Asked by Hayde
                When writing a function whose graph represents a combination of transformations, is the order not important?
            
            
        Answers
                    Answered by
            Reiny
            
    depends on the transformation.
e.g. suppose we have f: (x,y) ----> (x,-y) , a reflection in the x-axis , and
g: (x,y) ---> (-x,-y) , a reflection in the origin
if we do fg
(x,y) ---> (x,-y) ---> (-x,y)
and if we do gf
(x,y) = (-x,-y) ---> (-x,y) , we get the same result, so for that combination the order does not matter,
but,
if u: (x,y) --- (x+2, y-1), a translation
and v: (x,y) ---> (-x,-y) , a reflection in the origin
uv ---(x,y) ---> (x+2,y-1) ---> (-x-2, -y+1)
vu ---(x,y) ---> (-x,-y) ---> (-x+2, y+1) , which is totally different
So in the last case, the order does matter
    
e.g. suppose we have f: (x,y) ----> (x,-y) , a reflection in the x-axis , and
g: (x,y) ---> (-x,-y) , a reflection in the origin
if we do fg
(x,y) ---> (x,-y) ---> (-x,y)
and if we do gf
(x,y) = (-x,-y) ---> (-x,y) , we get the same result, so for that combination the order does not matter,
but,
if u: (x,y) --- (x+2, y-1), a translation
and v: (x,y) ---> (-x,-y) , a reflection in the origin
uv ---(x,y) ---> (x+2,y-1) ---> (-x-2, -y+1)
vu ---(x,y) ---> (-x,-y) ---> (-x+2, y+1) , which is totally different
So in the last case, the order does matter
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