For example, we know that one point has the $x$-value $50$ and the tangent has a slope of $30^\circ$ and a second point has the $y$-value $80$ and the tangent has a slope of $40^\circ$, how to find out, which quadratic function has these characteristics?
I've tried solving this with derivations:
$$2a\times50+b=\tan(30)$$
$$2ax_2 + b= \tan(40)$$
$$ax_2^2+bx_2+c=80$$
The general quadratic is $$f(x)=ax^2+bx+c\implies f'(x)=2ax+b$$ and its solutions are given by $$x=\frac{-b\pm\sqrt{b^2-4a(c-f(x))}}{2a}$$ Note that $f'(x)=\tan(\text{angle})$ so for the first one we plug in $x=50$ and $\text{angle}=30^\circ$ to get $$2a(50)+b=\tan(30^\circ)\implies 100a+b=\frac1{\sqrt3}\implies\boxed{b=\frac1{\sqrt3}-100a}$$
For the second one we plug in $x=\dfrac{-b\pm\sqrt{b^2-4a(c-80)}}{2a}$ and $\text{angle}=40^\circ$ to get $$\pm\sqrt{b^2-4a(c-80)}=\tan(40^\circ)\implies \boxed{c=\frac{b^2+320a-\tan(40^\circ)}{4a}}$$
So your desired quadratic is of the form $$f(x)=ax^2+\left(\frac1{\sqrt3}-100a\right)x+\frac{30000\sqrt3a^2+120(24-5\sqrt3)a+\sqrt3(1-\tan(40^\circ))}{12\sqrt3a}$$ for all real $a\neq0$.