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4 © 2010 Pearson Education, Inc. All rights reserved 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 k is the minimum value of f, and if a < 0, the parabola opens down and k is the maximum value of f.

5 © 2010 Pearson Education, Inc. All rights reserved 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 © 2010 Pearson Education, Inc. All rights reserved 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 © 2010 Pearson Education, Inc. All rights reserved PROCEDURE FOR GRAPHING f (x) = a(x – h) 2 + k Step 4Find the x-intercepts (if any). Set 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 © 2010 Pearson Education, Inc. All rights reserved 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. Replace x with 0. Then f (0) = ah 2 + k is the y-intercept.

9 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 2 Graphing a Quadratic Function in Standard Form Sketch the graph of Solution Step 1a = –3, h = –2, and k = 12 Step 2a = –3, a < 0, the parabola opens down. Step 3 (h, k) = (–2, 12); the function f has a maximum value 12 at x = –2.

10 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 2 Graphing a Quadratic Function in Standard Form Step 4Set f (x) = 0 and solve for x. Solution continued

13 © 2010 Pearson Education, Inc. All rights reserved 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:

14 © 2010 Pearson Education, Inc. All rights reserved Step 4Find the x-intercepts (if any). 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). PROCEDURE FOR GRAPHING f (x) = ax 2 + bx + c

15 © 2010 Pearson Education, Inc. All rights reserved 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. PROCEDURE FOR GRAPHING f (x) = ax 2 + bx + c

16 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution Sketch the graph of Step 1a = 2, b = 8, and c = –10 Step 2a = 2, a > 0, the parabola opens up.

17 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Step 3Find (h, k). Minimum value of –18 at x = –2 Solution continued

18 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution continued Step 4Let f (x) = 0. x-intercepts: –5 and 1

19 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution continued Step 5 Let x = 0. Step 6Axis of symmetry is x = –2. The symmetric image of (0, –10) with respect to the axis x = –2 is (–4, –10).

20 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 3 Graphing a Quadratic Function f (x) = ax 2 + bx + c Solution continued Step 7Sketch the parabola passing through the points found in Steps 3–6.

21 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 4 Identifying the Characteristics of a Quadratic Function from Its Graph The graph of f (x) = –2x 2 +8x – 5 is shown. Find the domain and range of f. Solution The domain is (–∞, ∞). The range is (–∞, 3].

22 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 5 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.)

23 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 5 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. 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).

24 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 5 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).

25 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 5 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.

26 © 2010 Pearson Education, Inc. All rights reserved EXAMPLE 5 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.