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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?

I'm Ingrid
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    If the three points are not collinear, then there exists a parabola that passes through all three. – Scene Jun 08 '23 at 22:10
  • @Scene that's what I was thinking, but the example $(0,0)$, $(0, 2)$ and $(2, 2)$ does not seems to work, even though the points are not collinear, and that's what is making me confused. – I'm Ingrid Jun 08 '23 at 22:13
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    Don’t assume that all parabolas have to take the form $y=ax^2+bx+c$ – David Quinn Jun 08 '23 at 22:14
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    The three points must have distinct $x$-coordinates if you’re looking for the graph of a function $y=f(x)$. – Ted Shifrin Jun 08 '23 at 22:15
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    There's a bit of an issue here in that "points lying on a parabola" is being equated with "coordinates satisfing a quadratic function of the form $y=ax^2+bx+c$ (with $a\neq0$)". Since $(0,0)$ and $(0,2)$ have the same $x$-coordinate, they won't both satisfy a function. That said, in the grand scheme of things, not all parabolas are graphs of functions. See, eg, my answer here, which shows how three non-collinear points (such as your example) actually determine infinitely-many parabolas ... almost-none of which are graphs of functions. – Blue Jun 08 '23 at 22:19
  • @Blue This makes a lot of sense, thank you!! The picture helped a lot. But what I still don't get is, if the parabola isn't always the graph of a function, what is a better way to solve the problem? Should I just prove the points are collinear without mentioning quadratic functions, or is my approach still fine anyway? – I'm Ingrid Jun 09 '23 at 00:05
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    @Ingrid: I can't tell if the problem intends "parabola" to mean "graph of a quadratic function". (This is often so when general conics have yet to be introduced. Like how, in lower grades, "number" means "positive whole number" until negatives, fractions, etc, come along.) If so, then your approach is pretty-much the approach when given three arbitrary pts. That said, it's worth noting that the given pts are "obviously" collinear (especially if graphed), which would've saved you time. (BTW: All of this ignores that, in certain contexts, lines are considered "degenerate parabolas". ;) – Blue Jun 09 '23 at 02:34
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    @Blue Got it, thank you!! This cleared up my confusion. Is it ok to leave the question unanswered? I don't think it needs an answer anymore, since the comments already answered everything. – I'm Ingrid Jun 09 '23 at 03:05
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    By inspection, the two last points are translates by (1,1) of the first. It follows, that they are on a line. There are no lines with more than two points on a quadric, sincea quadratic equation has at most two zeros if moving or rotating its graph around in the plane. – Roland F Jun 09 '23 at 08:35

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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.

I'm Ingrid
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