1. Splash Screen

2. Five-Minute Check (over Lesson 4–6) CCSS Then/Now New Vocabulary
Example 1: Write Functions in Vertex Form
Example 2: Standardized Test Example: Write an Equation Given a Graph
Concept Summary: Transformations of Quadratic Functions
Example 3: Graph Equations in Vertex Form
Lesson Menu

3. Find the value of the discriminant for the equation 5x 2 – x – 4 = 0.
B. –21
C. 21
D. 81
5-Minute Check 1

4. Find the value of the discriminant for the equation 5x 2 – x – 4 = 0.
B. –21
C. 21
D. 81
5-Minute Check 1

5. Describe the number and type of roots for the equation 5x 2 – x – 4 = 0.
A. 2 irrational roots
B. 2 real, rational roots
C. 1 real, rational root
D. 2 complex roots
5-Minute Check 2

6. Describe the number and type of roots for the equation 5x 2 – x – 4 = 0.
A. 2 irrational roots
B. 2 real, rational roots
C. 1 real, rational root
D. 2 complex roots
5-Minute Check 2

7. Find the exact solutions of 5x 2 – x – 4 = 0 by using the Quadratic Formula.
5-Minute Check 3

8. Find the exact solutions of 5x 2 – x – 4 = 0 by using the Quadratic Formula.
5-Minute Check 3

9. Solve x 2 – 9x + 21 = 0 by using the Quadratic Formula.
5-Minute Check 4

10. Solve x 2 – 9x + 21 = 0 by using the Quadratic Formula.
5-Minute Check 4

11. Solve 2x 2 + 4x = 10 by using the Quadratic Formula.
5-Minute Check 5

12. Solve 2x 2 + 4x = 10 by using the Quadratic Formula.
5-Minute Check 5

13. The function h(t) = –16t t + 6 models the height h in feet of an arrow shot into the air at time t in seconds. How long, to the nearest tenth, does it take for the arrow to hit the ground?
A. about 3.2 s
B. about 4.5 s
C. about 5.1 s
D. about 5.5 s
5-Minute Check 6

14. The function h(t) = –16t t + 6 models the height h in feet of an arrow shot into the air at time t in seconds. How long, to the nearest tenth, does it take for the arrow to hit the ground?
A. about 3.2 s
B. about 4.5 s
C. about 5.1 s
D. about 5.5 s
5-Minute Check 6

15. Mathematical Practices 7 Look for and make use of structure.
Content Standards
F.IF.8.a Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context.
F.BF.3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f (kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology.
Mathematical Practices
7 Look for and make use of structure.
CCSS

16. You transformed graphs of functions.
Write a quadratic function in the form y = a(x – h)2 + k.
Transform graphs of quadratic functions of the form y = a(x – h)2 + k.
Then/Now

17. vertex form
Vocabulary

18. A. Write y = x 2 – 2x + 4 in vertex form.
Write Functions in Vertex Form
A. Write y = x 2 – 2x + 4 in vertex form.
y = x2 – 2x + 4 Notice that x2 – 2x + 4 is not a perfect square.
y = (x2 – 2x + 1) + 4 – 1
Balance this addition by subtracting 1.
Example 1

19. y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square.
Write Functions in Vertex Form
y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square.
Answer:
Example 1

20. y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square.
Write Functions in Vertex Form
y = (x – 1)2 + 3 Write x2 – 2x + 1 as a perfect square.
Answer: y = (x – 1)2 + 3
Example 1

21. y = –3x2 – 18x + 10 Original equation
Write Functions in Vertex Form
B. Write y = –3x 2 – 18x + 10 in vertex form. Then analyze the function.
y = –3x2 – 18x + 10 Original equation
y = –3(x2 + 6x) + 10 Group ax2 + bx and factor, dividing by a.
y = –3(x2 + 6x + 9) + 10 – (–3)(9) Complete the square by adding 9 inside the parentheses.
Balance this addition by subtracting –3(9).
Example 1

