The graph of the original function is:
The question asks us to find the value of the new graph.
In the orignal graph, when x = 0, y = 1 i.e:
[tex]\begin{gathered} y=2^x \\ \text{when,} \\ x=0 \\ y=2^0=1 \end{gathered}[/tex]This is the value for the original graph.
In the question we are asked, the value of y when x = 0 is not 1. This means that the graph has been shifted to the right.
Whenever a graph has been shifted to the right by n units, you must subtract n from the x-term of the function.
Let us put this into mathematics:
[tex]\begin{gathered} y=2^{x-n} \\ \text{where n is the number of units the graph was shifted to the right} \\ \\ \end{gathered}[/tex]In order to find the value of n, we need to study the new graph and substitute a coordinate point from it into the new equation above.
From studying the new graph, when x = 1, y = 1 as well. Hence, we can find n using this coordinate
[tex]\begin{gathered} y=2^{x-n} \\ \text{when,} \\ x=1,y=1 \\ 1=2^{1-n} \\ 1=2\times2^{-n} \\ \text{divide both sides by 2} \\ \frac{1}{2}=2^{-n} \\ 2^{-1}=2^{-n} \\ \\ \therefore-1=-n\text{ (Divide both sides by -1)} \\ n=1 \end{gathered}[/tex]Therefore, the new graph is:
[tex]\begin{gathered} y=2^{x-1} \\ y=2^x\times2^{-1} \\ y=0.5(2^x)\text{ (Option A)} \end{gathered}[/tex]
