Copyright 2006 Pearson Education Inc Publishing as Pearson
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
3 3. 1 3. 2 3. 3 3. 4 Quadratic Functions and Equations Quadratic Functions and Models Quadratic Equations and Problem Solving Quadratic Inequalities Transformations of Graphs Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
3. 1 Quadratic Functions and Models Learn basic concepts about quadratic functions and their graphs. ♦ Complete the square and apply the vertex formula. ♦ Graph a quadratic function by hand. ♦ Solve applications and model data. ♦ Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Basic Concepts Recall that a linear function can be written as f(x) = ax + b (or f(x) = mx + b). The formula for a quadratic function is different from that of a linear function because it contains an x 2 term. f(x) = 3 x 2 + 3 x + 5 g(x) = 5 x 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4
Quadratic Function • • • The graph of a quadratic function is a parabola—a U shaped graph that opens either upward or downward. A parabola opens upward if a is positive and opens downward if a is negative. The highest point on a parabola that opens downward and the lowest point on a parabola that opens upward is called the vertex. The vertical line passing through the vertex is called the axis of symmetry. The leading coefficient a controls the width of the parabola. Larger values of |a| result in a narrower parabola, and smaller values of |a| result in a wider parabola. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5
Examples of different parabolas Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 6
Example Use the graph of the quadratic function shown to determine the sign of the leading coefficient, its vertex, and the equation of the axis of symmetry. Solution Leading coefficient: The graph opens downward, so the leading coefficient a is negative. Vertex: The vertex is the highest point on the graph and is located at (1, 3). Axis of symmetry: Vertical line through the vertex with equation x = 1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 7
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Example Write the formula f(x) = x 2 + 10 x + 23 in vertex form by completing the square. Solution Given formula Subtract 23 from each side. Let k = 10; add (10/2)2 = 25. Factor perfect square trinomial. Subtract 2. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 9
Example Find the vertex of the graph of symbolically. Support your answer graphically and numerically. Solution a = 1/2 , b = 4, and c = 8. x-coordinate of vertex: y-coordinate evaluate f(4): The vertex is (4, 0). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 10
Solution continued Graphically and Numerically axis of symmetry Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 11
Example Use the vertex formula to write f(x) = 3 x 2 3 x + 1 in vertex form. Solution 1. Begin by finding 2. Find y. the vertex. The vertex is: Vertex form: Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 12
Example Graph the quadratic equation g(x) = 3 x 2 + 24 x 49. Solution The formula is not in vertex form, but we can find the vertex. The y-coordinate of the vertex is: The vertex is at (4, 1). The axis of symmetry is x = 4, and the parabola opens downward because the leading coefficient is negative. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 13
Solution continued Graph: g(x) = 3 x 2 + 24 x 49 Table of Values x 2 3 4 y 13 4 1 5 6 4 13 vertex Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 14
Applications and Models Example A junior horticulture class decides to enclose a rectangular garden, using a side of the greenhouse as one side of the rectangle. If the class has 32 feet of fence, find the dimensions of the rectangle that give the maximum area for the garden. Solution Let w be the width and L be the W L length of the rectangle. Because the 32 -foot fence does not go along the greenhouse, if follows that W + L + W = 32 or L = 32 – 2 W Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 15
Solution continued The area of the garden is the length times the width. This is a parabola that opens downward, and by the vertex formula, the maximum area occurs when The corresponding length is L = 32 – 2 W = 32 – 2(8) = 16 feet. The dimensions are 8 feet by 16 feet. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 16
Example A model rocket is launched with an initial velocity of vo = 150 feet per second and leaves the platform with an initial height of ho = 10 feet. a) Write a formula s(t) that models the height of the rocket after t seconds. b) How high is the rocket after 3 seconds? c) Find the maximum height of the rocket. Support your answer graphically. Solution a) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 17
Solution continued b) The rocket is 316 feet high after 3 seconds. c) Because a is negative, the vertex is the highest point on the graph, with an t-coordinate of The y-coordinate is: The vertex is at (4. 7, 361. 6). Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 18
Solution continued Graphical support is shown below. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 19
