Finding correct symmetry axis 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.
 A: 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 $x = -2$
A: $$f(x)=2\left(x^2+4x+\frac{5}{2}\right)=2((x+2)^2\pm...)$$
So it is symmetric on $x=-2$ which is parallel to $y$-axis.
A: 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}
