|Completing the square is a process that can be used to solve an equation, find the vertex on a parabola, or to put an equation of a conic in standard form. On this page I am going to give several examples of completing the square that will allow us to find the vertex of a parabola.
BACKGROUND: The basic idea behind completing the square is that we are going to build an expression that will factor into a perfect square. Note that in the following,
x2 + 2bx + b2 = (x + b)2
x2 + 2bx + b2 is called a perfect square trinomial because it factors into (x + b)2, a squared binomial. In the process of completing the square we will build a perfect square trinomial that will factor into a squared binomial.
Note that in the above example the following is true:
First, I will give and example of completing the square and will then follow that with detailed steps of the process. In the following examples we will change the equation of a parabola (y = ax2 + bx + c) into the form y = a( x - h) 2 + k. Determining the vertex from this form will be discussed at the end of the discussion of completing the square.
EXAMPLE 1: Complete the square on x: y = x2 + 6x + 2
Add and subtract the square of (1/2) of the coefficient of the linear term
Now, here is a summary of the steps of the process:
EXAMPLE 2: Complete the square on x: y = 2x2 + 12x- 5
(1) The coefficient of the squared term, x2, must be 1
If the coefficient is not 1 then factor it out of the two variable terms
y = 2(x2 + 6x) - 5
(2) Add and subtract the square of (1/2) of the coefficient of the linear term
y = 2(x2 + 6x + 32 - 32) - 5
(3) Separate the perfect square trinomial from the constant terms
y = 2(x2 + 6x + 32) - 32(2) - 5
(4) Factor the perfect square trinomial
y = 2(x + 3)2 - 32(2) - 5
(5) Add the constant terms
y = 2(x + 3)2 - 23
You should take particular notice that in step (3) above, the constant term that came out of the parentheses had to be multiplied by 2.
EXAMPLE 3: Complete the square on t: x = -3t 2+ 3t - 1
x = -3( t 2- t) - 1
This process can be used even with numbers that would be considered unwiedy without a calculator - you just follow the same steps and pattern shown above. Here is such an example.
EXAMPLE 4: Complete the square on x: y = 42.94x 2+ 62.34x - 12.31
y = 42.94 (x 2+ 1.45x) - 12.31 [1.45 = 62.34/42.94]
Recall that in the above examples we changed the equation of a parabola (y = ax2 + bx + c) into the form y = a( x - h) 2 + k.In this latter form the vertex is identified as the point with coordinates (h,k). Here is why:
A parabola that opens upward or downward is either going to have a minimum value of y (if the parabola opens upward) or a maximum value of y (if the parabola opens downward).If a > 0 then the parabola opens upward and there is a minimum value for y. This will occur when a( x - h) 2 = 0 for that is the smallest value that a (x - h) 2 can take on.a( x - h) 2 = 0
If a < 0 then the parabola opens downward and there is a maximum value for y. This will occur when a( x - h) 2 = 0 for that is the largest value that a(x - h) 2 can take on (recall that a < 0).a ( x - h) 2 = 0
Now let's go back to each of the previous four examples and identify the vertex.
FROM EXAMPLE 1: y = (x + 3)2 - 7
Vertex ( - 3, - 7)
FROM EXAMPLE 2: y = 2(x + 3)2 - 23
Vertex ( - 3, - 23)
FROM EXAMPLE 3: x = -3( t - .5 )2 - .25
Vertex (.5, - . 25) [assuming that the ordered pairs are in the form, (t,x)]
FROM EXAMPLE 4: y = 42.94 (x + .73 ) 2 - 35.19
Vertex: ( - .73, - 35.19)
© 2000 Jo Steig