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.