On the symmetry of the origin: f(x)=-f(-x),

On the symmetry of y: f(x)=f(-x)

Axisymmetry of x: g(x)=-f(x), that is, when the x value is the same, the sign of the y value is opposite.

About linear symmetry, this is very troublesome. Let the linear equation be y=kx+b, the points (x 1, y 1) are points on f(x) and (x2, y2) are points on g(x), then when k ((x 1+x2)/.

Your last question is the first case, about the symmetry of the origin.

f(x)=lg((2+5x)/(2-5x))。

Verifiable f(x)=-f(-x)

The image is symmetrical about the origin, and f(x) is odd function.

note:

If you encounter something similar in the future, such as f (x) LG ((x+1)/(kloc-0/-x), there is no doubt that the image is symmetrical about the origin.

To say how to judge quickly, there is a simple method besides using the above identities.

That is to be familiar with the commonly used function images, such as the first y = 2 x and y = 2 (-x), which are all basic expressions of exponential functions, then we can immediately judge that they all pass through the fixed point (0, 1), and x takes a positive value and a negative value, so it is symmetrical about Y.

Typing for half a day, I hope it will help you!