



There
are some graphing tricks that use the quadratic formula.



To
use them, we first have to set up the equation to look like this: 





Y = AX^{2}
+ BX + C






Remember
that A, B, and C are just placeholders 


that
stand for any old number we might have in the equation. 





Now
look at the mess of an equation we got at the end of the last page: 











The
place where the parabola turns around (called the vertex) is at: 











That
means if we have the equation: 





Y = 3X^{2}
 6X + 4






A
= 3 B = 6 C = 4



(Don't
forget the signs go with the numbers!)






The
place where the parabola turns around is where ... 








So
the graph turns around where X = 1. 





Where
is Y when X = 1? 





The
messy equation from the last page says that at that point ... 











It's
true. And you can find it that way if you want. 


But
you can also just plug X = 1 back into the original problem 


and
see what you get for Y. 






Y = 3X^{2} 
6X + 4 



Y = 3(1)^{2}
 6(1) + 4 



Y = 3  6 + 4 



Y = 1 





So
the graph vertex is at X = 1, Y = 1. 





Does
the graph go up or down? 


Is
the graph wide or skinny? 





Look
at the messy equation from the last page. 


The
number that answers those questions is what we are calling
"A." 


It's
the number next to the X ^{2}. 


In
this equation, that number is 3. 





That
means: 


3
is positive, so the graph goes up. 


3
is bigger than 1 so the graph is skinny. 


And
from the stuff we did before, 


we
know the vertex is at X=1, Y= 1. 





That's
enough info to draw the graph ... 








Let's
try another one ... 





Example: 





What
does the graph of Y = X ^{2} + 4X 3 look like? 





The
first thing to do is figure out what A, B, and C are for this
equation. 


B
= 4 and C = 3. Those are no problem, but what about A? 





Here's
the deal, if there is no number next to the X ^{2} or X 


the
number is 1. 


It's
really there, we just don't need to write a "times one." 


If
there is a minus sign next to the X, the number is 1. 


So: 


A
= 1 B = 4 C = 3






And
the X value at the vertex: 








To
find Y when X = 2, 


just
substitute 2 into the original equation for X. 






Y = (2)^{2}
+ 4(2) 3 



Y = 4 + 8 3 



Y = 1 





So
the vertex is at X = 2, Y = 1. 





OK,
does the parabola open up or down? 


A
= 1, so it opens down. 


1
is still a 1, so the graph is not too thin or wide. 





So
draw it ... 








copyright 2005 Bruce Kirkpatrick 
