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I noticed something interesting while doing some homework on finding the vertex of a graph, and was wondering if anyone had an explanation for it.

Solving for $f(X) = A(X ^ 2) + BX$, I saw that for any numbers $A$ and $B$, plugging $X$ as the axis of symmetry would cause $A + B$ to equal $-A$

Random example: $f(X) = 7(X ^ 2) + 273X$

Plugging in the axis of symetry gives me $f(-19.5) = 7(-19.5 ^ 2) + 273(-19.5) = 2661.75 - 5323.5 = -2661.75$

I thought this seemed interesting :D

Yash Jain
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BOBONA
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  • you mean $f(-\frac{b}{2a}) = a\frac {b^2}{4a^2} - b\frac {b}{2a} = -\frac {b^2}{4a^2}$? – Doug M Feb 02 '18 at 02:08
  • I think, instead of $A + B = -A$, their example shows that they discovered that if $T$ is the $x$-coordinate of the axis of symmetry for a quadratic defined by $f(X) = AX^2 + BX$, we have that $f(T) = AT^2 + BT = -AT^2$ (it actually is a challenge to find nice-looking notation here) – pjs36 Feb 02 '18 at 02:31

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