Quadratic Functions Unit

Transcription

1 Quadratic Functions Unit (Level IV Academic Math) NSSAL (Draft) C. David Pilmer 009 (Last Updated: Dec, 011) Use our online math videos. YouTube: nsccalpmath

2 This resource is the intellectual property of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education. The following are permitted to use and reproduce this resource for classroom purposes. Nova Scotia instructors delivering the Nova Scotia Adult Learning Program Canadian public school teachers delivering public school curriculum Canadian nonprofit tuition-free adult basic education programs The following are not permitted to use or reproduce this resource without the written authorization of the Adult Education Division of the Nova Scotia Department of Labour and Advanced Education. Upgrading programs at post-secondary institutions Core programs at post-secondary institutions Public or private schools outside of Canada Basic adult education programs outside of Canada Individuals, not including teachers or instructors, are permitted to use this resource for their own learning. They are not permitted to make multiple copies of the resource for distribution. Nor are they permitted to use this resource under the direction of a teacher or instructor at a learning institution. Acknowledgments The Adult Education Division would like to thank the following university professors for reviewing this resource to ensure all mathematical concepts were presented correctly and in a manner that supported our learners. Dr. Genevieve Boulet (Mount Saint Vincent University) Dr. Robert Dawson (Saint Mary s University) The Adult Education Division would also like to thank the following NSCC instructors for piloting this resource and offering suggestions during its development. Charles Bailey (IT Campus) Elliott Churchill (Waterfront Campus) Barbara Gillis (Burridge Campus) Barbara Leck (Pictou Campus) Suzette Lowe (Lunenburg Campus) Floyd Porter (Strait Area Campus) Brian Rhodenizer (Kingstec Campus) Joan Ross (Annapolis Valley Campus) Jeff Vroom (Truro Campus)

3 Table of Contents Introduction.. i Negotiated Completion Date i The Big Picture ii Course Timelines. iii Introduction to Quadratic Functions 1 Using the Graphing Calculator to Interpret Quadratic Functions The Most Basic Quadratic Function 8 Quadratic Functions and Transformations... 9 State the Transformations: Quadratic Functions.. 1 Visualizing Quadratic Functions.. 15 Graphing Quadratic Functions Using Transformations Using Finite Differences to Identify Quadratic Functions... 6 Using Finite Differences to Determine the Equation... 1 Find the Equation Given Three Points. 0 Find the Equation Given the Verte and a Point. Putting It Together Part 1 50 Multiplying Polynomials. 5 Factoring Part 1 66 Factoring Part 7 Alternate Forms of Decomposition (Optional) 80 Factoring Part 85 Standard Form to Transformational Form Solving Quadratic Equations by Factoring.. 99 Solving Quadratic Equations Using the Quadratic Formula 105 Programming the Quadratic Formula into a TI-8 or TI-8 11 Word Problems Involving Quadratic Equations. 11 Finding the Verte Application Questions.. 1 Putting It Together Part. 19 Post-Unit Reflections Additional Practice: Multiplying Polynomials (Optional) Additional Practice: Factoring Polynomials (Optional) Additional Practice: Standard Form to Transformational Form (Optional) 15 Additional Practice: Word Problems Involving Quadratic Equations (Optional).. 15 Review Sheet: Types of Factoring Factoring Flow Chart Terminology. 158 Answers 159 NSSAL i Draft 009 C. D. Pilmer

4 Introduction In this unit we will learn about quadratic functions. The graphs of these functions form curves called parabolas. Their equations can be written in the transformation form y = k( c) d or in the standard form y = a b c. We will learn to work with both forms of the equations and see how this type of function can be used to model a variety of real world applications and be used to answer questions regarding those applications. It is important to note that different teachers and professors use different terminology to describe these two forms of the quadratic functions. Many refer to y = a b c as the general form and y = k( c) d as the standard form. Please be aware of this if you are using other resources to assist you with the concepts covered in this unit. This is a very lengthy unit because many smaller concepts have to be taught prior to learning the larger concepts directly connected to quadratic functions. For eample, before one can learn how to solve quadratic equations, one must know how to multiply polynomial epressions and factor polynomial epressions. Similarly, before one can learn how to graph quadratic functions, one must learn what transformations are and how they affect curves. The key to success in this unit is recognizing how all of the smaller concepts connect and give one a broad understanding of quadratic functions. In this unit learners will do the following. Interpret graphs of real world phenomena. All of the graphs will be quadratic functions. Graph quadratic functions of the form y = k( c) d using transformations. Determine the equation of a quadratic function in a variety of ways. Multiply and factor polynomial epressions. Change quadratic functions from their standard form to their transformational form. Solve quadratic equations by factoring and using the quadratic formula. Determine the coordinates of the verte of a quadratic function. Solve multi-step multi-concept application questions All of this will be accomplished while looking at a variety of real world applications of quadratic functions and revisiting concepts covered in prior units (e.g. domain, range, intercepts, ). Negotiated Completion Date After working for a few days on this unit, sit down with your instructor and negotiate a completion date for this unit. Start Date: Completion Date: Instructor Signature: Student Signature: NSSAL ii Draft 009 C. D. Pilmer

