MAC Rev.S Learning Objectives. Learning Objectives (Cont.) Module 4 Quadratic Functions and Equations

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1 MAC 1140 Module 4 Quadratic Functions and Equations Learning Objectives Upon completing this module, you should be able to 1. understand basic concepts about quadratic functions and their graphs.. complete the square and apply the vertex formula. 3. graph a quadratic function by hand. 4. solve applications and model data. 5. understand basic concepts about quadratic equations. 6. use factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equations. 7. understand the discriminant. 8. solve problems involving quadratic equations. Rev.S10 Learning Objectives (Cont.) 9. perform arithmetic operations on complex numbers. 10. solve quadratic equations having complex solutions. 11. solve quadratic inequalities graphically. 1. solve quadratic inequalities symbolically. 13. graph functions using vertical and horizontal translations. 14. graph functions using stretching and shrinking. 15. graph functions using reflections. 16. combine transformations. Rev.S10 3 1

2 Quadratic Functions and Equations There are five sections in this module: 3.1 Quadratic Functions and Models 3. Quadratic Equations and Problem- Solving 3.3 Complex Numbers 3.4 Quadratic Inequalities 3.5 Transformations of Graphs Rev.S10 4 Let s get started by looking at some basic concepts about quadratic functions and graphs. Rev.S10 5 Quadratic Functions 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 term. f(x) = 3x + 3x + 5 g(x) = 5 x Rev.S10 6

3 Quadratic Functions (Cont.) 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. Rev.S10 7 Quadratic Functions (Cont.) 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. Rev.S10 8 Example of Different Parabolas Note: A parabola opens upward if a is positive and opens downward if a is negative. Rev.S10 9 3

4 Example of Different Parabolas (Cont.) Note: 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. Rev.S10 10 Graph of the Quadratic Function Now, let s 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. 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. Rev.S10 11 Let s Look at the Vertex Form of a Quadratic Function We can write the formula f(x) = x + 10x + 3 in vertex form by completing the square. Rev.S10 1 4

5 Let s Write the Vertex Form of a Quadratic Function by Completing the Square y x x = y! 3 = x + 10x y 3 5 x 0x! + = y + = ( x + 5) y = +! ( x 5) Subtract 3 from each side. Let k = 10; add (10/) = 5. Factor perfect square trinomial. Vertex Form Rev.S10 13 How to Use Vertex Formula to Write a Quadratic Function in Vertex Form? Use the vertex formula to write f(x) = 3x 3x + 1 in vertex form. : Since a = -3, b = -3 and c = 1, we just need to substitute them into the vertex formula. Mainly, you need to know the vertex formula for x; once you have solved for x, you can solve for y. 1. Begin by finding. Find y. the vertex f! " 3! # " 3! # b $ # % = # $ % # $ " % + 1 = & ' & ' & ' 4 x =! a The vertex is:! 1 " (! 3), 7 # $ % & 4 ' =! (! 3)! 1 " 7 1 Vertex form: f ( x) = # 3$ x + % + =! & ' 4 Rev.S10 14 Example Graph the quadratic equation g(x) = 3x + 4x 49. The formula is not in vertex form, but we can find the vertex. b 4 x =! =! = 4 a (! 3) The y-coordinate of the vertex is: g (4) =! 3(4) + 4(4)! 49 =! 1 The vertex is at (4, 1). The axis of symmetry is x = 4, and the parabola opens downward because the leading coefficient is negative. Rev.S

6 Graph: g(x) = 3x + 4x 49 Table of Values Example (Cont.) x y vertex Rev.S10 16 Example of Application 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 3 feet of fence, find the dimensions of the rectangle that give the maximum area for the garden. (Think about using the vertex formula.) Let w be the width and L be the length of the rectangle. L W Because the 3-foot fence does not go along the greenhouse, if follows that W + L + W = 3 or L = 3 W Rev.S10 17 Example of Application (Cont.) The area of the garden is the length times the width. A = LW = (3! W ) W L W = 3W! W This is a parabola that opens downward, and by the vertex formula, the maximum area occurs when 3 W =! = 8 feet (! ) The corresponding length is L = 3 W = 3 (8) = 16 feet. The dimensions are 8 feet by 16 feet. Rev.S

7 Another Example A model rocket is launched with an initial velocity of v o = 150 feet per second and leaves the platform with an initial height of h o = 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. a) s( t) =! 16t + v o t + h o =! 16t + 150t + 10 Rev.S10 19 Another Example (Cont.) b) s (3) =! 16(3) + 150(3) + 10 = 316 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: b 150 t =! =! = " 4.7 a (! 16) s (4.7) =! 16(4.7) + 150(4.7) + 10 = feet The vertex is at (4.7, 361.6). Rev.S10 0 How to Solve Quadratic Equations? The are four basic symbolic strategies in which quadratic equations can be solved. Factoring Square root property Completing the square Quadratic formula Rev.S10 1 7

