```Question 118862
What are the procedures to write a quadratic equation in vertex form?
<pre><font size = 4><b>
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 {{{1/2}}}.
-6({{{1/2}}}) = -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:

{{{drawing(400,400,-3,10,-3,10,
graph(400,400,-3,10,-3,10),
locate(1.63,7.3,o),locate(4.07,7.3,o), locate (2.85,4.32,o) )}}}

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

{{{drawing(400,400,-3,10,-3,10,
graph(400,400,-3,10,-3,10,2(x - 3)^2 + 4),
locate(1.63,7.3,o),locate(4.07,7.3,o), locate (2.85,4.32,o) )}}}

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

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 {{{1/2}}}.
8({{{1/2}}}) = +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:

{{{drawing(400,400,-10,3,-5,5,
graph(400,400,-10,3,-5,5),
locate(-5.1,2.2,o),locate(-3.1,2.2,o), locate(-4.1,3.22,o) )}}}
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:

{{{drawing(400,400,-10,3,-5,5,
graph(400,400,-10,3,-5,5,-1(x + 4)^2 + 3),
locate(-5.1,2.2,o),locate(-3.1,2.2,o), locate(-4.1,3.22,o) )}}}

Edwin</pre>```