Graphing f(x) = (x - h) + k 3.3A 2 Chapter 3 Quadratic Functions

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Presentation transcript:

Graphing f(x) = (x - h) + k 3.3A 2 Chapter 3 Quadratic Functions MATHPOWERTM 11, WESTERN EDITION 3.3A.1

Graphing f(x) = (x - h)2 Comparing f(x) = (x - h)2 with f(x) = x 2: (-2, 0) (2, 0) x y -1 1 -2 0 -3 1 f(x) = (x - 2)2 f(x) = (x + 2)2 f(x) = x2 Vertex is (-2, 0) Axis of symmetry is x + 2 = 0 Minimum value of y = 0, when x = -2 Vertex is (2, 0) Axis of symmetry is x - 2 = 0 Minimum value of y = 0, when x = 2 3.3A.2

Graphing f(x) = (x - h)2 [cont’d] Graph f(x) = (x - 4)2. The vertex is (4, 0). The axis of symmetry is x - 4 = 0. The graph opens upwards. It has a minimum value of y = 0, when x = 4. 1 2 3 4 f(x) = x2 f(x) = (x - 4)2 The domain is all real numbers. The y-intercept is 16. The range is y > 0. The x-intercept is (4, 0). 3.3A.3

f(x) = a(x - h)2 + k The Standard Form of the Quadratic Function Horizontal shift Stretch factor Vertical shift f(x) = a(x - h)2 + k Indicates direction of opening Coordinates of the vertex are (h, k) If a > 0, the graph opens up and there is a minimum value of y. If a < 0, the graph opens down and there is a maximum value of y. Axis of symmetry is x - h = 0 3.3A.4

Graphing f(x) = a(x - h)2 + k When a > 0: Move over 1 and up 2, to account for the stretch factor of 2. Graph f(x) = 2(x - 3)2 + 4. The vertex will be the point (3, 4). 1. Think of the critical point for the parabola f(x) = x2. 4 3 Move 4 units up. f(x) = 2(x - 3)2 + 4 2 2. Move that point to the new vertex (h, k). 1 (0, 0) 1 2 3 3. Consider the “a” value and plot your next reference points. Move 3 units to the right. 3.3A.5

Graphing f(x) = a(x - h)2 + k [cont’d] When a < 0: The new vertex is (3, 4). Graph f(x) = -2(x - 3)2 + 4. f(x) = -2(x - 3)2 + 4 Move over 1 and down 2, to account for the stretch factor and the fact that a < 0. Move 4 units up. (0, 0) Move 3 units to the right. 3.3A.6

Finding the Intercepts Given the function y = 2(x - 4)2 - 12: y = 2(0 - 4)2 - 12 y = 2(16) - 12 y = 20 Find the y-intercept. The y -intercept is 20. Let x = 0. 2(x - 4)2 - 12 = 0 2(x - 4)2 = 12 (x - 4)2 = 6 x - 4 = + √ 6 Find the x-intercepts. The x-intercepts are 4 + √ 6 and 4 - √ 6 . Let y = 0. x = 4 + √ 6 3.3A.7