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before I ask for anything I must admit I'm working hard to understand this beautiful subject. Thanks in advance. $$ f(x)= 2(x)^2+8x+5 $$ Acoording to the graph of this function, there is a x-axis symmetry. The problem is I can not prove it algebraically. Thanks again.

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Do this $$f(x) = 2(x^2 + 4x) + 5 = 2(x^2 + 4x + 4) + 5 - 8 = 2(x + 2)^2 - 3.$$ Note that $f$ is insensitive to the sign of $x + 2$ so $f$ is symmetric about the line $x = -2$

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    Thanks, this forum is a great learning source.2012-10-23
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$$f(x)=2\left(x^2+4x+\frac{5}{2}\right)=2((x+2)^2\pm...)$$

So it is symmetric on $x=-2$ which is parallel to $y$-axis.

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I think the requestor was looking for proof of the general expression. BTW -- this is a duplicate question (others have asked this in various other posts).

\noindent Let the point $x_0$ define the axis of symmetry. By definition, $x_0$ has the property that any deviation $\delta x$ from $x_0$, irrespective of whether it is positive or negative, will give you the same value of $y$. That is \begin{equation} y (x + \delta x) = y (x - \delta x) \end{equation} A positive deviation, $\delta x$, from $x_0$ can be expressed as \begin{equation} y^{+} = a (x_0 + \delta x)^2 + b (x_0 + \delta x) + c \end{equation} Similarly, a negative deviation, $- \delta x$, from $x_0$ can be expressed as \begin{equation} y^{-} = a (x_0 - \delta x)^2 + b (x_0 - \delta x) + c \end{equation} In order to find the value of $x_o$ for the axis of symmetry, we set these two expressions equal to one another \begin{equation} \begin{aligned} a (x_0 + \delta x)^2 + b (x_0 + \delta x) + c & = a (x_0 - \delta x)^2 + b (x_0 - \delta x) + c \nonumber \\ a (x_0^2 + 2 x_0 \delta + \delta^2) + b (x_0 + \delta ) + c & = a (x_0^2 - 2 x_0 \delta + \delta^2) + b (x_0 - \delta ) + c \end{aligned} \end{equation} Now we just simplify to get \begin{equation} \begin{aligned} 2 a x_0 \delta + b \delta & = - 2 a x_0 \delta - b \delta \\ 4 a x_0 \delta + 2 b \delta & = 0 \\ 4 a x_0 \delta = - 2 b \delta \\ x_0 = - \dfrac{2b}{4a} \nonumber \end{aligned} \end{equation} or \begin{equation} x_0 = - \dfrac{b}{2a} \label{eq:axis_of_symmetry_parabola} \end{equation}