Showing a set of parabolas is uncountable I was given this question to prepare for an exam:
Show that the set of all functions $f(x)$ such that $f''(x)$ = -3 on ($-\infty$, $\infty$) is uncountable.
I know that this gives me a set of parabolas $f(x) = -\frac{3}{2}x^{2} + ax + b$, but I'm unsure of how to show this set is uncountable. I thought I could find a correspondence between the roots of the parabola and the real numbers, but there's no guarantee there are any roots. Would a and b determine a unique parabola, or is this an insufficient number of points? 
Could anyone give me a hint to figure this out, but not the solution?
edit: Thank you for your help, I have constructed the following bijection to show the set of these functions is equivalent to the real plane.
Let z : $\mathbb{R^2}$ $\to$ A, where A is the set of these functions and z(a,b) = $ -\frac{3}{2}x^{2} + ax + b$. Then z is clearly a bijection and thus A is uncountable.
 A: The parabola determined by $a,b$ as in your formula is unique. This is trivially established by the fact that a degree two polynomial is determined by its values at three points (and you have uncountably many to choose from). 
A: You could set $a = 0$, then the $y$-intercept of the parabola depends only on $b$ and gives a bijection with choices of $b$.  That two choices of $b$ give distinct parabolas is straightforward -- all of these are vertical translates of each other and all these parabolas are functions.  You have uncountably many distinct options for $b$, so there at least that many parabolas in the collection.
This is generally true -- the set of vertical translates of a function cover the vertical strip containing the domain of the function with uncountably many translated copies.  You can see this by picking a point in the domain and noticing that every pair with nonzero translation has distinct values at that point.
A: You have $f(x) = \frac{-3}{3}x^2+ax+b$. Note that every pair of $a,b$ gives you a different function $f$. Why? Because a polynomial function is uniquely determined by its coefficients i.e., if you have two polynomials $f$ and $g$ and they have same values every where,then they have the same coefficients. This is true for   polynomials with coefficients in an infinite field. So, now you have a bijection from $\mathbb R^2$ to your set of polynomials. Since, $\mathbb R^2$ is uncountable. Your set of polynomial is also uncountable.
