
fails the vertical line test since $y=2$ and $y=-2$ are both solutions.

As an exercise, see if you can prove that if $y$ is a function of $x$, such as $y=g(x)$, then the equation $f(x,y)=y-g(x)$ will always pass the vertical line test :)
A: To say that a "relation" (or oftentimes "graph") is a "function" from A to B (A is the domain and B is the range) is to say that for any number a (in A) (so any number we take a "y value" at), there is exactly one number that this a gets mapped to. In practice, this typically amounts to checking how the "relation" is defined, and comparing it with this "exactly one" condition. Let's say our "relation" is f(x) = x (so the "identity function", sends numbers to themselves), we start with a number a, and f sends this a to f(a), which is just a in this case, so we start with a, and this a gets sent just to a, so where the function sends any number a (to itself) is definitely unique. $f(x) = x^2$ is another function, to check this is a function amounts to just checking that when you take any number x and square it, the output is unique (there is a single output, for example, -3 gets sent to the unique output $(-3)^2 = 9$) In this way, $x^2$ is a function. This process doesn't have to be "algebraic" it is just "clear" (or "rigorous") (more so than looking at a picture at least).
