



So what
do you do if someone hands you something like this to graph:






Y = 3X^{2}
 6X + 5






You can
make a T table like we did on the last page, 


but that
might take a long time. 





It would
be great if we could do something to this equation 


so we
could just look at it and see what the graph looked like. 





WE CAN! 





To do it,
we use a special type of completing the square. 


We just
use the 3X ^{2}  6X part of the equation 


and leave
the rest where it is. 








We will
do a special complete the square on this piece of the equation. 


There is
one special rule. 


Since we
are only working on a piece of the equation and not the whole thing, 


we can't
do anything that would change the value. 


That
means we can only add or subtract 0, 


and we
can only multiply or divide by 1. 





Hey, that
doesn't leave us a whole heck of a lot. 





No it
doesn't, but with a couple of new tricks we will pick up here, 


it will
be enough. 





We can't
divide through by 3 to get rid of the coefficient on the X
^{2}, 


but we
can factor it to the outside of a set of parenthesis. 





3X^{2}
 6X



3(X^{2}
 2X)






The part
inside the parenthesis is a good start 


on the complete the square
process. 


The next
step is to add a constant that is equal 


to half the coefficient on
the X term, squared. 





Half of
2 is 1, square that and you get 1. 


So all we
need to do is add 1. 


The
problem is, we are only allowed to add zero. 


OK, so we
add 1 and subtract 1, that adds up to zero... 





3(X^{2}
 2X + 1  1)






The first
three terms inside the parenthesis 


are what
we need to complete the square. 





3((X^{2}
 2X + 1)  1)



3((X^{2}
 1)^{2}  1)






Now we
get rid of the outer parenthesis, 


by
multiplying the inside terms by the three. 








Now we
are ready to put this piece back inside the equation. 











Combine
the 3 and the +5 and look what we have ... 





Y
= 3(X  1)^{2} + 2






Now we
can get somewhere, 


Since the
coefficient is positive, the "U" points up. 


Since the
coefficient is greater than 1, the "U" is narrow. 


Since
there is a 1 inside the parenthesis, the "U" is moved 1
unit to the right. 


Since
there is a + 2 outside the parenthesis, the "U" is moved 2
units up. 











Take a
minute and think about what we just did. 


I'm going
to go through it again, 


but this
time, I'm going to use the letters A, B, and C in the equation 


to stand
for any old numbers we might have. 





Y
= AX^{2} + BX + C






We zeroed
in on the AX ^{2} + BX part ... 











We
factored away the number that was next to the X ^{2} ... 











We took
the number that was next to the X (it's B/A) and divided it in half
... 











Then we
multiplied that times itself ... 








Then we
added and subtracted this number inside the parenthesis ... 











Then we
took the first three terms inside the parenthesis, 


and
factored them. 








Multiply
through by the A on the outside ... 








Simplify
the term on the right (cancel an A top and bottom) ... 











Then we
stuck this piece back into the equation ... 











From here
we can see what the parabola (the "U") does ... 





If A is
positive, the "U" goes up. 


If A is
negative, the "U" goes down. 


The
bottom point of the parabola (called the vertex) is at: 








copyright 2005 Bruce Kirkpatrick 
