If the function f (x) defined on R satisfies f (x + y) = f (x) + F (y) + 2XY (x, y belongs to R), f (1) = 2, then f (- 2) is equal to

If the function f (x) defined on R satisfies f (x + y) = f (x) + F (y) + 2XY (x, y belongs to R), f (1) = 2, then f (- 2) is equal to

f(0)=f(0+0)=f(0)+f(0)+0 =2f(0)
So f (0) = 0
f(2)=f(1+1)=f(1)+f(1)+2 = 2+2+2=6
And 0 = f (0) = f (2-2) = f (2) + F (- 2) - 8 = 6 + F (- 2) - 8 = f (- 2) - 2
So f (- 2) = 2

The function f (x) = (1 + 2x) / (1 + x), the image of function y = g (x) and the image of function y = f ^ - 1 (x + 1) are symmetric with respect to the straight line y = x, then the value of G (2) is equal to
(A) - 1 (b) - 2 (c) - 4 / 5 (d) - 2 / 5 please describe the process

Y = f ^ - 1 (x + 1) = (1 + X + 1) / [1 + 2 (x + 1)] = (x + 2) / (2x + 3) its symmetric image with respect to y = x is g (x), which is to find its inverse function (you know)
To require g (2) is to find the value of x where y = (x + 2) / (2x + 3) = 2
Solution equation (x + 2) / (2x + 3) = 2
x=-4/3
I can't find the answer,

Given the function y = a ^ x-3 - 2 (a > 0, a is not equal to 1), the image of constant crossing point P, P point coordinates are?

When x = 3, x-3 = 0, a ^ (x-3) = 1, which has nothing to do with a, then y = - 1
So the image passes through (3, - 1)