Algebra 2 Finding Parabolic Verticies by Completing the Square
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Another Factor
Finding Parabolic Verticies by Completing the Square

 

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

 

 Y = 3X2 - 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.
 

 3X2 - 6X

 3(X2 - 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(X2 - 2X + 1 - 1)

 
 The first three terms inside the parenthesis 
 are what we need to complete the square.
 

  3((X2 - 2X + 1) - 1)

  3((X2 - 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 = AX2 + 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

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