In my experience, this is a very unusual question and I found it difficult.
Five obvious points on the f(a)=b graph are
(-4,4) (-1,0) (0,2) (2,-1) (4,-4)
I have graphed these points and the new g function is in red.
I am not sure what the correct notation is to name it....
Now
\(f(a)=b\qquad g(3a+1)=\frac{b}{2}\\ g(3a-1)=\frac{b}{2}\\ g(3a-1)=\frac{f(a)}{2}\\ let\;\;\;x=3a-1\\ then\;\;\; a=\frac{x+1}{3}\\~\\ g(x)=\displaystyle \frac{f(\frac{x+1}{3})}{2} \)
Check
\(If \;\;x=-4 \;\;then\;\; g(-4)=\frac{f(-1)}{2}=0\quad true\\ If \;\;x=-1 \;\;then\;\; g(-1)=\frac{f(0)}{2}=\frac{2}{2}=1\quad true\\ etc\)
I do not know about describing the transformation. That is all I have for now.
If you get an answer or work out how to add to this, please share it with me and with others that may also be interested.
LaTex:
f(a)=b\qquad g(3a+1)=\frac{b}{2}\\
g(3a-1)=\frac{b}{2}\\
g(3a-1)=\frac{f(a)}{2}\\
let\;\;\;x=3a-1\\
then\;\;\; a=\frac{x+1}{3}\\~\\
g(x)=\frac{f(\frac{x+1}{3})}{2}
