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At school we are studying the parabola and our teacher said that the formula for the axis of a parabola is $x=-\frac{b}{2a}$ without giving us the demonstration; so I tried to come up with a nice prove for the equation by myself.

Here's what I thought:

The axis of a parabola passes from his vertex $v$ that is the point equidistant from both the solution of the equation.

$$v=\frac{x_2-x_1}{2}\rightarrow v=\frac12\left(\frac{-b+\sqrt{b^2-4ac}+b+\sqrt{b^2-4ac}}{2a}\right)=\frac{\sqrt{b^2-4ac}}{2a}$$

Where did I go wrong?

PunkZebra
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  • It is plus, not minus: the middle points of the line segment determined by $;(a,b),,,,(x,y);$ is $$\left(\frac{a+x}2;,;;\frac{b+y}2\right)$$ – Timbuc Oct 09 '14 at 17:18

2 Answers2

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The middle point is with a plus, not a minus:

The solutions are

$$x_1=\frac{-b+\sqrt{\Delta}}{2a}\;,\;\;x_2=\frac{-b-\sqrt{\Delta}}{2a}\;\;,\;\;\;\Delta:=b^2-4ac\implies$$

$$\text{Middle point}\,:\;\;\frac{x_1+x_2}2=-\frac b{2a}$$

Timbuc
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  • yes but why? I know that the distance of two points in a plane is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and now $x_2- x_1$ – PunkZebra Oct 09 '14 at 17:20
  • The distance yes, not the middle point. Just try it with some simples on the $;x$- axis to convince yourself. :) – Timbuc Oct 09 '14 at 17:27
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The symmetry of the parabola $y=ax^2+bx+c$ about $x=-\frac{b}{2a}$ is evident from:

$$ax^2+bx+c=a\left(x-\left(-\frac{b}{2a}\right)\right)^2+c-\frac{b^2}{4a}.$$

JP McCarthy
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