Master Algebra 1 Word Problems with our Comprehensive Practice

How to figure out the formula of axis of simmetry in quadratic functions
Well, this is what I did. Assuming that that quadratic equation or function is given by the general formula $f(x)=a{x}^2+bx+c$.
So, assuming that there exist an axis of simmetry (eje de simetria en español), I had to find two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ so that $x_1\ne <!--\; \ne \; -->x_2$ but $f(x_1)=f(x_2)$. To make the typing job a bit easier $x_1=x$ and $x_2=d$.
So $f(x)=a{x}^2+bx+c$ and $f(d)=a{d}^2+bd+c$ and $f(x)=f(d)$:
$a{x}^2+bx+c=a{d}^2+bd+c$
$a{x}^2+bx=a{d}^2+bd$
$a{x}^2-<!--\; -\; -->a{d}^2=bd-<!--\; -\; -->bx$
$a(x^2-<!--\; -\; -->d^2)=b(d-<!--\; -\; -->x)$
$a(x+d)(x-<!--\; -\; -->d)=-<!--\; -\; -->(x-<!--\; -\; -->d)b$
$a(x+d)=-<!--\; -\; -->b$
Because we are talking about quadratic functions b can be 0. So if we $b=0$, we have:
$a(x+d)=0$
and again, if it is a quadractic function a can´t be 0, so we´d have that:
$x+d=0$ and $x=-<!--\; -\; -->d$, which is true. For example in the function $f(x)=x^2-<!--\; -\; -->16$, $f(4)=f(-<!--\; -\; -->4)=0$.
Well, continuing with the calculations we´d have:
$x+d=\frac{-<!--\; -\; -->b}{a}$
$d=\frac{-<!--\; -\; -->b}{a}-<!--\; -\; -->x$
$d=\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a}$
Because it´s an axis of simmetry, it´s in the middle, so I´ll have to use this formula:
$\frac{x_1+x_2}{2}$
So: $x_1+x_2=x+d$ and $d=\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a}$, so I would have.
$x+d=x+(\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a})$
$=\frac{xa}{a}+(\frac{-<!--\; -\; -->b-<!--\; -\; -->xa}{a})$
$=\frac{xa-<!--\; -\; -->b-<!--\; -\; -->xa}{a}$
$x+d=\frac{-<!--\; -\; -->b}{a}$
So:
$\frac{x+d}{2}=\frac{-<!--\; -\; -->b}{a}÷2$
$=\frac{-<!--\; -\; -->b}{2a}$
That´s what I´ve done so far, but what I probably want to know is why there exists and axis of symmetry (which I think will be the line $x=\frac{-<!--\; -\; -->b}{2a}$). I thought I could prove it using that fact but I got nowhere that´s why I posted this forum. (there might be some grammar mistakes, that´s because I am not an english speaker, you can correct where you think I made a mistake.

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Let $\text{Let}z=x+yi$
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