IB STUDIES Graphing Quadratic Functions

1. IB STUDIESGraphing Quadratic Functions

2. y vertex x Let a, b, and c be real numbers a 0. The function f(x) = ax2 + bx + c is called a quadratic function. 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. f(x) = ax2 + bx + c axis

3. y x vertex minimum y x vertex maximum Leading Coefficient The leading coefficient of ax2 + bx + c is a. a > 0 opens upward When the leading coefficientis positive, the parabola opens upward and the vertex is a minimum. f(x) = ax2 + bx + c When the leading coefficient is negative, the parabola opens downwardand the vertex is a maximum. f(x) = ax2 + bx + c a < 0 opens downward

4. Example: Compare the graphs of , and y 5 x -5 5 The simplest quadratic functions are of the form f(x) = ax2 (a  0) Simple Quadratic Functions These are most easily graphed by comparing them with the graph ofy = x2.

5. y 4 (3, 2) vertex x -4 4 Example: Graph f(x) = (x –3)2 + 2 and find the vertex and axis. Example: f(x) = (x –3)2 + 2 f(x) = (x –3)2 + 2is 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

6. Example: Graph the parabola f(x) = 2x2 + 4x – 1 and find the axis of symmetry, x and y-intercepts and vertex. Quadratic Function in Standard Form f(x) = 2x2 + 4x – 1 Use your GDC to find the TP axis of symmetry x-intercepts (-2.22,0) and (0.225,0) y-intercept (0,-1)

7. Example: Graph and find the vertex, axis of symmetry and x and y-intercepts of f(x) = –x2 +6x + 7. Vertex and x-Intercepts a< 0parabola opens downward. vertex (3,16) y-intercept (0,7) x-intercepts (7, 0), (–1, 0) axis of symmetry x=3 equivalent forms:

8. The vertex of the graph of f(x) = ax2 + bx + c (a0) At the vertex, Vertex of a Parabola 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 So, the vertex is (5, -3).

9. At the vertex, 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: Example: Basketball The path is a parabola opening downward. The maximum height occurs at the vertex. So, the vertex is (9, 15). The maximum height of the ball is 15 feet.

10. barn corral x x 120 – 2x The maximum occurs at the vertex where a = –2 and b = 120 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? Example: Maximum Area 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. 120 – 2x = 120 – 2(30) = 60 The maximum area occurs when the width is 30 feet and the length is 60 feet.

11. y (0, 1) x (2, –1) Example: Find an equation for the parabola with vertex (2, –1) passing through the point (0, 1). Example: Parabola y = f(x) Substitute point (0,1) to find c.