Math help please?

y = (1/2)^x …… [1]
y = 2^x …… [2]

[1] = [2] :
(1/2)^x = 2^x
[2^(-1)]^x = 2^x
2^(-x) = 2^x
-x = x
x + x = 0
2x = 0
x = 0

Substitute x = 0 into [2] :
y = 2^0
y = 1

Hence, (x, y) = (0, 1)

One tends to infinitely the other to zero

I will suggest an answer to your question even when I do not understand what you want.

The function y = ( 1/2 )^x can be written y = 2 ^ ( - x ) ............................... 1/2 = 2^( -1)

The graph of y = 2^ (-x) shows that as x -> oo , y -> 0; and when x -> - oo ; y -> 0. The graph of y = 2^x shows that when x => oo , y => oo and when x => - oo , y => 0.
The relationship between the two functions is : y = 2^ (-x) is the Reflexion of y = 2^x about the y-axis

If you chose the function y = SR ( x) whose domain is x = [ 0, oo ). then the function y = SR ( -x ) whose domain is: x = ( - oo , 0 ] have the same relationship as the two functions given. ............................................... SR = Square Root

I hope this the solution to the problem.

hi