Homework Corrections (Page 1 of 2) A #24 / Holt 9-2 #19 – 33, 41, 42, 47 – 52 19. no zeros 20. x = 0 21. x = -8, -2 22. x = -1 23. x = 6 24. x = 25. x = 26. x = 27. x = 5 28. x = 29. (-3.5, -12.25) 30. (4, 32) 31. (-2, 5)

Homework Corrections (Page 2 of 2) A #24 / Holt 9-2 #19 – 33, 41, 42, 47 – 52 48. y = 3x – 2 49. y = -4x + 2 50. no 51. yes 52. yes 32. 33. (1, 4.5) 41. B 42. J 47. y = -2x + 3

Lesson Objective: I will be able to … Graph a quadratic function in the form y = ax2 + bx + c Language Objective: I will be able to … Read, write, and listen about vocabulary, key concepts, and examples

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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 = ax2 + bx + c, when x = 0, y = c. So the y-intercept of a quadratic function is c.

Example 1: Graphing a Quadratic Function Pages 15 – 16 Graph y = 3x2 – 6x + 1. Step 1 Find the axis of symmetry. Use x = . Substitute 3 for a and –6 for b. = 1 Simplify. The axis of symmetry is x = 1. Step 2 Find the vertex. y = 3x2 – 6x + 1 The x-coordinate of the vertex is 1. Substitute 1 for x. = 3(1)2 – 6(1) + 1 = 3 – 6 + 1 Simplify. = –2 The y-coordinate is –2. The vertex is (1, –2).

Step 3 Find the y-intercept. y = 3x2 – 6x + 1 y = 3x2 – 6x + 1 Pages 15 – 16 Example 1 Continued Graph y = 3x2 – 6x + 1. Step 3 Find the y-intercept. y = 3x2 – 6x + 1 y = 3x2 – 6x + 1 Identify c. The y-intercept is 1; the graph passes through (0, 1). 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. Substitute x-coordinates. Let x = –1. Let x = –2. y = 3(–1)2 – 6(–1) + 1 y = 3(–2)2 – 6(–2) + 1 = 3 + 6 + 1 = 10 = 12 + 12 + 1 = 25 Simplify. Two other points are (–1, 10) and (–2, 25).

Example 1 Continued Graph y = 3x2 – 6x + 1. Pages 15 – 16 Graph y = 3x2 – 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. x = 1 (–2, 25) (–1, 10) (0, 1) (1, –2) x = 1 (–1, 10) (0, 1) (1, –2) (–2, 25)

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 Make sure that the quadratic equation is in standard y = ax2 + bx + c form before following the steps needed to graph the equation. Remember!

Example 2: Sports Application Page 17 The height in feet of a basketball that is thrown can be modeled by f(x) = –16x2 + 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. Step 1 The y-coordinate of the vertex represents the maximum height and the x-coordinate of the vertex represents the time it takes the basketball to reach the maximum height. Find the vertex. Use x = . Substitute –16 for a and 32 for b.

Step 2 Find the y-intercept. Example 2 Continued Page 17 x = 1 f(x) = –16x2 + 32x Substitute 1 for x. = –16(1)2 + 32(1) = –16(1) + 32 = –16 + 32 = 16 The vertex is (1, 16). Step 2 Find the y-intercept. f(x) = –16x2 + 32x + 0 Identify c. The y-intercept is 0; the graph passes through (0, 0).

Page 17 Example 2 Continued (0, 0) (1, 16) (2, 0) Step 3 Graph the axis of symmetry, the vertex, and the point containing the y-intercept. Then reflect the point across the axis of symmetry. Connect the points with a smooth curve. The vertex is (1, 16). So it takes 1 second for the basketball to reach 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.

Classwork Assignment #25 9-3 Reading Strategies / Practice A Worksheet

Homework Assignment #25 Holt 9-3 #8 – 14, 28, 36 – 39 KIN 9-4