I have a curve $y_1=ax^2+bx+c$ and another curve $y_2=dx^2+ex+f$ and a point $(x,y)$. This point is in between both curves and $y_1>y_2$. How do I calculate whether the point $(x,y)$ is closer to $y_1$ or closer to $y_2$?
First solution: I plugged in a given $x$ to find $y_1$ and $y_2$ at that $x$. Then I looked to see if $y_1-y>y-y_2$; if so, then the point is closer to $y_2$, otherwise the point is closer to $y_1$.
Is there any other method to calculate this? Maybe using vectors or inner products?
In a plane, the square of the distance between a point $(x_0,y_0)$ and a parabola $$y=ax^2+bx+c$$ is given by $$D^2=(x-x_0)^2+(y-y_0)^2=(x-x_0)^2+(ax^2+bx+c-y_0)^2\tag 1$$ To obtain the minimum distance, you need to solve for $x$ the simple equation $$\frac {d\,D^2}{dx}=0$$
Computing and rearranging gives the cubic equation $$2 a^2 x^3+3 a b x^2+ \left(2 a c-2 a y_0+b^2+1\right)x+(b c-b y_0-x_0)=0\tag 2$$ which does not make any problem. To avoid nasty results and formulae, use the trigonometric solution for three real roots or the hyperbolic solution for one real root (have a look here); never use Cardano method).
When $(2)$ is solved, go back to $(1)$ to have the distance.
Example
Consider the two parabolas $$y_1=x^2+2x+3 \qquad \text{and} \qquad y_2=4x^2+5x+6$$ and the point $(20,600)$.
So, for the first parabola, we need to solve first $$2x^3+6x^2-1189x-1214=0$$
Following the steps given in the linked page, we have three reals roots given by $$x_k=\sqrt{\frac{2390}{3}} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{63\sqrt 3}{1195 \sqrt{2390}}\right)\right)\right)-1$$ with $k=0,1,2$.
Converted to decimals, they are $$x_1=-25.435022 \qquad x_2=-1.0175732 \qquad x_3=23.452595$$
We do not have to be great wizards saying that $x_3$ is the solution to keep. So, we have $$\large\color{red}{D_1^2=11.92540}$$
Doing the same for the second parabola, we now need to solve $$32x^3+60x^2-4726x-2990=0$$ Three real roots again $$x_k=\frac{1}{4} \sqrt{\frac{9527}{3}} \cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(\frac{495\sqrt 3}{9527 \sqrt{9527}}\right)\right)-\frac{5}{8}$$ which are $$x_1=-12.823629 \qquad x_2=-0.62932980 \qquad x_3=11.577958$$ Same guess t get
$$\large\color{red}{D_2^2=70.93823}$$
Just conclude and $\cdots\cdots$ check the result !