Answer:
[tex]\large\boxed{f(x)=(1/2)2^x}[/tex]
Explanation:
You can just verify the outputs of the four functions for x = -1, x = 0.25, and x = 1.
1. [tex]f(x)=2(1/2)^x[/tex]
[tex]x=-1\implies f(x)=2(1/2)^{-1}=2(2)=4[/tex]
Then, this function does not go through (-1, 0.25)
2. [tex]f(x)=(1/2)2^x[/tex]
[tex]x=-1\implies f(x)=(1/2)(2)^{-1}=(1/2)(1/2)=1/4=0.25\\ \\ x=0\implies f(x)=(1/2)(2)^0=1/2=0.5\\ \\ x=1\implies f(x)=(1/2)(2)^1=(1/2)(2)=1[/tex]
Hence, this function goes through the 3 points.
Also, you can verify that it approches y = 0 in quadrant 2, because when x approaches a very large negative number ( - ∞), [tex]2^x[/tex] becomes very small ( approaches zero). Therefore, this function meets all the requirements: it approaches y = 0 in quadrant 2, increases into quadrant 2, and goes through the three given points.