3. 2 Quadratic Equations and Problem Solving Understand basic concepts about quadratic equations ♦ Use factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equations ♦ Understand the discriminant ♦ Solve problems involving quadratic equations ♦ Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
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Solving Quadratic Equations The are four basic symbolic strategies in which quadratic equations can be solved. • Factoring • Square root property • Completing the square • Quadratic formula Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 22
Factoring A common technique used to solve equations that is based on the zero-product property. Example Solution Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 23
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Example A rescue helicopter hovers 68 feet above a jet ski in distress and drops a life raft. The height in feet of the raft above the water is given by Determine how long it will take for the raft to hit the water after being dropped from the helicopter. Solution (continued on next slide) The raft will hit the water when its height is 0 feet above the water. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 25
Solution continued The life raft will hit about 2. 1 seconds after it is dropped. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 26
Completing the Square Completing the square is useful when solving quadratic equations that do not factor easily. If a quadratic equation can be written in the form where k and d are constants, then the equation can be solved using Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 27
Example Solve 2 x 2 + 6 x = 7. Solution Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 28
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Example Solve the equation Solution Let a = 2, b = 5, and c = 9. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 30
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Example Use the discriminant to determine the number of solutions to the quadratic equation Solution Since solutions. the equation has two real Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 32
Modeling Projectile Motion Example The following table shows the height of a toy rocket launched in the air. a) b) Height of a toy rocket t (sec) 0 1 2 s(t) feet 28 12 36 Use to model the data. After how many seconds did the toy rocket strike the ground? Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 33
Solution continued a) If t = 0, then s(0) = 12, so The value of vo can be found by noting that when t = 2, s(2) = 28. Substituting gives the following result. Thus s(t) = 16 t 2 + 40 t + 12 models the height of the toy rocket. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 34
Solution continued b) The rocket strikes the ground when s(t) = 0, or when – 16 t 2 + 40 t + 12 = 0. Using the quadratic formula, where a = 4, b = – 10 and c = – 3 we find that • Only the positive solution is possible, so the toy rocket reaches the ground after approximately 2. 8 seconds. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 35
Example A box is is being constructed by cutting 2 inch squares from the corners of a rectangular sheet of metal that is 10 inches longer than it is wide. If the box has a volume of 238 cubic inches, find the dimensions of the metal sheet. Solution Step 1: Let x be the width and x + 10 be the length. Step 2: Draw a picture. x-4 x x+6 x + 10 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 36
Solution continued Since the height times the width times the length must equal the volume, or 238 cubic inches, the following can be written Step 3: Write the quadratic equation in the form ax 2 + bx + c = 0 and factor. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 37
Solution continued The dimensions can not be negative, so the width is 11 inches and the length is 10 inches more, or 21 inches. Step 4: After the 2 square inch pieces are cut out, the dimensions of the bottom of the box are 11 – 4 = 7 inches by 21 – 4 = 17 inches. The volume of the box is then , which checks. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 38
3. 3 Quadratic Inequalities ♦ Solve quadratic inequalities graphically ♦ Solve quadratic inequalities symbolically Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Quadratic Inequalities If an equals sign is replaced by >, , <, or ≤, a quadratic inequality results. A first step in solving a quadratic inequality is to determine the x-values where equality occurs. These x-values are the boundary numbers. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 40
Graphical Solutions The graph of a quadratic function opens either upward or downward. a = 1 parabola opens up x-intercepts: 1 and 2 Parabola lies below the x-axis between the intercepts for the equation x 2 x 2 = 0 Solutions to x 2 x 2 < 0, is the solution set {x| 1 < x < 2} or ( 1, 2) in interval notation. Solutions to x 2 x 2 > 0 include x-values either left of x = 1 or right of x = 2, where the parabola is above the x-axis {x|x < 1 or x > 2}. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 41
