1. Quadratic Functions Vertex & Axis of Symmetry
Dr. Fowler  AFM  Unit 2-3
Quadratic Functions Vertex & Axis of Symmetry
Transformations
Maximizing Revenue

2. Graphing Quadratics – Introduction and Important Terms:

4. S

10. Without graphing, locate the vertex and axis of symmetry of the parabola defined by Does it open up or down?

11. Since a = 2 > 0 the parabola opens up and therefore will have no x-intercepts.

12. The domain of f is the set of all real numbers.

13. GRAPHING
Standard form: y = ax² + bx + c
Vertex form: y = a(x-h)² + k

14. From Standard to Vertex Form
• What's the pattern? + x 6 x2 6x 36
(x + 6)2
x2 + 12x + 36
• How about these?
x2 + 4x ______
(x _____ )2
+ 4
+ 2
x x ______
(x _____ )2
+ 25
+ 5
x2 – 14x ______
(x _____ )2
+ 49
– 7

15. From Standard to Vertex Form
• Converting from standard form to vertex form can be easy…
x2 + 6x + 9
(x + 3)2
x2 – 2x =
(x – 1)2
x2 + 8x =
(x + 4)2
x2 + 20x =
(x + 10)2
… but we're not always so lucky

16. From Standard to Vertex Form
• The following equation requires a bit of work to get it into vertex form.
y = x2 + 8x + 10
y = (x2 + 8x ) + 10
+ 16
– 16
16 is added to complete the square. 16 is sub-tracted to maintain the balance of the equation.
y = (x + 4)2 – 6
The vertex of this parabola is located at ( –4, –6 ).

17. From Standard to Vertex Form
• Lets do another. This time the x2 term is negative.
y = –x2 + 12x – 5
Un-distribute a negative so that when can complete the square
y = (–x2 + 12x ) – 5
y = –(x2 – 12x ) – 5
y = –(x2 – 12x ) – 5
+ 36
+ 36
The 36 in parentheses becomes negative so we must add 36 to keep the equation balanced.
y = – (x – 6)2 + 31
The vertex of this parabola is located at ( 6, 31 ).

18. From Standard to Vertex Form
• The vertex is important, but it's not the only important
point on a parabola
y-intercept at (0, 10)
x-intercepts at (1,0) and (5, 0)
Vertex at (3, -8)

19. Standard Form to Vertex Form
Write the equation in vertex form:
y = 3x2 – 6x + 8

20. Standard Form to Vertex Form
Write the equation in vertex form:
y = 3x2 – 6x + 8
y = 3(x2 – 2x ) + 8 factor out “a”

21. Standard Form to Vertex Form
Write the equation in vertex form:
y = 3x2 – 6x + 8
y = 3(x2 – 2x ) + 8 factor out “a”
y = 3(x2 – 2x + 1) + 8 – 3 complete the square

22. Standard Form to Vertex Form
Write the equation in vertex form:
y = 3x2 – 6x + 8
y = 3(x2 – 2x ) + 8 factor out “a”
y = 3(x2 – 2x + 1) + 8 – 3 complete the square
y = 3(x – 1) factor and combine

23. Example 1: Graph y = (x + 2)2 + 1
Step 1 Plot the vertex (-2 , 1)
Step 2 Draw the axis of symmetry, x = -2.
Step 3 Find and plot two points on one side, such as (-1, 2) and (0 , 5).
Step 4 Use symmetry to complete the graph, or find two points on the left side of the vertex.

24. Video Angry Birds Parabalos

25. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

26. Since a = 2 > 0 the parabola opens up.

27. The domain of f is the set of all real numbers.

28. Since a is negative, the parabola opens down.

29. The domain of f is the set of all real numbers.

30. Since a is negative, the graph of f opens down so the function will have a maximum value.

32. Excellent Job !!! Well Done