How to explain this formula : 2b-f(2a-t)=f(t) stating the condition under which the graph of a function is symmetric about the point (a,b)?

Question

[ Edited. The original question contained a mistake in the formula]

I'm reading a precalculus book telling me that if :

• f is a function

• (a,b) is a point

• t and 2a-t belong to the domain of f

then the grapf of f ( or, more precisely, the curve representing f) is symmetric relatively to the point (a,b) iff

                2b-f(2a-t) =  f(t). 

How to justify this formula analytically? Also, which intuitive/graphical explanation could be given for this formula ?

Remark. The book I am reading is some sort of AZ to highschool math that gives very few explanations, actually no explanation for the formula in question.

Answer 1

Take a look at the following picture :

Symmetry can be expressed by the fact that point $(a;b)$ is the midpoint of

$$(a-t,f(a-t)) \ \ \text{and} \ \ (a+t,f(a+t))$$

Thus, the arithmetic mean of the ordinates of these points must be equal to $b$ :

$$\dfrac12 (f(a-t)+f(a+t)) = b \ \ \ \iff \ \ \ f(a-t)=2b-f(a+t)\tag{1}$$

Setting $T=a-t \ \ \iff \ \ t=a-T$, we can transform (1) into the final relationship :

$$f(T)=2b-f(2a-T)$$

and not $f(T)=2b-f(2a-b)$ as was written.