1. Slide 3.1- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Quadratic Functions Learn to graph a quadratic function in standard form. Learn to graph a quadratic function. Learn to solve problems modeled by quadratic functions. SECTION 3.1 1 2 3

3. Slide 3.1- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley QUADRATIC FUNCTION A function that can be defined by an equation of the form where a, b, and c, are real numbers with a ≠ 0, is called a quadratic function.

4. Slide 3.1- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley THE STANDARD FORM OF A QUADRATIC FUNCTION The quadratic function is in standard form. The graph of f is a parabola with vertex (h, k). The parabola is symmetric with respect to the line x = h, called the axis of the parabola. If a > 0, the parabola opens up, and if a < 0, the parabola opens down.

5. Slide 3.1- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding a Quadratic Function Find the standard form of the quadratic function whose graph has vertex (–3, 4) and passes through the point (–4, –7). Let y = f (x) be the quadratic function. Solution

6. Slide 3.1- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING f (x) = a(x – h) 2 + k Step 1The graph is a parabola. Identify a, h, and k. Step 2Determine how the parabola opens. If a > 0, the parabola opens up. If a < 0, the parabola opens down. Step 3Find the vertex. The vertex is (h, k). If a > 0 (or a < 0), the function f has a minimum (or a maximum) value k at x = h.

7. Slide 3.1- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING f (x) = a(x – h) 2 + k Step 4Find the x–intercepts. Find the x–intercepts (if any) by setting f (x) = 0 and solving the equation a(x – h) 2 + k = 0 for x. If the solutions are real numbers, they are the x–intercepts. If not, the parabola either lies above the x–axis (when a > 0) or below the x–axis (when a < 0).

8. Slide 3.1- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING f (x) = a(x – h) 2 + k Step 6Sketch the graph. Plot the points found in Steps 3–5 and join them by a parabola. Show the axis x = h of the parabola by drawing a dashed line. Step 5Find the y–intercept. Find the y–intercept by replacing x with 0. Then f (0) = ah 2 + k is the y – intercept.

9. Slide 3.1- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing a Quadratic Function in Standard Form Graph the quadratic function Solution Step 1a = 2, h = 3, and k = –8 Step 2a = 2, a > 0, the parabola opens up. Step 3 (h, k) = (3, –8); the function f has a minimum value –8 at x = 3. Step 4Set f (x) = 0 and solve for x.

10. Slide 3.1- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing a Quadratic Function in Standard Form Solution continued Step 5Replace x with 0. Step 6axis: x = 3, opens up, vertex: (3, –8), passes through (1, 0), (5, 0) and (0, 10), the graph is y = 2x 2 shifted three units right and eight units down.

11. Slide 3.1- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing a Quadratic Function in Standard Form Solution continued

12. Slide 3.1- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING f (x) = ax 2 + bx + c Step 1The graph is a parabola. Identify a, b, and c. Step 2Determine how the parabola opens. If a > 0, the parabola opens up. If a < 0, the parabola opens down. Step 3Find the vertex (h, k). Use the formula:

13. Slide 3.1- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING f (x) = a(x – h) 2 + k Step 4Find the x–intercepts. Let y = f (x) = 0. Find x by solving the equation ax 2 + bx + c = 0. If the solutions are real numbers, they are the x–intercepts. If not, the parabola either lies above the x–axis (when a > 0) or below the x–axis (when a < 0).

14. Slide 3.1- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING f (x) = a(x – h) 2 + k Step 5Find the y–intercept. Let x = 0. The result f (0) = c is the y – intercept. Step 7Draw a parabola through the points found in Steps 3–6. Step 6The parabola is symmetric with respect to its axis, Use this symmetry to find additional points.

15. Slide 3.1- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution Graph the quadratic function Step 1a = –2, b = 8, and c = –5 Step 2a = –2, a < 0, the parabola opens down. Step 3Find (h, k). Maximum value of 3 at x = 2

16. Slide 3.1- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution continued Step 4Let f (x) = 0. Step 5 Let x = 0.

17. Slide 3.1- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution continued Step 6Axis of symmetry is x = 2. Let x = 1, then the point (1, 1) is on the graph, the symmetric image of (1, 1) with respect to the axis x = 2 is (3, 1). The symmetric image of the y–intercept (0, –5) with respect to the axis x = 2 is (4, –5). Step 7The parabola passing through the points found in Steps 3–6 is sketched on the next slide.

18. Slide 3.1- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution continued

19. Slide 3.1- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Identifying the Characteristics of a Quadratic Function from Its Graph The graph of f (x) = –2x 2 +8x – 5 is shown. a.Find the domain and range of f. b.Solve the inequality –2x 2 +8x – 5 > 0. b.The graph is above the x-axis in the interval Solution a.The domain is (–∞, ∞). The range is (–∞, 3].

20. Slide 3.1- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Graphing a Quadratic Inequality Graph the quadratic function f (x) = x 2 + 2x + 2 and solve each quadratic inequality. a. x 2 + 2x + 2 > 0b. x 2 + 2x + 2 < 0 Solution Step 1a = 1, b = 2, and c = 2 Step 2a = 1 > 0, the parabola opens up. Step 3Find (h, k).

21. Slide 3.1- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Graphing a Quadratic Inequality Solution continued Minimum value of 1 at x = –1 Step 4Set f (x) = 0. Step 5 Set x = 0. real, the graph does not intersect the x–axis. The solutions are not

22. Slide 3.1- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Graphing a Quadratic Inequality Solution continued Step 6Axis of symmetry is x = –1. Step 7The points (–2, 2), and (0, 2) are symmetric with respect to the axis of symmetry. Here’s the sketch of the graph.

23. Slide 3.1- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Graphing a Quadratic Inequality Solution continued The entire graph lies above the x–axis. Thus, y is always > 0. a.x 2 +2x + 2 > 0 is always true, the solution is (–∞,∞) b. x 2 +2x + 2 < 0 is never true, the solution is

24. Slide 3.1- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Yours, Mine, and Ours A widower with 10 children marries a widow who also has children. After their marriage, they have their own children. If the total number of children is 24, and we assume that the children of the same parents do not fight, find the maximum possible number of fights among the children. (In this example, a fight between Sean and Misty, no matter how many times they fight, is considered as one fight.)

25. Slide 3.1- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Yours, Mine, and Ours Solution Suppose the widow had x number of children before marriage. Then the couple has 24 – 10 – x = 14 – x additional children after their marriage. Since the children of the same parents do not fight, there are no fights among the 10 children the widower brought into the marriage, among the x children the widow brought into the marriage, or among the 14 – x children of the couple (widower and widow).

26. Slide 3.1- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Yours, Mine, Ours Solution continued The possible number of fights among the children of (i)the widower (10 children) and the widow (x children) is 10x. (ii)the widower (10 children) and the couple (14 – x children) is 10(14 – x), and (iii)the widow (x children) and the couple (14 – x children) is x(14 – x).

27. Slide 3.1- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Yours, Mine, Ours Solution continued The possible number y of all fights is given by In the quadratic function y = f (x) = –x 2 +14x + 140, we have a = –1, b = 14, and c = 140.

28. Slide 3.1- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Yours, Mine, Ours Solution continued The vertex (h, k) is given by Since, a = –1 < 0, the function f has maximum value k. Hence, the maximum possible number of fights among the children is 189.