 Why the graph of quadratic function is a parabola

Consider the function $f(x)=ax^2+bx+c$, where the coefficients $a,b,c$ are constant real numbers and $a$ is not zero. The extrema point of this function is located at $\left(\frac{-b}{2a},f\big(\frac{-b}{2a}\big)\right)$. Now to ease notation let us denote: $\begin{array}{ll} u=\frac{-b}{2a} \\ t=f\big(\frac{-b}{2a}\big) \end{array}$ Let $m$ be a real constant. We denote $y=t-m$ and $F=(u,t+m)$. Notice that $(u,t)$ is a point on the graph of $f$ and the distance from $(u,t)$ to the line $y$ is equal to the distance from $(u,t)$ to the point $F$. Now we can use Pythagorean theorem to calculate the distance from any point $(x,f(x))$ to the point $F$ and the distance from $(x,f(x))$ to the line $y$. If we can find a constant $m$ that satisfy the following equality: \begin{equation} \tag{1} \sqrt{(x-u)^2+\big(t+m-f(x)\big)^2} = \big| t-m - f(x)\big| \end{equation} then $f$ is a parabola. From the last equality we get: $\begin{array}{lll} (x-u)^2+\big(t+m-f(x)\big)^2 = \Big(t-m-f(x)\Big)^2 \\ (x-u)^2+(t+m)^2-2(t+m)f(x)+f^2(x)=(t-m)^2-2(t-m)f(x)+f^2(x) \\ (x-u)^2+4tm-4mf(x)=0 \\ x^2-2xu+u^2=4m\big(f(x)-t\big) \\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2} = 4m\big(ax^2+bx+c-\frac{b^2}{4a}+\frac{b^2}{2a}-c\big) \\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2} = 4m\big(ax^2+bx+\frac{b^2}{4a}\big) \\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2} = 4ma\big(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\big) \\ 1=4am \\ m=\frac{1}{4a} \end{array}$ So $m$ must satisfy equality (1). Hence, the graph of $f$ must be a parabola with focal $F=\left(\frac{-b}{2a},f(\frac{-b}{2a})+\frac{1}{4a}\right)$ and directrix $y(x)=f(\frac{-b}{2a})-\frac{1}{4a}$. We get $\begin{array}{ll} y(x)=c-\frac{b^2+1}{4a} \\ F=\left(\frac{-b}{2a},c-\frac{b^2-1}{4a}\right). \end{array}$ Applying the same method on the function $g(x)=x^4$ reveals that the graph of $g$ is not a parabola.