3.3 Analyzing Graphs of Quadratic Functions Find the vertex, the axis of symmetry, and the maximum or minimum value of a quadratic function using the method of completing the square. Graph quadratic functions. Solve applied problems involving maximum and minimum function values. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Graphing Quadratic Functions of the Type f (x) = a(x  h)2 + k The graph of a quadratic function is called a parabola. The point (h, k) at which the graph turns is called the vertex. The maximum or minimum value of f(x) occurs at the vertex. Each graph has a line x = h that is called the axis of symmetry. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Find the vertex, the axis of symmetry, and the maximum or minimum value of f (x) = x2 + 10x + 23. Solution: Complete the square. Vertex: (–5, –2) Axis of symmetry: x = –5 Minimum value of the function: 2 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) Graph f (x) = x2 + 10x + 23 Vertex Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Find the vertex, the axis of symmetry, and the maximum or minimum value of Solution: Complete the square. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example continued Graph: Vertex: (4, 0) Axis of symmetry: x = 4 Minimum value of the function: 0 Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Vertex of a Parabola The vertex of the graph of f (x) = ax2 + bx + c is We calculate the x-coordinate. We substitute to find the y-coordinate. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example For the function f(x) = x2 + 14x  47: a) Find the vertex. b) Determine whether there is a maximum or minimum value and find that value. c) Find the range. d) On what intervals is the function increasing? decreasing? Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) Solution a) f (x) = x2 + 14x  47 The x-coordinate of the vertex is: Since f (7) = 72 + 14 • 7  47 = 2, the vertex is (7, 2). b) Since a is negative (a = –1), the graph opens down, so the second coordinate of the vertex, 2, is the maximum value of the function. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Solution continued c) The range is (∞, 2]. d) Since the graph opens down, function values increase as we approach the vertex from the left and decrease as we move to the right of the vertex. Thus the function is increasing on the interval (∞, 7) and decreasing on (7, ∞). Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Application - Example A stonemason has enough stones to enclose a rectangular patio with 60 ft of stone wall. If the house forms one side of the rectangle, what is the maximum area that the mason can enclose? What should the dimensions of the patio be in order to yield this area? Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 1. Familiarize. Make a drawing of the situation, using w to represent the width of the fencing. 2. Translate. Since the area of a rectangle is given by length times width, we have A(w) = (60  2w)w =  2w2 + 60w. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 3. Carry out. We need to find the maximum value of A(w) and find the dimensions for which that maximum occurs. The maximum will occur at the vertex of the parabola, the first coordinate is Thus, if w = 15 ft, then the length l = 60  2 • 15 = 30 ft and the area is 15 • 30 = 450 ft2. 4. Check. (15 + 15 + 30) = 60 feet of fencing. 5. State. The maximum possible area is 450 ft2 when the patio is 15 feet wide and 30 feet long. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley