# Proof of axis of symmetry equation [closed]

Because quadratic functions are symmetrical how do you prove the axis of symmetry equation.

$x=(-b/(2a))$

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## closed as off-topic by user91500, Claude Leibovici, Hakim, Grigory M, JSchlatherJun 5 at 7:08

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Suppose that the quadratic has equation $y=ax^2+bx+c$. By doing the algebra you can check that taking $$x=-\frac{b}{2a}+t\quad\hbox{and}\quad x=-\frac{b}{2a}-t$$ both give the same $y$ value. This shows that the graph of the quadratic is symmetric about the line $x=-b/(2a)$.

You can do the algebra in the obvious straightforward way, or to make it a little easier, rewrite the equation in the form $$y=a\Bigl(x+\frac{b}{2a}\Bigr)^2-\frac{b^2-4ac}{4a}\ .$$

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Given a quadratic equation (which describes the graph of a parabola) of a general form $ax^2+bx+c=0$, we can take the derivative and set it equal to $0$ to find the critical $x$-value that gives either the minimum or maximum of a parabola.