Example Solve the inequality. Write the solution set for each in interval notation. a) 3 x 2 + x 4 = 0 b) 3 x 2 + x 4 < 0 c) 3 x 2 + x 4 > 0 Solution a) Factoring (3 x + 4)(x – 1) = 0 x = – 4/3 x = 1 The solutions are – 4/3 and 1. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 42
Solution continued b) 3 x 2 + x 4 < 0 Parabola opening upward. x-intercepts are 4/3 and 1 Below the x-axis (y < 0) Solution set: (– 4/3, 1) c) 3 x 2 + x 4 > 0 Above the x axis (y > 0) Solution set: ( , 4/3) (1, ) y<0 y>0 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley y>0 43
Example The quadratic inequality can be used to compute stopping distances in feet for a car traveling x miles per hour on dry, level pavement. Solve the inequality shown below to determine safe speeds on a curve where a driver can see the road ahead for at most 150 feet. Solution We locate the point of intersection where x is positive. This occurs when x 31. 23. Safe speeds are less than 31 mph. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 44
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Example Solve x 2 > 7 x – 10 symbolically. Write the solutions in interval notation. Solution Step 1: Rewrite the inequality as x 2 – 7 x + 10 > 0 Step 2: Solve x 2 – 7 x + 10 = 0 (x – 5)(x – 2) = 0 x = 5 or x = 2 Step 3: These two boundary numbers separate the number line into three disjoint intervals. (– , 2), (2, 5), and (5, ) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 46
Solution continued Step 4: Choose test values. Interval (– , 2) (2, 5) (5, ) Test Value x x 2 – 7 x + 10 Positive or Negative? 0 3 6 10 – 2 4 Positive Negative Positive The expression is positive when x < 2 or x > 5. The solution set is (– , 2) (5, ). The boundary numbers are not included because the inequality involves >. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 47
3. 4 Transformations of Graphs Graph functions using vertical and horizontal translations ♦ Graph function using stretching and shrinking ♦ Graph function using reflections ♦ Combine transformations ♦ Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Vertical Shifts A graph is shifted up or down. The shape of the graph is not changed—only its position. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 49
Horizontal Shifts A graph is shifted left or right. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 50
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Example Shifts can be combined to translate a graph of y = f(x) both vertically and horizontally. Shift the graph of y = x 2 to the left 3 units and downward 2 units. y = x 2 y = (x + 3)2 2 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 52
Example Find an equation that shifts the graph of f(x) = x 2 2 x + 3 left 4 units and down 3 units. Solution To shift the graph left 4 units, replace x with (x + 4) in the formula for f(x). y = f(x + 4) = (x + 4)2 – 2(x + 4) + 3 To shift the graph down 3 units, subtract 3 from the formula. y = f(x + 4) 3 = (x + 4)2 – 2(x + 4) + (3 3) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 53
Stretching and Shrinking Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 54
Horizontal Stretching and Shrinking Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 55
Example Use the graph of y = f(x) to sketch the graph of each equation. a) y = 2 f(x) b) Solution a) Vertical stretching Multiply each y-coordinate on the graph by 2. ( 2, 1 2) = ( 2, 2) (0, 2 2) = (0, 4) (2, 1 2) = (2, 2) y = f(x) ( 2, 1) Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley (2, 1) (0, 2) 56
Solution continued b) y = f(x) ( 2, 1) Horizontal stretching Divide each x-coordinate by ½. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley (2, 1) (0, 2) 57
Reflections of Graphs Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 58
Example For the function f(x) = x 2 + x 2 graph its reflection across the x-axis and across the y-axis. Solution The graph is a parabola with x-intercepts 2 and 1. To obtain its reflection across the x-axis, graph y = f(x), or y = (x 2 + x 2). The x-intercepts have not changed. To obtain the reflection across the y-axis let y = f( x), or y = ( x)2 x 2. The x-intercepts have changed to 1 and 2. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 59
Combining Transformations of graphs can be combined to create new graphs. For example the graph of y = 3(x + 3)2 + 1 can be obtained by performing four transformations on the graph of y = x 2. 1. Shift of the graph 3 units left: y = (x + 3)2 2. Vertically stretch the graph by a factor of 3: y = 3(x + 3)2 3. Reflect the graph across the x-axis: y = 3(x + 3)2 4. Shift the graph upward 1 unit: y = 3(x + 3)2 + 1 Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 60
Combining Transformations continued Stretch vertically by a factor of 3 y = 3(x + Shift to the left 3 units. 2 3) +1 Reflect across the x-axis. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Shift upward 1 unit. 61
Example Describe how the graph of the equation can be obtained by transforming the graph of y = |x|. Then graph the equation. Solution Reflect the graph across the y-axis. Shift the graph left 3 units. Shift the graph down 2 units. Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 62
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