I think I was clear I want to know how to figure out the formula of axis of simmetry in quadratic functions. I found I way to do it but that was assuming that there was an axis of simmetry in this functions. I think that what I am asking is too much but anyway probably some of you . If you want I´ll type here how I deduced the formula. Thanks in advance.
Well, this is what I did. Assuming that that quadratic equation or function is given by the general formula $f(x)=ax^2+bx+c$.
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 $x_1 \neq x_2$ but $f(x_1)=f(x_2)$. To make the typing job a bit easier $x_1=x$ and $x_2=d$
So $f(x)=ax^2+bx+c$ and $f(d)=ad^2+bd+c$ and $f(x)=f(d)$: $$ax^2+bx+c=ad^2+bd+c$$ $$ax^2+bx=ad^2+bd$$ $$ax^2 -ad^2= bd-bx$$ $$ a(x^2 -d^2)=b(d-x)$$ $$ a(x+d)(x-d)=-(x-d)b$$ $$a(x+d)=-b$$
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. Thanks!).
Let us talk about the two improvements that can be made.
1) If you want to prove a fact is true in general, there is no need to discuss the particular case of “Because we are talking about quadratic functions b can be 0. So if we (let) b=0, …..”.
2) From “ continuing with the calculations we´d have: $x + d = \dfrac {−b}{a}$”, we can arrive at the conclusion of $\dfrac {x+d}{2}=\frac {-b}{2a}$ immediately by dividing both sides by 2.
Regarding your query of “why there exists and axis of symmetry”, I try to explain it in the following way.
Firstly, it is a provable fact that $y =x^2$ is symmetric (and therefore the said axis exists). [The proof is more or less similar to what you did in the first part of your post.]
Next, by completing square, $f(x) = ax^2 + bx + c$ can be re-written as $f(x) = a(x – h)^2 + k$; where h and k are functions of a, b and c.
Note that (i) The “a” controls the opening (upward or downward) of the curve and the degree (how wide) of opening of the quadratic curve; (ii) The “h” and “k” are respectively horizontal and vertical shifts of the function. All these are under the topic of transformations.
In short, y = f(x) can be transformed equivalently to $y = x^2$.
Added:- In other words, any quadratic function is just the transformed image of the basic $y = x^2$ through (series of translations, reflections, amplifications...).