1. 9-3 Graphing Quadratic Functions Warm Up Warm Up Lesson Presentation Lesson Presentation California Standards California StandardsPreview

2. 9-3 Graphing Quadratic Functions Warm Up Find the axis of symmetry. 1. y = 4x 2 – 72. y = x 2 – 3x + 1 3. y = –2x 2 + 4x + 34. y = –2x 2 + 3x – 1 Find the vertex. 5. y = x 2 + 4x + 56. y = 3x 2 + 2 7. y = 2x 2 + 2x – 8 x = 0 x = 1 (–2, 1) (0, 2)

3. 9-3 Graphing Quadratic Functions California Standards 21.0 Students graph quadratic functions and know that their roots are the x-intercepts. 23.0 Students apply quadratic equations to physical problems, such as the motion of an object under the force of gravity.

4. 9-3 Graphing Quadratic Functions Recall that a y-intercept is the y-coordinate of the point where a graph intersects the y-axis. The x-coordinate of this point is always 0. For a quadratic function written in the form y = ax 2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is c.

5. 9-3 Graphing Quadratic Functions In the previous lesson, you found the axis of symmetry and vertex of a parabola. You can use these characteristics, the y-intercept, and symmetry to graph a quadratic function.

6. 9-3 Graphing Quadratic Functions Additional Example 1: Graphing a Quadratic Function Graph y = 3x 2 – 6x + 1. Step 1 Find the axis of symmetry. = 1 The axis of symmetry is x = 1. Simplify. Use x =. Substitute 3 for a and –6 for b. Step 2 Find the vertex. y = 3x 2 – 6x + 1 = 3(1) 2 – 6(1) + 1 = 3 – 6 + 1 = –2= –2 The vertex is (1, –2). The x-coordinate of the vertex is 1. Substitute 1 for x. Simplify. The y-coordinate of the vertex is –2.

7. 9-3 Graphing Quadratic Functions Additional Example 1 Continued Step 3 Find the y-intercept. y = 3x 2 – 6x + 1 The y-intercept is 1; the graph passes through (0, 1). Identify c.

8. 9-3 Graphing Quadratic Functions Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 1, choose x- values less than 1. Let x = –1. y = 3(–1) 2 – 6(–1) + 1 = 3 + 6 + 1 = 10 Let x = –2. y = 3(–2) 2 – 6(–2) + 1 = 12 + 12 + 1 = 25 Substitute x-coordinates. Simplify. Two other points are (–1, 10) and (–2, 25). Additional Example 1 Continued

9. 9-3 Graphing Quadratic Functions Graph y = 3x 2 – 6x + 1. Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Additional Example 1 Continued x = 1 (–2, 25) (–1, 10) (0, 1) (1, –2) x = 1 (–1, 10) (0, 1) (1, –2) (–2, 25)

10. 9-3 Graphing Quadratic Functions Because a parabola is symmetrical, each point is the same number of units away from the axis of symmetry as its reflected point. Helpful Hint

11. 9-3 Graphing Quadratic Functions Check It Out! Example 1a Graph the quadratic function. y = 2x 2 + 6x + 2 Step 1 Find the axis of symmetry. Simplify. Use x =. Substitute 2 for a and 6 for b. The axis of symmetry is x.

12. 9-3 Graphing Quadratic Functions Step 2 Find the vertex. y = 2x 2 + 6x + 2 Simplify. Check It Out! Example 1a Continued = 4 – 9 + 2= –2= –2 The x-coordinate of the vertex is. Substitute for x. The y-coordinate of the vertex is. The vertex is.

13. 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. y = 2x 2 + 6x + 2 The y-intercept is 2; the graph passes through (0, 2). Identify c. Check It Out! Example 1a Continued

14. 9-3 Graphing Quadratic Functions Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Let x = –1 y = 2(–1) 2 + 6(–1) + 1 = 2 – 6 + 2 = –2 Let x = 1 y = 2(1) 2 + 6(1) + 2 = 2 + 6 + 2 = 10 Substitute x-coordinates. Simplify. Two other points are (–1, –2) and (1, 10). Check It Out! Example 1a Continued Since the axis of symmetry is x = –1, choose x values greater than –1.

