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Chapter 8 Quadratic Equations and Functions. Martin-Gay, Intermediate Algebra, 5ed 22 Math 083 Bianco 11/18/09  Turn in HW  Pick up survey, complete,

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Presentation on theme: "Chapter 8 Quadratic Equations and Functions. Martin-Gay, Intermediate Algebra, 5ed 22 Math 083 Bianco 11/18/09  Turn in HW  Pick up survey, complete,"— Presentation transcript:

1 Chapter 8 Quadratic Equations and Functions

2 Martin-Gay, Intermediate Algebra, 5ed 22 Math 083 Bianco 11/18/09  Turn in HW  Pick up survey, complete, and turn in  Monday quiz on 8.1, 8.2, and 8.6  Wednesday no class  92113 crn #

3 Martin-Gay, Intermediate Algebra, 5ed 33 § 8.6 Further Graphing of Quadratic Functions

4 Martin-Gay, Intermediate Algebra, 5ed 44 Graphs of Quadratic Functions In the previous section, we discovered that if the equation of a quadratic is written in the right format, we know a lot of information about the graph 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 Martin-Gay, Intermediate Algebra, 5ed 55 Since we can find out a lot about a quadratic function before we ever graph it, it would be in our best interest to get the quadratic into the appropriate form so we can easily find that information. The techniques we use would be similar to completing the square. Graphs of Quadratic Functions

6 Martin-Gay, Intermediate Algebra, 5ed 66 Graph f(x) = -3x 2 + 6x + 4. Find the vertex, axis, and any intercepts. Before graphing, we rewrite the equation into the form that will communicate information about the graph. Graphs of Quadratic Functions Example: Continued

7 Martin-Gay, Intermediate Algebra, 5ed 77 The vertex is at (1,7). The axis of symmetry is the line x = 1 Graphs of Quadratic Functions Example continued: Continued

8 Martin-Gay, Intermediate Algebra, 5ed 88 Intercepts for f(x) = -3(x – 1) 2 + 7 occur when x = 0 or y = 0. y-intercept occurs when we 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). Graphs of Quadratic Functions Example continued: Continued

9 Martin-Gay, Intermediate Algebra, 5ed 99 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 Graphs of Quadratic Functions Continued Example continued:

10 Martin-Gay, Intermediate Algebra, 5ed 10 x y The vertex is at (1,7). The axis of symmetry is the line x = 1. y-intercept is the point (0,4). x-intercepts are at the two points and Graphs of Quadratic Functions Example continued:

11 Martin-Gay, Intermediate Algebra, 5ed 11 In many applications, you are not interested in the entire graph of the quadratic function, but merely the vertex (the highest or lowest point of the graph). In that case, it is not necessary to convert the equation of the quadratic into the previous form that gives you much more information than just the vertex. There is a formula (derived in the text, if you are interested) for finding the vertex of the parabola using standard form. Finding a Maximum or Minimum

12 Martin-Gay, Intermediate Algebra, 5ed 12 The vertex of f(x) = ax 2 + bx + c, is at The maximum or minimum value of the parabola occurs at this vertex. The maximum or minimum value will be the second coordinate of the vertex. Finding a Maximum or Minimum

13 Martin-Gay, Intermediate Algebra, 5ed 13 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? Since the maximum value will occur at the vertex, we find the coordinates according to the previous formula. a = -1 and b = 360, so Finding a Maximum or Minimum Example: Continued

14 Martin-Gay, Intermediate Algebra, 5ed 14 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 Finding a Maximum or Minimum Example continued:

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22 Martin-Gay, Intermediate Algebra, 5ed 22 Homework  8.6 #1-33 odd and 45-53 odd


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