SOLUTION: If I was given a graph of y=f(x) and the graph has points: (-4,0), (0,2), (2,-2), (3,0) connected in that order. How would I sketch the graph of y = f(x-1)? What about y=|f(x)|?

If you had, say this graph:





f(x-1) means that you are subtracting 1 from the value of the variable,
therefore you must choose x as 1 GREATER to compensate for the 1 that
is subtracted from the variable.  This may seem confusing, but if you
think about it, when you subtract from the variable, you must choose a 
larger value of x that will overcome the subtraction. Therefore you would 
add 1 to each x-coordinate and shift each point to the right by 1 unit and 
the new points would be (-3,0), (1,2), (3,-2), (4,0).  

A larger value of x must be chosen to produce the same value of y that x
would have produce if 1 had not been subtracted from the variable x. 
  


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What about y=|f(x)|?

The absolute value never becomes negative, and so any part of the graph 
that drops below the x-axis must be reflected above the x axis.  That
means change any negative y-coordinates to positive.  The points with
positive y-coordinates will not change.  So the points will be 
(-4,0), (0,2), (2,2), (3,0). All the poins above the x-axis will be the
same and in this case only the one point (2,-2) has a negative y value,
and so the others are the same and only (2,-2) changes to (2,2). Notice 
below that the portion of the original graph that contains (2,-2) got 
reflected across the x-axis, and the rest of the graph is the same as
the original.  If the original graph goes down below the x-axis to the
left of (-2,0), as this one does, then that part will be reflected above 
the x-axis.  IOW the part of the graph that falls below the x-axis will
be reflected above the x-axis.   






What about y=f|x|?

This will ignore anything on the left of the y-axis in the original graph,
since all negative values of x become positive, and so it amounts to not 
substituting any negative values of x in the original equation f(x), then
refecting what's on the right of the y-axis into the y-axis.  Note that 
the point(-4,0) is not used or reflected at all because it cannot be a value 
of |x| to produce any value for f(|x|) at all.  Also notice that (2,-2)
reflects across the y-axis into the point (-2,-2). The point that is on
the y-axis (0,2) reflects into itself, just as points on a mirror reflect
into themselves.  



Edwin