Cardinality of an infinite set of functions

Question

There is a set A:

A = {a,b,c belongs to R | f(x) = ax^2 + bx + c, x belongs to R, a is not 0}

And set C is:

C = {f belongs to A | f[Q] contained in Q}

Is the set C from cardinality aleph or aleph nor?

This is what I came up with:

C is aleph:

1. f[Q] belongs to P(Q), thus C is aleph, as there are P(Q) different functions in this group
2. for h:P(Q) --> C, h(x) = f: R--> R, f[Q] = x, f(x) if x not belongs to Q, x. h is one to one thus has the same cardinality as P(Q)

C is aleph nor:

1. Both a, b, c must be from Q thus C is Q * Q * Q thus aleph nor
2. There can be functions in A where rationals will produce non rational, and specific sets of rationals are cannot be created using any type of function in A, such as Q itself.

I believe it to be aleph, but did not manage to find any proof to that. which cardinality does this set belongs to? And what is the proof for that?

Answer 1

Since$$c=f(0),\,a+b=f(1)-f(0),\,3a+b=f(2)-f(1)$$implies$$a=\frac{f(0)+f(2)}{2}-f(1),\,b=\frac{f(0)-f(2)}{2}+f(1),$$specifying the coefficients of $f\in C$ is equivalent to specifying arbitrary rational values of $f(0),\,f(1),\,f(2)$. So $|C|=|\Bbb Q|^3=|\Bbb N|$, which English-speaking mathematicians usually denote $\aleph_0$ so that $|R|=2^{\aleph_0}$.