8 Factoring A common technique used to solve equations that is based on the zero-product property. Example x! 4x! 5 = 1 x! 4x! 6 = 0 x! x! 3 = 0 ( x )( x )! = 0 x! 3 = 0 or x + 1 = 0 x = 3 or x =! 1 Rev.S10 Square Root Property Example:! + = 16t 68 0! 16t =! 68! 68 t =! t = 4 17 t = ± 4 t " ±.1 Rev.S10 3 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! k "! k " x + kx + = x + # $ # $. % & % & Rev.S10 4 8

9 Completing the Square (Cont.) Solve x + 6x = 7. x + 6x = 7 7 x + 3x =! 3 " 7! 3 " x + 3x + # $ = + # $ % & % &! 3 " 3 # x + $ = % & x + = ± x = ' ± = ' ± 4 Rev.S10 5 Quadratic Formula Solve the equation x! 5x! 9 = 0. Let a =, b = 5, and c = 9.! b ± b! 4ac x = a 5 ± (! 5)! 4( )(! 9) x = ( ) 5 ± 97 x = 4 Rev.S10 6 Quadratic Equations and the Discriminant Use the discriminant to determine the number of solutions to the quadratic equation b! 4ac =! 6! 4! 3 15 = 16 ( ) ( )( ) Since b! 4ac > 0, the equation has two real solutions. Rev.S10 7 9

10 Example of a Modeling a Projectile Motion The following table shows the height of a toy rocket launched in the air. t (sec) s(t) feet ( ) Height of a toy rocket 1 36 a) Use s t =! 16t + vot + ho to model the data. b) After how many seconds did the toy rocket strike the ground? Rev.S10 8 Example (Cont.) ( ) a) If t = 0, then s(0) = 1, so s t =! 16t + v0t + 1. The value of v o can be found by noting that when t =, s() = 8. Substituting gives the following result.! 16( ) + v0 ( ) + 1 = 8 v = 80 Thus s(t) = 16t + 40t + 1 models the height of the toy rocket. b) The rocket strikes the ground when s(t) = 0, or when 16t + 40t + 1 = 0.! 16t + 40t + 1 = 0 4t! 10t! 3 = 0 Using the quadratic formula, where a = 4, b = 10 and c = 3 we find that x!.8 or x! " 0.3 Only the positive solution is possible, so the toy rocket reaches the ground after approximately.8 seconds. Rev.S v = 40 0 One More Example A box is is being constructed by cutting 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 38 cubic inches, find the dimensions of the metal sheet. Step 1: Let x be the width and x + 10 be the length. Step : Draw a picture. x - 4 x + 6 x Since the height times the width times the length must equal the volume, or 38 cubic inches, the following can be written Rev.S

11 One More Example (Cont.) ( x! 4)( x + 6) = 38 or ( x! 4)( x + 6) = 119 Step 3: Write the quadratic equation in the form ax + bx + c = 0 and factor. x + x! 4 = 119 x + x! 143 = 0 ( x + 13)( x! 11) = 0 x =! 13 or x = 11 The dimensions can not be negative, so the width is 11 inches and the length is 10 inches more, or 1 inches. Step 4: After the square inch pieces are cut out, the dimensions of the bottom of the box are 11 4 = 7 inches by 1 4 = 17 inches. The volume of the box is then x 7 x 17 = 38, which checks. Rev.S10 31 Complex Numbers A complex number can be written in standard form as a + bi, where a and b are real numbers. The real part is a and the imaginary part is b. Rev.S10 3 Examples Write each expression in standard form. Support your results using a calculator. a) ( 4 + i) + (6 3i) b) ( 9i) (4 7i) a) ( 4 + i) + (6 3i) = i 3i = i b) ( 9i) (4 7i) = 4 9i + 7i = 4 i Rev.S

12 More Examples Write each expression in standard form. Support your results using a calculator. 16 a) ( + 5i) b) 3 + i a) ( + 5i) = ( + 5i)( + 5i) = 4 10i 10i + 5i = 4 0i + 5( 1) = 1 0i b) =! 3 + i 3+ i 3" i 48 " 16i = 9 " i 48 " 16i 4 8 = = ± i Rev.S10 34 Quadratic Equations with Complex s We can use the quadratic formula to solve quadratic equations, even if the discriminant is negative. However, there are no real solutions, and the graph does not intersect the x-axis. The solutions can be expressed as imaginary numbers. Rev.S10 35 Example Solve the quadratic equation 4x 1x = 11. Rewrite the equation: 4x 1x + 11 = 0 a = 4, b = 1, c = 11! b ± b! 4ac x = a 1 ± (1)! 4(4)(11) 1 ±! 3 = 8 (4) 1 ± 4i 3 i = = ± 8 Rev.S10 36 = 1

