Is every 2-nd degree polynomial symmetric about its extremum? I took the general 2-nd degree polynomial $P(x) = ax^2 + bx + c$, computed $P'(x_0) = 0 => x_0 = -b/2a$, but then $P(x_0 + d)$ and $P(x_0 - d)$ are not equal.
Is this evidence against my hypothesis? It seems like it is, however I might have done a mistake, and in any case my intuition tells me that symmetry indeed exists. Like, every 2-nd degree polynomial I have seen in my life was symmetric. Can someone confirm or disprove this?
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You're right, a parabola is symmetric with respect to vertical reflections about the line $x = -b/2a$.
In fact, the computation you suggested works...
$\begin{align*} P(x_0 + d) - P(x_0 - d) &= a((x_0+d)^2 - (x_0-d)^2) + b(2d) \\ &= a((x_0+d) - (x_0-d))((x_0+d) + (x_0 - d)) + 2bd \\ &= a(2d)(2x_0) + 2bd \\ &= -2bd + 2bd \\ &=0 \end{align*}$
We can prove this by writing $$P(x) = ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 -\frac{b^2}{4a^2} + c$$ Getting rid of the vertical shift and scaling, this is just $$\left(x+\frac{b}{2a}\right)^2$$ Which is indeed symmetric about the axis $x=-\frac{b}{2a}$