How to obtain more than two abcissas for one ordinate? I'm reading a book on trigonometry and the author shown that some of the examples contain one abcissa for one ordinate and he showed that a parabola has two abscissas for every positive ordinates, I got curious: 


*

*Is it possible to have more than two abscissas for one and only one ordinate?

*Is it possible to have more than two abscissas for every element of a specific set of numbers?

 A: The answer to both questions is easily yes if you don’t require the function to be continuous. Let 
$$f(x)=\begin{cases}
2,&\text{if }x\in\{0,1,2\}\\
\tan^{-1}x,&\text{otherwise}\;
\end{cases}$$
The function $\tan^{-1}x$ is one-to-one and takes values between $-\frac{\pi}2$ and $\frac{\pi}2$, so every ordinate except $1$ corresponds to at most one abscissa, while $1$ corresponds to three abscissae.
The same basic idea can be used to provide a positive answer to the second question. Let $Y$ be the set of ordinates that are to have more than two abscissae. For each $y\in Y$ pick a set $X_y$ of three real numbers, making sure that the sets $X_y$ are pairwise disjoint and that $\left|\Bbb R\setminus\bigcup_{y\in Y}X_y\right|=|\Bbb R\setminus Y|$. (This is always possible.) Let $$g:\Bbb R\setminus\bigcup_{y\in Y}X_y\to\Bbb R\setminus Y$$ be any bijection, and let
$$f(x)=\begin{cases}
y,&\text{if }x\in X_y\\
g(x),&\text{otherwise}\;;
\end{cases}$$
this function $f$ has the desired properties.
The answer to the first question is still yes even if you require the function to be continuous; for instance, take 
$$f(x)=\begin{cases}
x,&\text{if }x\le 0\\
0,&\text{if }0\le x\le 1\\
x-1,&\text{if }x\ge 1\;.
\end{cases}$$
Apart from the trivial example of a constant function, however, you cannot get such an example if $f$ is required to be a polynomial.
If $F$ is any finite set of real numbers, you can easily modify the last example to get a continuous function that has a whole interval of abscissae for each ordinate in $F$ and is one-to-one everywhere else. It's also possible to modify it to get examples for countably infinite sets $F$, but that’s just a bit tricky.
If you want $F$ to be a closed interval, you can do it with a polynomial of degree $5$. For instance, to get $F=[a,b]$ you can use a fifth degree polynomial with two local minima at height $a$ and two local maxima at height $b$. More generally, with a polynomial of degree $4n+1$ you can get $F$ to be the union of $n$ pairwise disjoint closed intervals.
A: The sine function keeps repeating.  It's a wave.  Therefore you have infinitely many abscissas for each ordinate that actually occurs.  (If the ordinate is more than $1$ or less than $-1$, then it doesn't "actually occur" (unless you bring in imaginary numbers).)  Look at it's graph and you'll see this.
