Assume that . Name a point that must be on the graph of .
So, for some "x," f (x) = 4 ..... and this "x" = 3
So "x/2" would be 3/2 .... and the "1/2" in front would make the original graph half as high at this point.
So...the point would be (3/2, 2)
Sorr, but this was wrong
If x = 6, then y = (1/2)f(6/2) = (1/2)f(3) = (1/2)*4 = 2
so (6, 2) is on the graph of y = (1/2)f(x/2)
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