Let be $P_0(x_0,y_0)\in\mathcal P : \, y=ax^2+bx+c$. We know that the $m$ angular coefficient of the straight line tangent $t$ in $P_0$ is:
$$m=2ax_0+b \tag 1$$
To find the $(1)$ I before consider the system,
$$\begin{cases} y=ax^2+bx+c & \\ y-y_0=m(x-x_0)\end{cases} \iff ax^2+(b-m)x+c+mx_0-y_0=0 \tag 2$$
If $P_0\in\mathcal P$ we have $x=x_0$ it is a zero of the $(2)$; being $t$ is tangent to $\mathcal P$ in $P_0$
$$2x_0=x_0+x_0=-\frac{b-m}{a} \iff m=2ax_0+b$$
Without the use of the derivatives that would be immediate, is there a more immediate alternative proof?