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Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

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Presentation on theme: "Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall."— Presentation transcript:

1 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall

2 Chapter 11 Graphing Quadratic Functions, Rational Functions, and Conic Sections

3 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall 11.2 Further Graphing of Quadratic Functions

4 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall If the quadratic is written in the form f(x) = a(x – h) 2 + k, then we can find the vertex, axis of symmetry, whether it opens up or down, and the width.

5 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example Graph f(x) =  3x 2 + 6x + 4. Find the vertex, axis, and any intercepts. Solution Complete the square on x to write the equation in the form y = a(x – h) 2 + k. continue

6 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Graph f(x) =  3x 2 + 6x + 4. Find the vertex, axis, and any intercepts. Vertex: (1, 7) Axis: x = 1 Intercepts occur when x = 0 or y = 0. y-intercept: set x = 0. y = –3(0 – 1) 2 + 7 y = –3(–1) 2 + 7 y = –3(1) + 7 y = 4 y-intercept is the point (0, 4). continue

7 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall x-intercept(s) occur when we set y = 0 0 = –3(x – 1) 2 + 7 3(x – 1) 2 = 7 (x – 1) 2 = 7/3 Note that the square root of 7/3 is about 1.7, so x-intercepts are at about ( − 0.7, 0) and (2.7, 0) when we sketch the graph of the quadratic equation. x-intercepts are at the two points, and

8 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Vertex: (1, 7) Axis: x = 1 y-intercept is the point (0,4) x-intercepts are at the two points x y

9 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Vertex Formula The graph of f(x) = ax 2 + bx + c, when a ≠ 0, is a parabola with vertex

10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall Example The Utah Ski Club sells calendars to raise money. The profit P, in cents, from selling x calendars is given by the equation P(x) = 360x – x 2. What is the maximum profit the club will earn? Solution Since the maximum value will occur at the vertex, we find the coordinates according to the previous formula. a =  1 and b = 360

11 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall The Utah Ski Club sells calendars to raise money. The profit P, in cents, from selling x calendars is given by the equation P(x) = 360x – x 2. What is the maximum profit the club will earn? The maximum profit will occur when the club sells 180 calendars. We substitute that number into the profit formula to find P(180) = 360(180) – (180) 2 = 64800 – 32400 = 32400 cents Remember to read the problem carefully. = $324


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