Download COMPLETING THE SQUARE

COMPLETING THE SQUARE
Note the difference between using completing the square to find a solution to a
quadratic equation and completing the square to express a quadratic function in the form
of y = a ( x – h ) 2 + k , which allows one to graph the parabola using a series of
transformations on the basic graph of y = x 2.
Example 1:
Solve for x:
3 x 2 + 6x = 18
Divide both sides of the equation by 3.
x2 + 2x = 6
Create a trinomial square by adding 1 to
both sides of the equation.
x2 + 2x + 1= 6 + 1
Create the equivalent square of a binomial.
(x + 1)2 = 7
Take the square root of both sides of the
equation.
x + 1 = 
Solve for x.
x = -1 
7
7
Example 2:
Given f(x) = 3 x 2 + 6 x - 18
Express in the form of
f(x) = a ( x – h ) 2 + k.
f(x) = 3 ( x 2 + 2 x + ______ ) - 18 + ________
Create a trinomial square but
remember to keep the function
balanced (equivalent).
f(x) = 3 ( x 2 + 2 x + 1 ) - 18 - 3(1)
Create the square of the binomial.
f(x) = 3 ( x + 1 ) 2 - 21
Note: When graphing this function
the axis of symmetry will be x = - 1
and the vertex will be ( -1, - 21 ).
SOLVING QUADRATIC FUNCTIONS FOR ZEROS OF THE FUNCTION:
METHOD # 1:
f ( x) = x 2 + 2 x - 8
x2 + 2x - 8 = 0
(x+4)(x–2) = 0
x = -4,2
Let f(x) = 0
These are the x-intercepts, the Real
zeros of the function.
Therefore the coordinate of the x-intercepts are ( -4, 0 ) and ( 2, 0 ).
b
2

 1
2a
2 1
This is the x-value for the axis of
f( - 1 ) = ( - 1 ) 2 + 2 ( - 1 ) - 8 = - 9
symmetry. The axis of symmetry is
the line x = - 1.
 b
 b 
, f
The vertex is 
 .
 2 a 
 2a
Therefore, the vertex is ( - 1, - 9 ).
f(0) = (0)2 +2(0) - 8 = -8
The y-intercept is – 8. Therefore the
coordinates of the y-intercept are
( 0, - 8 ).
METHOD #2: COMPLETING THE SQUARE
f ( x) = x 2 + 2 x - 8
f (x) = x 2 + 2 x + 1 - 8 - 1
f (x) = (x + 1 ) 2 - 9
The function is now in the form of
f(x) = a ( x – h ) 2 + k, where
( h, k ) is the vertex and x = h is the
axis of symmetry.
Therefore, the axis of symmetry is x = - 1.
The vertex is ( - 1, - 9 ).
The parabola opens upward since “a” is positive and therefore the y value of the vertex is
a minimum value.
To find the zeros of the function, let f(x) = 0 and solve for x:
(x + 1)2 - 9 = 0
(x + 1)2 = 9
 x  1
2
 9
x + 1 = 3
x = - 1 3
x = - 4 or 2
If the zeros are real numbers, then they are the x-intercepts.
To find the y-intercept, let x = 0 and solve for f(0).
f (0) = (0) 2 + 2 (0) - 8 = - 8