22. y = –3(x + 3)2 + 37 Write x2 + 6x + 9 as a perfect square.
Write Functions in Vertex Form
y = –3(x + 3) Write x2 + 6x + 9 as a perfect square.
Answer:
Example 1

23. y = –3(x + 3)2 + 37 Write x2 + 6x + 9 as a perfect square.
Write Functions in Vertex Form
y = –3(x + 3) Write x2 + 6x + 9 as a perfect square.
Answer: y = –3(x + 3)2 + 37
Example 1

24. A. Write y = x 2 + 6x + 5 in vertex form.
A. y = (x + 6)2 – 31
B. y = (x – 3)2 + 14
C. y = (x + 3)2 + 14
D. y = (x + 3)2 – 4
Example 1

25. A. Write y = x 2 + 6x + 5 in vertex form.
A. y = (x + 6)2 – 31
B. y = (x – 3)2 + 14
C. y = (x + 3)2 + 14
D. y = (x + 3)2 – 4
Example 1

26. B. Write y = –3x 2 – 18x + 4 in vertex form.
A. y = –3(x + 3)2 – 23
B. y = –3(x + 3)2 + 31
C. y = –3(x – 3)2 – 23
D. y = –3(x – 3)2 + 31
Example 1

27. B. Write y = –3x 2 – 18x + 4 in vertex form.
A. y = –3(x + 3)2 – 23
B. y = –3(x + 3)2 + 31
C. y = –3(x – 3)2 – 23
D. y = –3(x – 3)2 + 31
Example 1

28. Which is an equation of the function shown in the graph?
Write an Equation Given a Graph
Which is an equation of the function shown in the graph?
Example 2

29. Write an Equation Given a Graph
Read the Test Item You are given a graph of a parabola. You need to find an equation of the parabola.
Solve the Test Item The vertex of the parabola is at (2, –3), so h = 2 and k = –3. Since the graph passes through (0, –1), let x = 0 and y = –1. Substitute these values into the vertex form of the equation and solve for a.
Example 2

30. Substitute –1 for y, 0 for x, 2 for h, and –3 for k.
Write an Equation Given a Graph
Vertex form
Substitute –1 for y, 0 for x, 2 for h, and –3 for k.
Simplify.
Add 3 to each side.
Divide each side by 4.
Example 2

31. The equation of the parabola in vertex form is
Write an Equation Given a Graph
The equation of the parabola in vertex form is
Answer:
Example 2

32. The equation of the parabola in vertex form is
Write an Equation Given a Graph
The equation of the parabola in vertex form is
Answer: The answer is B.
Example 2

33. Which is an equation of the function shown in the graph?B. C. D.
Example 2

34. Which is an equation of the function shown in the graph?B. C. D.
Example 2

35. Concept

36. Step 1 Rewrite the equation in vertex form.
Graph Equations in Vertex Form
Graph y = –2x 2 + 4x + 1.
Step 1 Rewrite the equation in vertex form.
y = –2x 2 + 4x + 1 Original equation
y = –2(x 2 – 2x) + 1 Distributive Property
y = –2(x 2 – 2x + 1) + 1 – (–2)(1) Complete the square.
y = –2(x – 1)2 + 3 Simplify.
Example 3

37. Step 3 Plot additional points to help you complete the graph.
Graph Equations in Vertex Form
Step 2 The vertex is at (1, 3). The axis of symmetry is x = 1. Because a = –2, the graph opens down and is narrower than the graph of y = –x 2.
Step 3 Plot additional points to help you complete the graph.
Answer:
Example 3

38. Step 3 Plot additional points to help you complete the graph.
Graph Equations in Vertex Form
Step 2 The vertex is at (1, 3). The axis of symmetry is x = 1. Because a = –2, the graph opens down and is narrower than the graph of y = –x 2.
Step 3 Plot additional points to help you complete the graph.
Answer:
Example 3

39. Graph y = 3x x + 8.
A. B.
C. D.
Example 3

40. Graph y = 3x x + 8.
A. B.
C. D.
Example 3

41. End of the Lesson