Question 66204
<pre><font size = 5><b>
the graph of y=f(x). graph each of the following
y=absolutevalue(f(x))
y=-f(x)
y=f(-x)
y=f(x+1)+1

You forgot to give the equation for f(x).

Suppose this red graph is the graph of f(x)

{{{ graph( 200, 200, -10, 10, -10, 10, .1(x-2)^3 + 2) }}} 

Then this green graph is f(-x) 

{{{ graph( 200, 200, -10, 10, -10, 10, .1(x-2)^3 + 2, .1(-x-2)^3 + 2) }}}

Notice that the green graph f(-x) is the 
reflection of the red graph in the y-axis.

The rule: replacing x by -x in the right 
side of the equation for f(x) results in 
a reflection of the graph into (or across)
the y-axis

-----------------------------

y = f(x+1)+1

If the red graph below is of f(x)

{{{ graph( 200, 200, -10, 10, -10, 10, .1(x-2)^3 + 2) }}} 

Then the green graph below is f(x+1) + 1

{{{ graph( 200, 200, -10, 10, -10, 10, .1(x-2)^3 + 2, .1(x+1-2)^3 + 2+1) }}} 
 
Two things were done to f(x).  First x was 
replaced by (x+1) which moves the graph LEFT 
by 1 unit.

That is y = f(x+1) would have looked like this:

{{{ graph( 200, 200, -10, 10, -10, 10, .1(x-2)^3 + 2, .1(x+1-2)^3 + 2) }}}

Rule: Replacing x by x+p moves the graph p 
unit left.  Replacing x by x-p moves the 
graph p units right.

Then adding +1 to that raised the green 
graph 1 unit

{{{ graph( 200, 200, -10, 10, -10, 10, .1(x-2)^3 + 2, .1(x+1-2)^3 + 2+1) }}}

Rule: Adding +p to the right side moves the
graph p units up. Adding -p to the right
side moves the graph p units down.

Edwin</pre>