1. Splash Screen

2. Lesson Menu 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

3. Over Lesson 4–6 5-Minute Check 1 A.–81 B.–21 C.21 D.81 Find the value of the discriminant for the equation 5x 2 – x – 4 = 0.

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

5. Over Lesson 4–6 5-Minute Check 3 Find the exact solutions of 5x 2 – x – 4 = 0 by using the Quadratic Formula. A.1, – B. C.–1, D.

6. Over Lesson 4–6 A. B. C. D. 5-Minute Check 4 Solve x 2 – 9x + 21 = 0 by using the Quadratic Formula.

7. Over Lesson 4–6 5-Minute Check 5 Solve 2x 2 + 4x = 10 by using the Quadratic Formula. A. B. C. D.

8. Over Lesson 4–6 5-Minute Check 6 A.about 3.2 s B.about 4.5 s C.about 5.1 s D.about 5.5 s The function h(t) = –16t 2 + 70t + 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?

9. CCSS 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.

10. Then/Now 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.

11. Vocabulary vertex form

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

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

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

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

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

17. Example 1 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

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

19. Example 2 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. Write an Equation Given a Graph

20. Example 2 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. Write an Equation Given a Graph

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

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

23. Concept

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

25. Example 3 Graph Equations in Vertex Form Step 2The 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 3Plot additional points to help you complete the graph. Answer:

26. Example 3 Graph y = 3x 2 + 12x + 8. A.B. C.D.

27. End of the Lesson