1 / 11

IB STUDIES Graphing Quadratic Functions

IB STUDIES Graphing Quadratic Functions. y. vertex. x. Let a , b , and c be real numbers a  0. The function f ( x ) = ax 2 + bx + c is called a quadratic function. Quadratic function. The graph of a quadratic function is a parabola.

esheila
Download Presentation

IB STUDIES Graphing Quadratic Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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.

More Related