Question 1159886
<pre>
This is the graph of y = f(x)

{{{drawing(400,3200/11,-7,4,-4,4,grid(1),
circle(-6,0,.16), circle(-3,3,.16), circle(0,0,.16), circle(1,-1,.16), circle(3,-1,.16),
circle(-6,0,.1), circle(-3,3,.1), circle(0,0,.1), circle(1,-1,.1), circle(3,-1,.1)
)}}}

First we form the graph of f(3x) which, since 3 is greater than 1, SHRINKS the
graph HORIZONTALLY by a factor of the reciprocal of 3.  It doesn't affect the 
y-values, but divides the x-coordinates by 3

(-6,0) --> (-2,0) 
(-3,3) --> (-1,3)
 (0,0) -->  (0,0) 
(1,-1) --> (-1/3,3) 
(3,-1) --> (1,-1)

Here is the graph of y = f(3x):

{{{drawing(400,3200/11,-7,4,-4,4,grid(1),
circle(-6/3,0,.16), circle(-3/3,3,.16), circle(0/3,0,.16), circle(1/3,-1,.16), circle(3/3,-1,.16),
circle(-6/3,0,.1), circle(-3/3,3,.1), circle(0/3,0,.1), circle(1/3,-1,.1), circle(3/3,-1,.1)
)}}}

Next we form the graph of -f(3x) which REFLECTS the graph across the x-axis.  It
doesn't affect the x-values, but multiplies the y-coordinates by -1, which is to
say, it changes their signs.

(-2,0) --> (-2,0) 
(-1,3) --> (-1,-3)
 (0,0) -->  (0,0) 
(-1/3,-1) --> (-1/3,1) 
(1/3,-1) --> (1/3,1)

{{{drawing(400,3200/11,-7,4,-4,4,grid(1),
circle(-6/3,-0,.16), circle(-3/3,-3,.16), circle(0/3,-0,.16), circle(1/3,1,.16), circle(3/3,+1,.16),
circle(-6/3,-0,.1), circle(-3/3,-3,.1), circle(0/3,-0,.1), circle(1/3,1,.1), circle(3/3,1,.1)
)}}}

finally we form the graph of y=-f(3x)+1 which SHIFTS the graph 1 unit upward. It
doesn't affect the x-values, but ADDS 1 to all the y-coordinates. We relabel
this final graph as P(x) = -f(3x)+1

(-2,0) --> (-2,1) 
(-1,-3) --> (-1,-2)
 (0,0) -->  (0,1) 
(-1/3,1) --> (-1/3,2) 
(1/3,1) --> (1/3,2)

{{{drawing(400,3200/11,-7,4,-4,4,grid(1),
circle(-6/3,-0+1,.16), circle(-3/3,-3+1,.16), circle(0/3,-0+1,.16), circle(1/3,1+1,.16), circle(3/3,+1+1,.16),
circle(-6/3,-0+1,.1), circle(-3/3,-3+1,.1), circle(0/3,-0+1,.1), circle(1/3,1+1,.1), circle(3/3,1+1,.1)
)}}}

Edwin</pre>