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What you do when you are given a quadratic function in standard form and you need it to be in vertex form.
Title: Sep 19 ­ 3:53 PM (1 of 9)
Example:
2
x + 6x + 12 = 15
2
1) Get x and x alone on one side of the equation
2
x + 6x = 15 ­ 12
2
x + 6x = 3
2) Off to the side, compute: 2
( )
b
2
Title: Sep 19 ­ 3:59 PM (2 of 9)
b = 6, so...
2
( )
6
2
2
(3)
9
3) Add this number to both sides of the equation
2
x + 6x + 9 = 3 + 9
*Remeber, you can do whatever you want to an equation as long as you do the same thing to both sides!
4) Factor on the left b
using , simplify on the right 2
2
x + 6x + 9 = 12
b
(x + )
2
2
2
*If the term is positive, add it, if it's negative, subtract it.
(x + 3) = 12
2
5) Move the "loose" term back to the left side of the equation
(x + 3) ­ 12 = 0
6) Find the vertex
(x + 3) ­ 12 = 0
2
Vertex: (­3, ­12)
Title: Sep 19 ­ 3:59 PM (3 of 9)
Example:
2
x + 3x ­ 8 = 0
b = 3
2
x + 3x = 8
3
2
2
( )
9
4
2
x + 3x + = 8 + 9
4
Keep this a fraction
Make this a decimal
9
4
2
x + 3x + = 8 + 2.25
2
3
(x+ ) = 10.25
2
2
3
(x+ ) ­ 10.25 = 0
2
Vertex:
Title: Sep 19 ­ 3:59 PM (4 of 9)
(­
3
2
, ­10.25
)
= 9
4
Example:
2
x ­ 6x + 12 = 9
Vertex: (3, ­6)
Title: Sep 19 ­ 3:59 PM (5 of 9)
Example:
2
3x ­ 18x + 8= 0
2
3(x ­ 6x ) + 8= 0
2
3(x ­ 6x ) = ­8
2
3(x ­ 6x + 9 ) = ­8 + 27
2
3(x ­ 3) = 19
2
3(x ­ 3) ­19 = 0
Vertex: (3, ­19)
Title: Sep 19 ­ 3:59 PM (6 of 9)
Example:
2
4x ­ 16x + 9= 0
Vertex: (2, ­7)
Title: Sep 19 ­ 3:59 PM (7 of 9)
Find the x­intercepts:
2
4x ­ 16x + 9= 0
2
4(x ­ 2) ­ 7= 0
2
4(x ­ 2) =7
2
(x ­ 2) = 7/4
(x ­ 2) = 7/4
+
x = 2 7/4
­
Title: Sep 19 ­ 3:59 PM (8 of 9)
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Title: Sep 19 ­ 3:59 PM (9 of 9)