5 The Big Picture The following flow chart shows the optional bridging unit and the eight required units in Level IV Academic Math. These have been presented in a suggested order. Bridging Unit (Recommended) Solving Equations and Linear Functions Describing Relations Unit Relations, Functions, Domain, Range, Intercepts, Symmetry Systems of Equations Unit by Systems, Plane in -Space, by Systems Trigonometry Unit Pythagorean Theorem, Trigonometric Ratios, Law of Sines, Law of Cosines Sinusoidal Functions Unit Periodic Functions, Sinusoidal Functions, Graphing Using Transformations, Determining the Equation, Applications Quadratic Functions Unit Graphing using Transformations, Determining the Equation, Factoring, Solving Quadratic Equations, Verte Formula, Applications Rational Epressions and Radicals Unit Operations with and Simplification of Radicals and Rational Epressions Eponential Functions and Logarithms Unit Graphing using Transformations, Determining the Equation, Solving Eponential Equations, Laws of Logarithms, Solving Logarithmic Equations, Applications Inferential Statistics Unit Population, Sample, Standard Deviation, Normal Distribution, Central Limit Theorem, Confidence Intervals NSSAL iii Draft 009 C. D. Pilmer

6 Course Timelines Academic Level IV Math is a two credit course within the Adult Learning Program. As a two credit course, learners are epected to complete 00 hours of course material. Since most ALP math classes meet for 6 hours each week, the course should be completed within 5 weeks. The curriculum developers have worked diligently to ensure that the course can be completed within this time span. Below you will find a chart containing the unit names and suggested completion times. The hours listed are classroom hours. In an academic course, there is an epectation that some work will be completed outside of regular class time. Unit Name Minimum Completion Time in Hours Maimum Completion Time in Hours Bridging Unit (optional) 0 0 Describing Relations Unit 6 8 Systems of Equations Unit 18 Trigonometry Unit 18 0 Sinusoidal Functions Unit 0 Quadratic Functions Unit 6 Rational Epressions and Radicals Unit 1 16 Eponential Functions and Logarithms Unit 0 Inferential Statistics Unit 0 Total: 150 hours Total: 00 hours As one can see, this course covers numerous topics and for this reason may seem daunting. You can complete this course in a timely manner if you manage your time wisely, remain focused, and seek assistance from your instructor when needed. NSSAL iv Draft 009 C. D. Pilmer

7 Introduction to Quadratic Functions In this unit we will be looking at quadratic functions. Quadratic functions are of the form y = a b c (standard form) or can also be written in the form y = k( c) d (transformational form). We will learn about these two different forms at a later date. Note that in both cases the function uses the second power (square) as the highest power of the unknown. When quadratic functions are graphed, they form curves called parabolas. Rather than starting with all the algebra associated with quadratic functions, let s start with some real world applications. One of the most common applications is projectile motion. When a person throws a baseball to another person, the trajectory or path of the ball is referred to the motion of the projectile. As you will see, there are other applications of quadratic functions, many associated with business where one is attempting to minimize cost or maimize profit. Eample 1 A soccer ball is kicked from ground level. Its flight path is shown on the graph. The height of the ball, in metres, is on the vertical ais. The distance the ball travels horizontally, in metres, is on the horizontal ais. (a) What is the maimum height reached by the ball? (b) How far does the ball travel horizontally when it reaches this maimum height? (c) What s the initial height of the soccer ball? (d) How far will the ball travel horizontally before it hits the ground? (e) Approimate the height of the ball after it has traveled metres horizontally. (f) Determine the horizontal distances that correspond to a height of metres. (g) State the domain and range. Answers: (a) The ball reaches a maimum height of metres. (b) To reach its maimum height, the ball must travel 6 metres horizontally. (c) The initial height of the soccer ball is 0 metres. (d) The ball travels 1 metres horizontally before it hits the ground. (e) After the ball has traveled metres horizontally, it reaches an approimate height of. metres. (f) The ball is at a height of metres at two instances; as the ball goes up and as it comes down. This will occur when the horizontal distances are metres and 9 metres. R 0 d 1 hε R 0 h (g) Domain: { dε } Range: { } Horizontal Distance in Metres Note: The flight paths of projectiles are not truly parabolic when we factor in wind resistance. For our purposes, using a quadratic function to model these situations is acceptable as long as we recognize the limitations of this mathematical model. Height in Metres 5 1 NSSAL 1 Draft 009 C. D. Pilmer

