Let's find g(x) using the given points:
[tex]\begin{gathered} g(-2)=6 \\ g(1)=0 \\ so\colon \\ (x1,y1)=(-2,6) \\ (x2,y2)=(1,0) \end{gathered}[/tex]Let's find the slope:
[tex]m1=\frac{y2-y1}{x2-x1}=\frac{0-6}{1-(-2)}=\frac{-6}{3}=-2[/tex]Using the point-slope equation:
[tex]\begin{gathered} y-y1=m1(x-x2) \\ y-6=-2(x-(-2)) \\ y-6=-2x-4 \\ y=-2x+2 \\ g(x)=-2x+2 \end{gathered}[/tex]Since g(x) and h(x) are parallel, we can conclude:
[tex]\begin{gathered} m1\cdot m2=-1 \\ -2\cdot m2=-1 \\ m2=\frac{1}{2} \end{gathered}[/tex]Using the given point for h(x):
[tex]\begin{gathered} h(4)=-6 \\ so\colon \\ (x1,y1)=(4,-6) \end{gathered}[/tex]Using the point-slope equation again:
[tex]\begin{gathered} y-y1=m(x-x1) \\ y-(-6)=\frac{1}{2}(x-4) \\ y+6=\frac{1}{2}(x-4) \\ y+6=\frac{1}{2}x-2 \\ y=\frac{1}{2}x-8 \\ h(x)=\frac{1}{2}x-8 \end{gathered}[/tex]