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Section 5

Chapter 9. Section 5. Graphs of Quadratic Functions. Graph a quadratic function. Graph parabolas with horizontal and vertical shifts. Use the coefficient of x 2 to predict the shape and direction in which a parabola opens. Find a quadratic function to model data. 9.5.

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Section 5

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  1. Chapter 9 Section 5

  2. Graphs of Quadratic Functions Graph a quadratic function. Graph parabolas with horizontal and vertical shifts. Use the coefficient of x2 to predict the shape and direction in which a parabola opens. Find a quadratic function to model data. 9.5

  3. Graph a quadratic function. Objective 1 Slide 9.5- 3

  4. The graph shown below is a graph of the simplest quadraticfunction,defined by y = x2. This graph is called a parabola. Graph a quadratic function. The point (0, 0), the lowest point on the curve, is the vertex. Slide 9.5- 4

  5. The vertical line through the vertex is the axisof the parabola, here x = 0. A parabola is symmetric about its axis. Graph a quadratic function. Slide 9.5- 5

  6. We use the variable y and function notation f (x) interchangeably. Although we use the letter f most often to name quadratic functions, other letters can be used. We use the capital letter F to distinguish between different parabolas graphed on the same coordinate axes. Graph parabolas with horizontal and vertical shifts. The graph of any quadratic function is a parabola with a vertical axis. Slide 9.5- 6

  7. Graph parabolas with horizontal and vertical shifts. Objective 2 Slide 9.5- 7

  8. Parabolas do not need to have their vertices at the origin. The graph of F(x) = x2 + k is shifted, or translated k units vertically compared to f(x) = x2. Graph parabolas with horizontal and vertical shifts. Slide 9.5- 8

  9. Graph f(x) = x2 + 3. Give the vertex, domain, and range. The graph has the same shape as f(x) = x2, but shifted up 3 units. Make a table of points. CLASSROOM EXAMPLE 1 Graphing a Parabola (Vertical Shift) Solution: vertex (0, 3) domain: (, ) range: [3, ) Slide 9.5- 9

  10. Graph parabolas with horizontal and vertical shifts. Slide 9.5- 10

  11. Graph f(x) = (x + 2)2. Give the vertex, axis, domain, and range. The graph has the same shape as f(x) = x2, but shifted 2 units to the left. Make a table of points. CLASSROOM EXAMPLE 2 Graphing a Parabola (Horizontal Shift) Solution: vertex (2, 0) axis x = 2 domain: (, ) range: [0, ) Slide 9.5- 11

  12. Graph parabolas with horizontal and vertical shifts. Slide 9.5- 12

  13. Graph f(x) = (x 2)2 + 1. Give the vertex, axis, domain, and range. The graph has the same shape as f(x) = x2, but shifted 2 units to the right and 3 unit up. Make a table of points. CLASSROOM EXAMPLE 3 Graphing a Parabola (Horizontal and Vertical Shifts) Solution: vertex (2, 1) axis x = 2 domain: (, ) range: [1, ) Slide 9.5- 13

  14. Graph parabolas with horizontal and vertical shifts. Slide 9.5- 14

  15. Use the coefficient of x2 to predict the shape and direction in which a parabola opens. Objective 3 Slide 9.5- 15

  16. Graph f(x) = 2x2 3. Give the vertex, axis, domain, and range. The coefficient (2) affects the shape of the graph; the 2 makes the parabola narrower. The negative sign makes the parabola open down. The graph is shifted down 3 units. CLASSROOM EXAMPLE 4 Graphing a Parabola That Opens Down Solution: Slide 9.5- 16

  17. Graph f(x) = 2x2 3. CLASSROOM EXAMPLE 4 Graphing a Parabola That Opens Down (cont’d) vertex (0, 3) axis x = 0 domain: (, ) range: (, 3] Slide 9.5- 17

  18. Use the coefficient of x2 to predict the shape and direction in which a parabola opens. Slide 9.5- 18

  19. Graph Parabola opens up. Narrower than f(x) = x2 Vertex: (2, 1) CLASSROOM EXAMPLE 5 Using the General Characteristics to Graph a Parabola Solution: axis x = 2 domain: (, ) range: [1, ) Slide 9.5- 19

  20. Find a quadratic function to model data. Objective 4 Slide 9.5- 20

  21. The number of higher-order multiple births (triplets or more) in the United States has declined in recent years as shown by the data in the table below. Here, x represents the number of years since 1995 and y represents the number of higher-order multiple births. Using the data points (1, 5939), (6, 7471), and (10, 6694), find another quadratic model for the data on higher-order multiple births. Use f(x) = ax2 + bx + c f(1) = a(1)2 + b(1) + c = 5939 f(6) = a(6)2 + b(6) + c = 7471 f(10) = a(10)2 + b(10) + c = 6694 CLASSROOM EXAMPLE 6 Modeling the Number of Multiple Births Solution: Slide 9.5- 21

  22. Simplify the system: a + b + c = 5939 (1) 36a + 6b + c = 7471 (2) 100a + 10b + c = 6694 (3) To eliminate c, multiply equation (1) by –1 and add the result to equation (2). abc = 5939 1  (1) 36a + 6b + c = 7471 (2) 35a + 5b = 1532 (4) CLASSROOM EXAMPLE 6 Modeling the Number of Multiple Births (cont’d) Slide 9.5- 22

  23. To eliminate c again, multiply equation (2) by –1 and add the result to equation (3). 36a 6bc = 74711  (2) 100a + 10b + c = 6694 (3) 64a + 4b = −777 (5) To eliminate b, multiply equation (4) by –4 and equation (5) by 5, then add to get the result. 140a 20b = −6128 −4  (4) 320a + 20b = −3885 5  (5) 180a = −10013 a = −10013 /180 a = −55.63 CLASSROOM EXAMPLE 6 Modeling the Number of Multiple Births (cont’d) Slide 9.5- 23

  24. To find b, substitute −55.63 for a in equation (4). 35a + 5b = 1532 35(−55.63) + 5b = 1532 −1947.05 + 5b = 1532 5b = 3479.05 b = 695.80 To find c, use equation (1). a + b + c = 5939 (1) c = 5939 – a – b c = 5939 – (–55.63) – 695.80 = 5298.83 The quadratic model using the three points is: y = –55.63x2 + 695.80x + 5298.83 CLASSROOM EXAMPLE 6 Modeling the Number of Multiple Births (cont’d) Slide 9.5- 24

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