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


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

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2 Answers 2

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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Another way through taking the derivative of a quadratic equation (this probably only helps if you've studied calculus though).

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.

Let's take the derivative on both sides and see what happens: \begin{align} \frac{d}{dx} (ax^2+bx+c)&=\frac{d}{dx}0 \\ a\frac{d}{dx}x^2+b\frac{d}{dx}x+\frac{d}{dx}c&=\frac{d}{dx}0 \\ a\cdot 2x+b\cdot1+0&=0\\ 2ax+b&=0 \\ 2ax&=-b \\ x&=-\frac{b}{2a} \end{align}

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