Answer:
[tex]y(x)=(x-2)^2-2\text{ }\operatorname{\Rightarrow}\text{ Opt}\imaginaryI\text{on}\operatorname{\lparen}\text{A}\operatorname{\rparen}[/tex]Explanation: The provided is the graph of a parabola, therefore it can be modeled by the vertex form of the standard equation of the parabola, which is as follows:
[tex]\begin{gathered} y(x)=A(x-B)^2+C\Rightarrow(1) \\ \\ (B,C)=(x,y)\Rightarrow\text{ Vertex} \end{gathered}[/tex]According to the graph, the vertex of the parabola is:
[tex](x,y)=(2,-2)[/tex]Plugging it in equation (1) results in the following:
[tex]y(x)=A(x-2)^2-2[/tex]The value of the coefficient A can be determined by plugging in the coordinates of the point (4,2) in the above equation:
[tex]\begin{gathered} 2=A(4-2)^2-2 \\ \\ 4=A(2)^2 \\ \\ 4=4A \\ \\ A=1 \end{gathered}[/tex]Therefore the equation that describes the graph is as follows:
[tex]y(x)=(x-2)^2-2\Rightarrow\text{ Option\lparen A\rparen}[/tex]The graph confirmation: