



So on
the last page we had the equation:






Y
= 2X^{3}  3X^{2}  2X + 3






We
figured out that the graph of this 


crossed
the X axis when ... 








Since the
biggest exponent is a 3 and the number next to it is positive 


the graph
of this equation looks something like the graph of Y = X
^{3}. 








The
important thing is that the line goes up on the far right side, 


and the
line goes down on the far left side. 


Like the
graph of Y = X ^{3} does. 





So the
graph goes up out to the right and down out to the left. 


We also
know that Y = 0 when X = 1 and X = 1 and X = ^{3}/2. 


We can
draw in some pieces of the graph from this info ... 





Now just
connect all of the lines and dots ... 











And that
gets us very close to the real graph. 





We could
get even closer by figuring out a couple more points. 


Like say
where X = 0 (always a good one), 


and where
X = 1.25. 








WARNING!
WARNING! DANGER! DANGER! 


Even
though I drew those last two points 


as places
the graph turns around, 


they
might not be the exact places. 





The best
way to figure out where those places are 


is with
some Calculus. 


(Hey!
It's something to look forward to!) 





copyright 2005 Bruce Kirkpatrick 
