Solution:
Given the graph below:
Given that graph A is a parent function of graph B, the equation of a parabola is expressed as
[tex]\begin{gathered} f(x)=a(x-h)^2+k\text{ ---- equation 1} \\ where \\ (h,k)\Rightarrow coordinate\text{ of the }vertex\text{ of the parabola} \end{gathered}[/tex]
In graph B, we have the coordinate of the vertex to be (4, -2).
This implies that
[tex]\begin{gathered} h=4 \\ k=-2 \end{gathered}[/tex]
By substituting into equation 1, we have
[tex]\begin{gathered} f(x)=a(x-4)^2+(-2) \\ \Rightarrow f(x)=a(x-4)^2-2----\text{ equation 2} \end{gathered}[/tex]
To solve for a, we selected a point on the graph B. By selecting the point (2, 2), we have f(x) to be 2, and x to be 2.
Thus, we have
[tex]\begin{gathered} 2=a(2-4)^2-2 \\ \Rightarrow2=4a-2 \\ add\text{ 2 to both sides of the equation} \\ 2+2=4a-2+2 \\ \Rightarrow4=4a \\ divide\text{ both sides by the coefficient of a, which is 4} \\ \frac{4}{4}=\frac{4a}{4} \\ \Rightarrow a=1 \end{gathered}[/tex]
Substitute the value of 1 for a into equation 2, we have the equation of the transformed function in the graph to be expressed as
[tex]f(x)=(x-4)^2-2[/tex]
Hence, the correct option is
