2.1 graphing quadratic functions
- 1. Copyright © by Houghton Mifflin Company, Inc. All rights 1 Let a, b, and c be real numbers a ≠ 0. The function f(x) = ax2 + bx + c is called a quadratic function. The graph of a quadratic function is a parabola. Every parabola is symmetrical about a line called the axis (of symmetry). The intersection point of the parabola and the axis is called the vertex of the parabola. x y axis f(x) = ax2 + bx + c vertex
- 2. Copyright © by Houghton Mifflin Company, Inc. All rights 2 The leading coefficient of ax2 + bx + c is a. When the leading coefficient is positive, the parabola opens upward and the vertex is a minimum. When the leading coefficient is negative, the parabola opens downward and the vertex is a maximum. x y f(x) = ax2 + bx + c a > 0 opens upward vertex minimum x y f(x) = ax2 + bx + c a < 0 opens downward vertex maximum
- 3. Copyright © by Houghton Mifflin Company, Inc. All rights 3 x y 4 4 Example: Graph and find the vertex and x-intercepts of f(x) = –x2 + 6x + 7. vertex (3, 16) Find the x-intercepts by solving –x2 + 6x + 7 = 0. (–x + 7 )( x + 1) = 0 factor x = 7, x = –1 x-intercepts (7, 0), (–1, 0) x = 3 f(x) = –x2 + 6x + 7 (7, 0)(–1, 0) (3, 16)
- 4. Copyright © by Houghton Mifflin Company, Inc. All rights 4 Example: Graph f(x) = (x – 3)2 + 2 and find the vertex and axis. f(x) = (x – 3)2 + 2 is the same shape as the graph of g(x) = (x – 3)2 shifted upwards two units. g(x) = (x – 3)2 is the same shape as y = x2 shifted to the right three units. f(x) = (x – 3)2 + 2 g(x) = (x – 3)2 y = x2 -4 x y 4 4 vertex (3, 2)
- 5. Copyright © by Houghton Mifflin Company, Inc. All rights 5 Vertex of a Parabola Example: Find the vertex of the graph of f(x) = x2 – 10x + 22. f(x) = x2 – 10x + 22 original equation a = 1, b = –10, c = 22 The vertex of the graph of f(x) = ax2 + bx + c (a ≠ 0) is , 2 2 b b f a a − − At the vertex, 5 )1(2 10 2 = − = − = a b x So, the vertex is (5, -3). 322)5(105)5( 2 2 −=+−== − f a b f
- 6. Copyright © by Houghton Mifflin Company, Inc. All rights 6 Example: A basketball is thrown from the free throw line from a height of six feet. What is the maximum height of the ball if the path of the ball is: 21 2 6. 9 y x x= − + + The path is a parabola opening downward. The maximum height occurs at the vertex. 2, 9 1 62 9 1 2 = − =→++ − = baxxy .9 2 = − = a b xAt the vertex, ( ) 159 2 == − f a b f So, the vertex is (9, 15). The maximum height of the ball is 15 feet.
- 7. Copyright © by Houghton Mifflin Company, Inc. All rights 7 Example: A fence is to be built to form a rectangular corral along the side of a barn 65 feet long. If 120 feet of fencing are available, what are the dimensions of the corral of maximum area? barn corralx x 120 – 2x Let x represent the width of the corral and 120 – 2x the length. Area = A(x) = (120 – 2x)x = –2x2 + 120x The graph is a parabola and opens downward. The maximum occurs at the vertex where , 2a b x − = a = –2 and b = 120 .30 4 120 2 = − − = − =→ a b x 120 – 2x = 120 – 2(30) = 60 The maximum area occurs when the width is 30 feet and the length is 60 feet.