Go forward one by one.
① Central symmetry
Let any 1 point (x 1, f(x 1)) on y=f(x) also have another corresponding point (2a-x 1, f(2a-x 1)) in function.
Then, the abscissa of the midpoint of these two points is
X is equal to (x1+2a-x1)/2 = a.
The ordinate of the midpoint is
Y equals [f (x1)+f (2a-x1)]/2 = b.
That is, f(x 1)=2b-f(2a-x 1).
Since x 1 is an arbitrary point, that is, for the function f(x)
f(x)=2b-f(2a-x)
So f(x) is symmetric about (a, b) →f(x)=2b-f(2a-x).
The above proof is a reverse proof of necessity.
② Axisymmetric
This is about the same. Let the function image have any x=x 1 and its corresponding x=2a-x 1.
Then the abscissa x of the midpoint of these two points is = (x1+2a-x1)/2 = a.
If the vertical coordinates of these two points are equal, that is, f(x 1)=f(2a-x 1).
Then the f(x) image is symmetric about x = a.
In the thinking part, f(x+a)=f(-x) indicates that the function is symmetrical about x=a/2; F(x+a)=f(x) indicates that the function has periodicity, in which one period t = a.