Explanation:
To know the new coordinates for F', G', and H', we need to follow the rule, so:
[tex]\begin{gathered} (x,y)_{}\to(\frac{3}{4}x,\frac{3}{4}y) \\ F(-2,2)\to(\frac{3}{4}\cdot-2,\frac{3}{4}\cdot2)=F^{\prime}(\frac{-3}{2},\frac{3}{2}) \\ G(-2,-4)\to(\frac{3}{4}\cdot-2,\frac{3}{4}\cdot-4)=G^{\prime}(\frac{-3}{2},-3) \\ H(-4,-4)\to(\frac{3}{4}\cdot-4,\frac{3}{4}\cdot-4)=H^{\prime}(-3,-3) \end{gathered}[/tex]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:
