Step-by-step explanation:
To know the new coordinates for F', G', and H', we need to follow the rule, so:
![\begin{gathered} (x,y)_{}\to((3)/(4)x,(3)/(4)y) \\ F(-2,2)\to((3)/(4)\cdot-2,(3)/(4)\cdot2)=F^(\prime)((-3)/(2),(3)/(2)) \\ G(-2,-4)\to((3)/(4)\cdot-2,(3)/(4)\cdot-4)=G^(\prime)((-3)/(2),-3) \\ H(-4,-4)\to((3)/(4)\cdot-4,(3)/(4)\cdot-4)=H^(\prime)(-3,-3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ocgmu01mgmhblcicb09nxkfqe8ktghkpxy.png)
Then, to reflect over the x-axis, we need to use the following rule:
(x, y) ---> (x, -y)
So, the new coordinates F'', G'', and H" after the reflection are:
F'(-3/2, 3/2) ----> F"(-3/2, -3/2)
G'(-3/2, -3) -----> G"(-3/2, 3)
H'(-3, -3) --------> H"(-3, 3)
Therefore, the graph of figure F"G"H" is: