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If a quadratic in the form $y=ax^2+bx+c$ is reflected on the $y$ axis, why then must $b=0$, making the equation in the form of $y=ax^2+c$?

I can remember the rule, but the reasoning behind it has escaped me!

Thanks!

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Do you mean "symmetric on the y-axis"? If so, then it's because you want $f(x) = f(-x)$ for all $x$, which is impossible if $b \neq 0$. –  dsg Apr 7 '11 at 10:13
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The maximum of the generic $y = ax^2 + bx + c$ can be found via differentiation: $$x_\text{vertex}: 2ax_\text{vertex} + b = 0$$

If the shape is symmetric wrt the $y$ axis, then the vertex must lie on the $y$ axis itself, thus $x_\text{vertex} = 0$. This happens if and only if $b=0$.

Additionally, the intersections of the parabola with the $x$ axis must be such that $x_1 = -x_2 \implies$ $x_1 + x_2 = 0$, with $x_{1,2} = \frac{-b}{2a} \pm \overbrace{\frac{\sqrt{b^2-4ac}}{2a}}^\text{meh!}$. Thus, $x_1 + x_2 = \frac{b}{a}$, which is zero if and only if $b=0$.

Also see Wikipedia.

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Of course! Thanks! –  MathsStudent Apr 7 '11 at 11:35
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I think what you mean to say is that the quadratic is symmetric about the $y$-axis. What that implies formally is that $y(x) = y(-x)$. Try substituting the quadratic into that equation and see what happens.

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