When looking at a parabola in the form $y = f(x) = a x^2 + b x + c$, I noticed that, if the vertex of a parabola is $(3,0)$, and a known point is $(4,2)$, then the point $(2,2)$ will also be on the plot.
To show why this is the case, I set $$f\left(x + \frac{b}{2a}\right) = f\left(x - \frac{b}{2a}\right)$$ and got the result $x =-\dfrac{b}{2a}$. What does this mean?
With $$f(x)=ax^2+bx+c$$ $$\to a(x^2+\frac ba x+\frac ca)$$ $$\to a\bigg[(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac ca\bigg]$$ $$\to a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}$$ Hence the minimum is found at $\bigg(-\frac{b}{2a}, c-\frac{b^2}{4a}\bigg)$
To prove the statement you are going for, i.e. showing symmetry about $x=-\frac{b}{2a}$, show that for some $\alpha$:
$$f(\frac{b}{2a}+\alpha)=f(\frac{b}{2a}-\alpha)=a\alpha^2+c-\frac{b^2}{4a}$$