15. 9-3 Graphing Quadratic Functions Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. Check It Out! Example 1a Continued (–1, –2) (1, 10) (–1, –2) (1, 10)

16. 9-3 Graphing Quadratic Functions Check It Out! Example 1b Graph the quadratic function. y + 6x = x 2 + 9 Step 1 Find the axis of symmetry. Simplify. Use x =. Substitute 1 for a and –6 for b. The axis of symmetry is x = 3. = 3= 3 y = x 2 – 6x + 9 Rewrite in standard form.

17. 9-3 Graphing Quadratic Functions Step 2 Find the vertex. Simplify. Check It Out! Example 1b Continued = 9 – 18 + 9 = 0= 0 The vertex is (3, 0). The x-coordinate of the vertex is 3. Substitute 3 for x. The y-coordinate of the vertex is 0. y = x 2 – 6x + 9 y = 3 2 – 6(3) + 9

18. 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. y = x 2 – 6x + 9 The y-intercept is 9; the graph passes through (0, 9). Identify c. Check It Out! Example 1b Continued

19. 9-3 Graphing Quadratic Functions Step 4 Find two more points on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 3, choose x-values less than 3. Let x = 2 y = 1(2) 2 – 6(2) + 9 = 4 – 12 + 9 = 1 Let x = 1 y = 1(1) 2 – 6(1) + 9 = 1 – 6 + 9 = 4 Substitute x-coordinates. Simplify. Two other points are (2, 1) and (1, 4). Check It Out! Example 1b Continued

20. 9-3 Graphing Quadratic Functions Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and two other points. Step 6 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. y = x 2 – 6x + 9 Check It Out! Example 1b Continued x = 3 (3, 0) (0, 9) (2, 1) (1, 4) (0, 9) (1, 4) (2, 1) x = 3 (3, 0)

21. 9-3 Graphing Quadratic Functions Additional Example 2: Problem-Solving Application The height in feet of a basketball that is thrown can be modeled by f(x) = –16x 2 + 32x, where x is the time in seconds after it is thrown. Find the basketball’s maximum height and the time it takes the basketball to reach this height. Then find how long the basketball is in the air.

22. 9-3 Graphing Quadratic Functions Additional Example 2 Continued 1 Understand the Problem The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the ground. The function f(x) = –16x 2 + 32x models the height of the basketball after x seconds. List the important information:

23. 9-3 Graphing Quadratic Functions 2 Make a Plan Find the vertex of the graph because the maximum height of the basketball and the time it takes to reach it are the coordinates of the vertex. The basketball will hit the ground when its height is 0, so find the zeros of the function. You can do this by graphing. Additional Example 2 Continued

24. 9-3 Graphing Quadratic Functions Solve 3 Step 1 Find the axis of symmetry. Use x =. Substitute –16 for a and 32 for b. Simplify. The axis of symmetry is x = 1. Additional Example 2 Continued

25. 9-3 Graphing Quadratic Functions Step 2 Find the vertex. f(x) = –16x 2 + 32x = –16(1) 2 + 32(1) = –16(1) + 32 = –16 + 32 = 16 The vertex is (1, 16). The x-coordinate of the vertex is 1. Substitute 1 for x. Simplify. The y-coordinate of the vertex is 16. Additional Example 2 Continued

26. 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. Identify c. f(x) = –16x 2 + 32x + 0 The y-intercept is 0; the graph passes through (0, 0). Additional Example 2 Continued

27. 9-3 Graphing Quadratic Functions Additional Example 2 Continued Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 1, choose an x- value that is less than 1. Let x = 0.5 f(x) = –16(0.5) 2 + 32(0.5) = –4 + 16 = 12 Another point is (0.5, 12).

