SOLUTION: What are the procedures to write a quadratic equation in vertex form?

First of all, learn what the vertex form is,
and how to graph the parabola from that form:

The vertex form of 

y = a(x - h)² + k

has vertex (h, k) and goes through the two points,

(h-1,k+a) and (h+1,k+a) 

Example 1:
Suppose you started with

y = 2x² - 12x + 22

1. Factor the coefficient of x² out of the first
   two terms, using a bracket, so you can put
   parentheses inside of it: 

   y = 2[x² - 6x] + 22

2. Out to the side or on scratch paper, complete
   the square by:
   (a) Multiplying the coefficient of x by .
        -6() = -3    

   (b) Squaring the result of (a).
        (-3)² = +9

3.  Add, then subtract, the result of 2.(b) inside
    the brackets:

    y = 2[x² - 6x + 9 - 9] + 22

4. Factor the binomial consisting of the first three
   terms inside the brackets:

    y = 2[(x - 3)(x - 3) - 9] + 22

5. If everything has gone right, the two factors of 
   that binomial will be the same, making it a 
   perfect square, so write it as a binomial squared:

   y = 2[(x - 3)² - 9] + 22

6. Remove the brackets by distributing, remembering
   to leave the parentheses intact:

   y = 2(x - 3)² - 18 + 22

7. Combine the last two numerical terms:

   y = 2(x - 3)² + 4

Compare that to

   y = a(x - h)² + k

Then a = 2, h = 3 and k = 4. So by the
above, the graph has vertex (h, k) = (3,4) and 
it goes through the two points,

(h-1,k+a) = (3-1,4+3) = (2,7)

and

(h+1,k+a) = (3+1,4+3) = (4,7)

So plot those three points:

 

and draw a U-shaped graph through them, called a parabola:

  

----------------------

Example 2:
Suppose you started with

y = -x² - 8x - 13

y = -(x + 4)² + 3

1. Factor the coefficient of x² out of the first
   two terms, using a bracket, so you can put
   parentheses inside of it: 

   y = -1[x² + 8x] - 13

2. Out to the side or on scratch paper, complete
   the square by:
   (a) Multiplying the coefficient of x by .
         8() = +4    

   (b) Squaring the result of (a).
        (+4)² = +16

3.  Add, then subtract, the result of 2.(b) inside
    the brackets:

    y = -1[x² + 8x + 16 - 16] - 13

4. Factor the binomial consisting of the first three
   terms inside the brackets:

    y = -1[(x + 4)(x + 4) - 16] - 13

5. If everything has gone right, the two factors of 
   that binomial will be the same, making it a 
   perfect square, so write it as a binomial squared:

   y = -1[(x + 4)² - 16] - 13

6. Remove the brackets by distributing, remembering
   to leave the parentheses intact:

   y = -1(x + 4)² + 16 - 13

7. Combine the last two numerical terms:

   y = -1(x + 4)² + 3

Compare that to

   y = a(x - h)² + k

Then a = -1, h = -4 and k = 3. So by the
above, the graph has vertex (h, k) = (-4,3) and 
it goes through the two points,

(h-1,k+a) = ( -4-1, 3+(-1) ) = (-5,2)

and

(h+1,k+a) = ( -4+1, 3+(-1) ) = (-3,2)


So plot those three points:

 
and draw a U-shaped graph through them, called a parabola. This is
an upside down U graph which will always be the case whenever "a",
the coefficient of x² is a negative number:

  

Edwin