1. More about Graphing Quadratic Functions
Section 11.7
More about Graphing Quadratic Functions

2. New form
This section is still about graphing quadratic functions, but gives the function in a different format, the polynomial format f(x) = ax² + bx + c.
We will have two different ways to graph these function.
Taking the polynomial and completing the square so it becomes f(x) = ±a(x-h)² +k.
Creating definitions of words that are used to graph a polynomial.

3. Completing the Square Given f(x) = ax² + bx + c , complete the square.
This will be harder and different
f(x) = a(x² + [b/a]x ) + c
f(x) = a(x² + [b/a]x + b²/4a² ) + c – a(b²/4a²)
f(x) = a(x² + [b/a]x + b²/4a² ) + c – (b²/4a)
f(x) = a(x + b/2a)² + c – (b²/4a)
f(x) = a(x + b/2a)² + (4ac/4a) – (b²/4a)
f(x) = a(x + b/2a)² + (4ac-b²) / 4a

4. Completing the Square Let us fix f(x) = a(x + b/2a)² + (4ac-b²) / 4a
so it looks like f(x) = ±a(x-h)² +k
How do we do that?
We will use 2 let statements
Let h = -b/2a
Let k = (4ac-b²) / 4a
f(x) = a(x - [-b/2a] )² + (4ac-b²) / 4a
f(x) = a(x – h)² + k

5. Example Write f(x) = 2x² - 3x + 1 into f(x) = ±a(x-h)² +k

6. Graph the Previous Function

7. Example
Write f(x) = -4x² + x - 5 into f(x) = ±a(x-h)² +k

8. Terms/ Definitions
Section 11.6 used vertical and horizontal movements to graph. Those movements were called the vertex.
In Section 11.7 we are going to find that same vertex, through the completing the square
f(x) = a(x - [-b/2a] )² + (4ac-b²) / 4a
f(x) = a(x - h)² + k
Vertex (h , k) = (-b/2a , (4ac-b²) / 4a)

9. Example Given f(x) = 2x² - 3x + 1 find the vertex
Vertex (h , k) = (-b/2a , (4ac-b²) / 4a)
a = 2 b = -3 c = 1
h= -b/2a = --3 / [2(2)] = 3/4
k = (4ac-b²) / 4a) = [4(2)(1)-(-3)²] / [4(2)]
= [8 - 9] / 8 = -1 / 8

10. Example
Given f(x) = -4x² + x - 5 find the vertex

11. Third Option Use the definition for h h = -b / 2a
Then substitute value for h into all the x variables of the given function
The y variable will b displayed, which is the k
Put the two values together to create the vertex, (h, k)

12. Example Given f(x) = 2x² - 3x + 1 find the vertex h = -b/2a
a = 2 b = -3 c = 1
h= -b/2a = --3 / [2(2)] = ¾
F(3/4) = 2(¾)² – 3(¾) + 1
= 2(9/16)-3(¾) + 1 =(9/8)-(9/4)+1
= (9/8)-(18/8)+(8/8) = -1/8
Vertex = (¾, -1/8)

13. Example
Given f(x) = -4x² + x - 5 find the vertex

14. Other Definitions Axis of Symmetry
A vertical line the cuts the graph in half
x = h
X – Intercept
Points on the x -axis
Set y = 0 and solve for x. (x , 0)
Y – Intercept
Points on the y -axis
Set x = 0 and solve for y. (0 , y)

15. Example
Given f(x) = -4x² + x - 5 find the intercepts and the axis of symmetry

16. Homework
Section 11.7
#9, 13, 15, 21, 29, 31, 43, 47, 54