28. 9-3 Graphing Quadratic Functions Step 5 Graph the axis of symmetry, the vertex, and the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve. Additional Example 2 Continued (0, 0) (1, 16) (2, 0)  (0.5, 12) (1.5, 12)

29. 9-3 Graphing Quadratic Functions The vertex is (1, 16). So at 1 second, the basketball has reached its maximum height of 16 feet. The graph shows the zeros of the function are 0 and 2. At 0 seconds the basketball has not yet been thrown, and at 2 seconds it reaches the ground. The basketball is in the air for 2 seconds. Additional Example 2 Continued (0, 0) (1, 16) (2, 0)  (0.5, 12) (1.5, 12)

30. 9-3 Graphing Quadratic Functions Look Back 4 Check by substituting (1, 16) and (2, 0) into the function. Additional Example 2 Continued 16 = –16(1) 2 + 32(1) 0 = –16(2) 2 + 32(2) 16 16 –16 + 32 0 0 –64 + 64

31. 9-3 Graphing Quadratic Functions Check It Out! Example 2 As Molly dives into her pool, her height in feet above the water can be modeled by the function f(x) = –16x 2 + 24x, where x is the time in seconds after she begins diving. Find the maximum height of her dive and the time it takes Molly to reach this height. Then find how long it takes her to reach the pool.

32. 9-3 Graphing Quadratic Functions 1 Understand the Problem The answer includes three parts: the maximum height, the time to reach the maximum height, and the time to reach the pool. Check It Out! Example 2 Continued List the important information: The function f(x) = –16x 2 + 24x models the height of the dive after x seconds.

33. 9-3 Graphing Quadratic Functions 2 Make a Plan Find the vertex of the graph because the maximum height of the dive and the time it takes to reach it are the coordinates of the vertex. The diver will hit the water when its height is 0, so find the zeros of the function. You can do this by graphing. Check It Out! Example 2 Continued

34. 9-3 Graphing Quadratic Functions Solve 3 Step 1 Find the axis of symmetry. Use x =. Substitute –16 for a and 24 for b. Simplify. The axis of symmetry is x = 0.75. Check It Out! Example 2 Continued

35. 9-3 Graphing Quadratic Functions Step 2 Find the vertex. f(x) = –16x 2 + 24x = –16(0.75) 2 + 24(0.75) = –16(0.5625) + 18 = –9 + 18 = 9 The vertex is (0.75, 9). Simplify. The y-coordinate of the vertex is 9. The x-coordinate of the vertex is 0.75. Substitute 0.75 for x. Check It Out! Example 2 Continued

36. 9-3 Graphing Quadratic Functions Step 3 Find the y-intercept. Identify c. f(x) = –16x 2 + 24x + 0 The y-intercept is 0; the graph passes through (0, 0). Check It Out! Example 2 Continued

37. 9-3 Graphing Quadratic Functions Step 4 Find another point on the same side of the axis of symmetry as the point containing the y-intercept. Since the axis of symmetry is x = 0.75, choose an x-value that is less than 0.75. Let x = 0.5 f(x) = –16(0.5) 2 + 24(0.5) = –4 + 12 = 8 Another point is (0.5, 8). Substitute 0.5 for x. Simplify. Check It Out! Example 2 Continued

38. 9-3 Graphing Quadratic Functions Step 5 Graph the axis of symmetry, the vertex, the point containing the y-intercept, and the other point. Then reflect the points across the axis of symmetry. Connect the points with a smooth curve. (1.5, 0) (0.75, 9) (0, 0) (0.5, 8)(1, 8) Check It Out! Example 2 Continued

39. 9-3 Graphing Quadratic Functions The vertex is (0.75, 9). So at 0.75 seconds, Molly's dive has reached its maximum height of 9 feet. The graph shows the zeros of the function are 0 and 1.5. At 0 seconds the dive has not begun, and at 1.5 seconds she reaches the pool. Molly reaches the pool in 1.5 seconds. (1.5, 0) (0.75, 9) (0, 0) (0.5, 8)(1, 8) Check It Out! Example 2 Continued

40. 9-3 Graphing Quadratic Functions Look Back 4 Check by substituting (0.75, 9) and (1.5, 0) into the function. Check It Out! Example 2 Continued 9 = –16(0.75) 2 + 24(0.75) 0 = –16(1.5) 2 + 24(1.5) 9 9 –9 + 18 0 0 –36 + 36

41. 9-3 Graphing Quadratic Functions Lesson Quiz 1. Graph y = –2x 2 – 8x + 4. 2. The height in feet of a fireworks shell can be modeled by h(t) = –16t 2 + 224t, where t is the time in seconds after it is fired. Find the maximum height of the shell, the time it takes to reach its maximum height, and length of time the shell is in the air. 784 ft; 7 s; 14 s