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3.2 Quadratic Functions & Graphs. Quiz  Write out the general form of a quadratic equation. f(x) = _____________.

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Presentation on theme: "3.2 Quadratic Functions & Graphs. Quiz  Write out the general form of a quadratic equation. f(x) = _____________."— Presentation transcript:

1 3.2 Quadratic Functions & Graphs

2 Quiz  Write out the general form of a quadratic equation. f(x) = _____________

3 Quadratic Function General Form Parabola f(x) = ax 2 + bx + c (a ≠ 0) Standard( tranformation) Form f(x) = a(x - h) 2 + k (a ≠ 0)

4 Complete the Square General Form f(x) = ax 2 + bx + c General Form f(x) = ax 2 + bx + c Standard Form f(x) = a(x – h) 2 + k Standard Form f(x) = a(x – h) 2 + k Complete the Square x 2 + 2px + p 2 = (x + p) 2 x 2 - 2px + p 2 = (x - p) 2 x 2 + 2px + p 2 = (x + p) 2 x 2 - 2px + p 2 = (x - p) 2

5 Complete the Square  Example: Given f(x) = 2x 2 - 8x + 1, complete the square to put it into the form f(x) = a(x – h) 2 + k.  How about f(x) = x 2 - 3x + 1?

6 The Graph of a Quadratic a > 0a < 0 x y x y Vertex Minimum point Maximum point Axis of symmetry x = h Axis of symmetry x = h Axis of symmetry x = h Axis of symmetry x = h

7 Find vertex of a parabola Transformation form: f(x) = a(x – h) 2 + k Vertex : ( h, k ) Axis of symmetry: x = h Transformation form: f(x) = a(x – h) 2 + k Vertex : ( h, k ) Axis of symmetry: x = h Transformation form: f(x) = ax 2 + bx + c Vertex : ( - b/2a, f(- b/2a) ) Axis of symmetry: x = -b/2a Transformation form: f(x) = ax 2 + bx + c Vertex : ( - b/2a, f(- b/2a) ) Axis of symmetry: x = -b/2a

8 Graphing parabolas  Determine if the graph opens up or down  Determine the vertex ( h, k )  Find the y – intercept  Plot the vertex and at least 2 additional points on one side of the vertex  Use symmetry finish the other half Example: f(x) = 2x 2 + x - 3

9 Application  Write the equation of the parabola with vertex at (8, 3) passing through (10, 5). f(x) = a ( x – h ) 2 + k 83105

10 Height of a Projected Object  If air resistance is neglected, the height s ( in feet ) of an object projected directly upward from an initial height s 0 feet with initial velocity v 0 feet per second is s (t) = -16t 2 + v 0 t + s 0, where t is the number of seconds after the object is projected.

11 Application A ball is thrown directly upward from an initial height of 100 feet with an initial velocity of 80 feet per second. 1. Give the function that describes the height of the ball in terms of time t. 2. Graph this function so that the y-intercept, the positive x- intercept, and the vertex are visible. 3. If the point (4.8, ) lies on the graph of the function. What does this mean for this particular situation? 4. After how many seconds does the projectile reach its maximum height? What is the maximum height? Solve analytically and graphically. 5. For what interval of time is the height of the ball greater than 160 feet? Determine the answer graphically. 6. After how many seconds will the ball fall to the ground? Determine the answer graphically.

12 Homework  PG. 174: 3-48(M3), 17 instead of 18 PG. 175: 57 – 72 (M3)  Key: 6, 27, 36, 57, 63  Reading: 3.3 Quadratic Equation & Ineq.


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