I was trying to solve the following problem: Prove that a parabola does not pass through the points $(1, 2)$, $(3,4)$ and $(5,6)$.
My attempt: If these points lie on the same parabola, then we have a function $f(x) = ax^2 + bx + c$ such that
$f(1) = a + b + c = 2$,
$f(3) = 9a + 3b + c = 4$, and
$f(5) = 25a + 5b + c = 6$
Now if we do $f(3) - f(1)$, we find $2 = 8a + 2b$. Also if we do $f(5) - f(3)$, then we find $2 = 2b + 16a$. Then we have:
$f(3) - f(1) = f(5) - f(3)$
$8a + 2b = 16a + 2b$
$4a = 8a$
Which only makes sense if $a = 0$, so actually, this isn't a quadratic function, but a first degree one, so the three points are collinear and there is no parabola passing through them.
I was thinking that, everytime there's no parabola, then the points are collinear and I always would find $a = 0$ if I tried to solve it the way I did, but then I realized that if the points were $(0,0)$, $(0,2)$ and $(2,2)$, then my way of solving wouln't work anymore, because even though the points are not collinear, there isn't a parabola passing through them anyway. This is making me confused. When is there no parabola passing through three points? Is there a condition that should always be met? And what would be a better way of solving the question?
As stated in the comments, not all parabolas are graphs of functions. Since $(0,0)$ and $(0,2)$ have the same $x$-coordinate, they won't both satisfy a function, but this doesn't mean there's not a parabola passing through them. Three non-collinear points determine infinitely-many parabolas, so even if the points do not satisfy a function, they all lie on a parabola anyway.