Symmetrical parabolas about a point

Question

Let the functions $f_1, f_2: \mathbb{R} \to \mathbb{R}$, where $f_1(x) = x^2 - 2x + 4$ and $f_2(x) = -x^2 + 6x - 10$. We denote $P_1$ and $P_2$ the graphs of the two functions in the same $XY$ plane, and $V_1$ and $V_2$ the vertices of the parabolas $P_1$ and $P_2$, respectively. Prove that $P_1$ and $P_2$ are symmetrical about the midpoint of $V_1V_2$.

Answer 1

The midpoint, as you have found, is at $M=(2,1)$.

Now we need to show that given any point $P=(a,b)$ on the first parabola there's a point $Q=(c,d)$ on the second parabola that is symmetrical to $P$ with respect to $M$. That is to say, $P+Q=2M$ (so $M$ is the midpoint of the line segment joining $P$ and $Q$), which we can rewrite as $Q=2M-P$, which is $(c,d)=2(2,1)-(a,b)=(4-a,2-b)$.

So, we are assuming $b=a^2-2a+4$ (that just says $P$ is on the first parabola), and we want to show that $d=-c^2+6c-10$ (which just says $Q$ is on the second parabola). We have already seen that $c=4-a$ and $d=2-b$, so we want to show $$ 2-b=-(4-a)^2+6(4-a)-10. $$ Now if you multiply everything out in that equation, and combine like terms, you get exactly $b=a^2-2a+4$, and we're done.