I am a high school student who is beginning to look at proofs and I was wondering if this could be considered a proof for a property of multiplication of points on a parabola. I've seen this result before online but I haven't seen a formal proof of it using this method. This is my first time attempting a proof on my own so any advice would be appreciated.

Theorem: Given the function $f(x) = x^{2}$, the line which passes through points A and B which lie on the graph of $f$ has the $y$-intercept $-(A_{x}B_{x})$ if $A_{x} < B_{x}$

first it is clear that:

$$f(A_{x}) = A_{y} = A_x^{2} \quad \& \quad f(B_{x}) = B_{y} = B_x^{2}$$

The slope of the line joining A & B will be:

$$m = \frac{B_{x}^{2}-A_{x}^{2}}{B_{x}-A_{x}}$$

The numerator is a difference of squares; thus the whole expression can be simplified to:

$$m = B_{x}+A_{x}$$

Then choose one point (A or B) and sub this value into the equation of a line and solve for b:

$$A_{x}^{2} = (B_{x}+A_{x})A_{x}+b$$ $$-(A_{x}B_{x}) = b$$

and since $A_{x}$ will always be negative if non-zero and $B_{x}$ will always be positive if non-zero ; this is equivalent to saying $-A_{x}B_{x}$ QED.

  • $\begingroup$ We have $f(A_x) = A_y = A_x^2$. Is that what you mean by $A^2$? $\endgroup$ – user19405892 Dec 28 '15 at 21:37
  • $\begingroup$ To pick nits, this doesn't make sense: $\forall x \in \mathbb{R} \mid (A_{x} \leq 0 \wedge B_{x} \geq 0)$ because $A_x$ and $B_x$ are not functions of $x$. I believe you mean "if $A_x \le 0$ and $B_x \ge 0$". You further need the condition that $A_x$ or $B_x$ is nonzero. $\endgroup$ – Reinstate Monica Dec 28 '15 at 21:46
  • $\begingroup$ The proof appears watertight. In the statement of the theorem I would drop the $\forall x\in\mathbb R$ because it has no meaning (the index $x$ in $A_x$ indicates a component, not a functional dependency on $x$). I would also say that the points $A$ and $B$ lie on the graph of $f$, not on $f(x)$ which represents a value. $\endgroup$ – Justpassingby Dec 28 '15 at 21:46
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    $\begingroup$ Also, the theorem would be simpler if you just required $A_x \ne B_x$ and the theorem stated that the absolute value of the $y$-intercept of the line $AB$ is $|A_x B_x|$. $\endgroup$ – Reinstate Monica Dec 28 '15 at 21:48
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    $\begingroup$ Or just that the value of the $y$-intercept is $-A_xB_x$. $\endgroup$ – Alex S Dec 28 '15 at 21:50

This proof mostly looks good except for a few points. First, when you say $A$ is a point on $f(x)$, this doesn't mean much in standard mathematical vernacular. Usually, one would say, "$A$ is a point on the graph of $f$," or to be even more precise, $A\in\{(x,f(x)):x\in\mathbb R\}$. Next, you should always explain your notation. It didn't take me too long to determine that $A_x$ is the first coordinate of $A$, but it would be best for you, as the one introducing the notation, to explain this. When you say $f(A_x)=A_y=A^2$, this doesn't make sense because we cannot square the point $A$. What you mean is $f(A_x)=A_y=A_x^2$. Same for $B$. Finally, it is incorrect to state that the numerator is a perfect square. It most likely is not. What you mean is that the numerator is the difference of perfect squares.

One more thing: the hypothesis that $A_x\leq 0\leq B_x$ should be stated before the conclusion of the theorem. This will make your conclusion stand out better and make the statement of the theorem more clear.

You also need to deal with the degenerate case $A_x=B_x=0$ separately because $m$ is not well defined in this case.

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  • $\begingroup$ For the record, a suitable alternative to "$A$ is a point on $f(x)$" in this context is "$A$ is a point on the curve $y = f(x)$". $\endgroup$ – Unit Dec 30 '15 at 19:15

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