13 Quadratic Inequality If an equals sign is replaced by >,, <, or, a quadratic inequality results. The first step in solving a quadratic inequality is to determine the x- values where equality occurs. These x-values are the boundary numbers. Let s look at a quadratic inequality graphically. The graph of a quadratic function opens either upward or downward. In this case, a = 1, parabola opens up, and we have x-intercepts: 1 and y x x =!! Rev.S10 37 Quadratic Inequality (cont.) Note the parabola lies below the x-axis between the intercepts for the equation x x = 0 s to x x < 0, is the solution set {x 1 < x < } or ( 1, ) in interval notation. s to x x > 0 include x-values either left of x = 1 or right of x =, where the parabola is above the x-axis, and thus {x x < 1 or x > }. y x x =!! Rev.S10 38 Another Example Solve the inequality. Write the solution set for each in interval notation. a) 3x + x 4 = 0 b) 3x + x 4 < 0 c) 3x + x 4 > 0 a) Factoring (3x + 4)(x 1) = 0 x = 4/3 x = 1 The solutions are 4/3 and 1. Rev.S

14 Another Example (Cont.) b) 3x + x 4 < 0 Parabola opening upward. x-intercepts are 4/3 and 1 Below the x-axis (y < 0) set: ( 4/3, 1) y < 0 c) 3x + x 4 > 0 Above the x axis (y > 0) set: (, 4/3) (1, ) y > 0 y > 0 Rev.S10 40 Solving Quadratic Inequalities in 4 Steps Rev.S10 41 Transformation of Graphs: Vertical Shifts A graph is shifted up or down. The shape of the graph is not changed only its position. Rev.S

15 Transformation of Graphs: Horizontal Shifts A graph is shifted left or right. The shape of the graph is not changed only its position. Rev.S10 43 Transformation of Graphs Rev.S10 44 Example of Transformation of Graphs Shifts can be combined to translate a graph of y = f(x) both vertically and horizontally. Shift the graph of y = x to the left 3 units and downward units. y = x y = (x + 3) y = (x + 3) Rev.S

16 Another Example Find an equation that shifts the graph of f(x) = x x + 3 left 4 units and down 3 units. To shift the graph left 4 units, replace x with (x + 4) in the formula for f(x). y = f(x + 4) = (x + 4) (x + 4) + 3 To shift the graph down 3 units, subtract 3 to the formula. y = f(x + 4) 3 = (x + 4) (x + 4) Rev.S10 46 Vertical Stretching and Shrinking Rev.S10 47 Horizontal Stretching and Shrinking Rev.S

17 Reflection of Graphs Rev.S10 49 Reflection of Graphs For the function f(x) = x + x graph its reflection across the x-axis and across the y-axis. The graph is a parabola with x-intercepts and 1. To obtain its reflection across the x-axis, graph y = f(x), or y = (x + x ). The x-intercepts have not changed. To obtain the reflection across the y-axis let y = f( x), or y = ( x) x. The x-intercepts have changed to 1 and. Rev.S10 50 Combining Transformations Transformations of graphs can be combined to create new graphs. For example the graph of y = 3(x + 3) + 1 can be obtained by performing four transformations on the graph of y = x. 1. Shift of the graph 3 units left: y = (x + 3). Vertically stretch the graph by a factor of 3: y = 3(x + 3) 3. Reflect the graph across the x-axis: y = 3(x + 3) 4. Shift the graph upward 1 unit: y = 3(x + 3) + 1 Rev.S

18 Combining Transformations (Cont.) Stretch vertically by a factor of 3 Shift to the left 3 units. y = 3(x + 3) + 1 Reflect across the x-axis. Shift upward 1 unit. Rev.S10 5 What have we learned? We have learned to 1. understand basic concepts about quadratic functions and their graphs.. complete the square and apply the vertex formula. 3. graph a quadratic function by hand. 4. solve applications and model data. 5. understand basic concepts about quadratic equations. 6. use factoring, the square root property, completing the square, and the quadratic formula to solve quadratic equations. 7. understand the discriminant. 8. solve problems involving quadratic equations. Rev.S10 53 What have we learned? (Cont.) 9. perform arithmetic operations on complex numbers. 10. solve quadratic equations having complex solutions. 11. solve quadratic inequalities graphically. 1. solve quadratic inequalities symbolically. 13. graph functions using vertical and horizontal translations. 14. graph functions using stretching and shrinking. 15. graph functions using reflections. 16. combine transformations. Rev.S

19 Credit Some of these slides have been adapted/modified in part/whole from the slides of the following textbook: Rockswold, Gary, Precalculus with Modeling and Visualization, 3th Edition Rev.S