Graph a quadratic function from the vertex form Complete the square to transform a quadratic function to vertex form Derive the vertex formula

1. Quadratic Functions & Models
Chapter 2 Polynomial and Rational Functions

2. Concepts and Objectives
 Graph a quadratic function in vertex form
 Graph a quadratic function by completing the square
 Identifying the axis of symmetry and vertex of a
parabola using the vertex formula

3. Quadratic Functions
 A function f is a quadratic function if
where a, b, and c are real numbers, and a  0.
 The graph of a quadratic function is a parabola whose
shape and position are determined by a, b, and c.
   2
f x ax bx c

4. Vertex Form
 The graph of gx = ax2 is a parabola with vertex at the
origin that opens up if a is positive and down if a is
negative. The magnitude (or absolute value) of a
determines the width of the parabola.
 The vertex form of a quadratic function is written
 The graph of this function is the same as that of gx
translated h units horizontally and k units vertically.
This means that the vertex of F is at h, k and the axis of
symmetry is x = h.
     
2
F x a x h k

5. Vertex Form (cont.)
 Example: Graph the function and give its domain and
range.
      
21
4 3
2
F x x

6. Vertex Form (cont.)
 Example: Graph the function and give its domain and
range.
Compare to : h = 4 and k = 3 (Notice
the signs!)
Vertex: 4, 3, axis of symmetry x = 4
We can graph this function by graphing the base
function and then shifting it.
      
21
4 3
2
F x x
     
2
F x a x h k

7. Vertex Form (cont.)
 Example, cont.:
Let’s consider the graph of
 Vertex is at 0, 0
 Passes through 2, ‒2 and
4, ‒8.
 (I picked 2 and 4 because of
the half.)
   21
2
g x x

8. Vertex Form (cont.)
 Example, cont.:
To graph F, we just shift everything over 4 units to the
right and 3 units up.
Domain: ‒, 
Range: ‒, 3] or y  3

9. Completing the Square
 If we are given a function that is not in vertex form, we
can “complete the square” to transform it into vertex
form. We do this by taking advantage of the additive
identify property (a + 0 = a).
 For example, the function is not a
binomial square. We can add 0 in the form of 52 – 52
(5 is half of 10), and group the parts that factor to a
binomial square:
     22 2
10 305 5f x x x
   2
10 30f x x x
     2 2 2
10 5 5 30x x
   
2
5 5x

10. Completing the Square (cont.)
 Example: What is the vertex of the function?
   2
6 7f x x x

11. Completing the Square (cont.)
 Example: What is the vertex of the function?
The vertex is at 3, ‒2.
   2
6 7f x x x
    22 2
3 736x x
    
2
3 9 7x
   
2
3 2x
6
3
2


12. Practice
 Find the vertex by completing the square.
1. fx = x2 +8x + 5
2. gx = x2 – 5x + 8
3. hx = 3x2 + 12x – 5

13. Practice (cont.)
1. fx = x2 +8x + 5
The vertex is at ‒4, ‒11.
      22 2
8 4 4 5f x x x
    
2
4 16 5x
   
2
4 11x

14. Practice (cont.)
2. gx = x2 – 5x + 8
The vertex is at
 
 
    

   
    
  
2
2 2
5
2 2
5 8
5
g x x x
 
    
 
2
5 25
8
2 4
x  
 
 
32
8
4
 
    
 
2
5 25 32
2 4 4
x
 
   
 
2
5 7
2 4
x
 
 
 
5 7
,
2 4

15. Practice (cont.)
3. hx = 3x2 + 12x – 5
The vertex is at ‒2, ‒17.
   2
4 53h x x x  
   2 2 2
5243 3 2x x    
    
2
3 2 12 5x
   3 2 17x

16. Vertex Formula
 You may have noticed that we are doing the same
process each time we complete the square to produce
the vertex form of the function. We can generalize this
to create a formula for the vertex of a parabola.
 Starting with the general quadratic form, we can
manipulate it by completing the square and end up with
a version that gives us a formula for the coordinates of
the vertex.

17. Vertex Formula (cont.)
 Deriving the vertex formula:
  2
f x ax bx c  
2 2
2 1 1
2 2
b b b
a x x c a
a a a
    
             
2 2
2 4
b b
a x c
a a
 
    
 

18. Vertex Formula (cont.)
 Comparing this to the vertex form of
shows that
and
 We can simplify this further by substituting h for x:
 So for any quadratic function,
   2
f x a x h k  
2
b
h
a
 
2
4
b
k c
a
 
   2
f h a h h k k   
  2
,f x ax bx c  
 , , and is the axis of symmetry.
2
b
h k f h x h
a
   

19. Classwork
 College Algebra
 Page 313: 14-20 (even), page 242: 22-26, 36-38
(even), page 226: 46-50, 54-58 (even)