Set of vertices of $f(x) := \frac{(a - x)(x - b)}{2}$ trace out $-f(x)$?

Question

Consider the parabola $$f_{a,b}(x) := \frac{(x - a)(b - x)}{2},$$ which has the vertex $\frac{a + b}{2}$ for some fixed parameters $a,b \in \mathbb{R}$, i.e. $a = b = 1$ such that $f_{1,1} = \frac{(x - 1)(1 - x)}{2}$. I now plotted $-f_{1.1}(x) = \frac{(x - 1)^2}{2}$.

Next I defined the vertex point $\left( \frac{a+b}{2}, f_{a,b}\left(\frac{a+b}{2}\right)\right)$ as a function of the two parameters $a,b \in \mathbb{R}$ and noticed that when I leave one of the parameters at 1 and change the other parameter, the vertex traces out the graph of $-f_{1,1}$. Why is this the case?

My Ideas. The above observation implies $$ \left\{ \left( \frac{a+b}{2}, f_{a,b}\left(\frac{a+b}{2}\right)\right): a = 1,b \in \mathbb{R}\right\} = \left\{ \left(x, \frac{(x - 1)^2}{2}\right): x \in \mathbb{R}\right\}. $$ The LHS is equal to $$ \left\{ \left( \frac{1+b}{2}, \frac{(1-b)^2}{8}\right): b \in \mathbb{R}\right\} $$ Now setting $x := \frac{1 + b}{2}$ gives $$ \frac{(x - 1)^2}{2} = \frac{(1 - b -2)^2}{8} = \frac{(b + 1)^2}{8} \ne \frac{(1 - b)^2}{8}. $$ Where's my mistake?

Answer 1

Both sets are equal: if $x = \frac{1 + b}{2}$ for some $b \in \mathbb R$, then $$ \frac{(x - 1)^2}{2} = \frac{1}{2}\left(\frac{b - 1}{2}\right)^2 = \frac{(b - 1)^2}{8}. $$

Answer 2

$$f(x)=-x^2+(a+b)x-ab \Rightarrow f(\pm \infty) =-\infty, x_0=(a+b)/2$$ Where $x_0$ is the position of max so $$f_{max}=f_0=f[(a+b)/2]=\frac{(a+b)^2-4ab}{4}=\frac{(a-b)^2}{4}.$$ It is a vertical down waard parabola which cuts the x-axis at $x=a,b$.