Get equation for a curve which intersects x at seemingly randomly distributed points? Is there any type of function that when graphed would show a curve which intersects the x axis multiple times, with each point being an arbitrary distance from the last?
I mean, not like a trig function where each intersect is the same distance from the last. (2,0); (4,0); (6,0); (8,0). And not like a spiral where the distance gets bigger and bigger (or smaller) (2,0); (4,0); (8,0); (16,0);
But for example, some curve which intersects x-axis at (2,0); (6,0); (14,0); (15,0); (20,0); (122,0)...
Does that type function exist?
If so, is it possible to solve/get the equation, given only those intersect points?
I wouldn't need the exact equation of any particular curve. Just the equation of any curve that happens to intersect x-axis at whatever given arbitrary x values. Is that at least that possible to do?
 A: If you specify an arbitrary finite set of points $\{a_1, a_2, a_3, \dots, a_n\}$,  the polynomial function 
$$f(x)=(x-a_1)(x-a_2)(x-a_3) \cdots (x-a_n)$$
has the desired property.
A: If I'm interpreting your question correctly, you want to know if, given an increasing sequence of points $(a_n)$ in $\mathbb R$, we can define a continuous function $f:\mathbb R\to\mathbb R$ such that $f(x)=0$ if and only if $x=a_i$ for some $i$ (that is, the graph of $f$ intersects the $x$ axis precisely at the points $(a_i,0)$). The answer is yes, in fact there are infinitely many curves which will suffice. They might be hard to describe however. The simplest example I can think of is 
$$f(x)=(-1)^i\sin\left(\frac{(x-a_i)\pi}{a_{i+1}-a_i}\right)\text{ for } a_i\leq x\leq a_{i+1}$$
and $f(x)=0$ otherwise.
This is continuous but not differentiable, and has the advantage of actually crossing the $x$-axis at each $a_i$ rather than just touching it.
A: you can find the curves using matrix method by taking the columns as the x polynomials and multiply with the constant matrix which finally equated to given y. 
but on which does the degree of x which have to be taken. on what does that depend.