8 Questions: 1. A baseball is thrown. Its flight path is shown on the graph. The height of the ball, in metres, is on the vertical ais. The distance the ball travels horizontally, in metres, is on the horizontal ais. Height in Metres Horizontal Distance in Metres (a) (b) (c) (d) (e) (f) (g) (h) Approimately how far will the ball travel horizontally before it hits the ground? Approimate the height of the ball after it has traveled 1 metre horizontally. What is the maimum height reached by the ball? How far does the ball travel horizontally when it reaches this maimum height? From what height was the baseball thrown? Approimate the horizontal distances that correspond to a height of metres. State the domain of this function. State the range of this function. Answers. Akira s company manufactures alternators for a large international automotive company. If her company manufactures too few alternators, she can not make a profit. If the large international company requests too many alternators, her profits drop because she has to pay overtime to her workers, and hire eternal trucking companies to deliver the additional parts. There is also a limit to the number of alternators her company is capable of producing. The following graph shows the relationship between the monthly profits her company can make and the number of alternators she produces in a month for the large international company. The profits are measure in tens of thousands of dollars. The number of alternators is measured in thousands. Profit Number of Alternators Produced NSSAL Draft 009 C. D. Pilmer

9 (a) (b) (c) (d) (e) (f) (g) How many alternators should Akira s company produce in a month to maimize her company s profit. If her company does not produce any alternators, what will her losses be for that month? If no profit is made in a month, how many alternators must be produced? If they produce 6000 alternators in a month, what will be the epected monthly profit? If the monthly profit is $ , how many alternators were produced? State the domain. State the range. Answers. The largest radio telescope in the world is the Arecibo telescope located in Puerto Rico. The reflecting surface for the telescope is embedded in the ground. If one was to cut the reflecting surface in half, the resulting curve would be formed. This graph shows the depth in feet of the reflecting surface relative to the distance in feet from the upper edge of the surface depth (ft) horizontal distance (ft) (a) (b) (c) (d) (e) (f) (g) Approimately how wide is the reflective surface on the telescope? Approimately how far down is the lowest point on the reflective surface? At what horizontal distances is the reflective surface 0 feet below ground level? State the depth intercept. State the domain. State the range. Is this function odd, even or neither? Answers NSSAL Draft 009 C. D. Pilmer

10 Using the Graphing Calculator to Interpret Quadratic Functions Eample 1 A garden hose sprays a stream of water across a lawn. The quadratic function h = 0.5d d 1 describes the height, h, of the stream of water above the lawn in terms of the horizontal distance, d, from the hose nozzle. Both the height and horizontal distance are measured in metres. Using graphing technology, answer each of these questions. Use the following WINDOW settings. (a) What is the maimum height reached by the water? (b) What horizontal distance corresponds to the maimum height of the stream of water? (c) What height is the hose nozzle? (d) At what horizontal distance will the water hit the ground? (e) If the water pressure was increased significantly, what feature on the graph would not change? Answer: (a) You first have to enter the function and graph it on the calculator. Y = > (Enter the function.) > WINDOW > (Adjust window settings.) > GRAPH Using the TRACE command and the right and left arrows, you can move to the highest point on the curve. Coordinates are typically of the form (,y) however, in this situation they are really (d,h). Since we want the maimum height for this question, we ll use the number 5. The water stream reaches a maimum height of 5 m. (b) You don t have to adjust anything on the calculator to do this part of the question. For this question you want the horizontal distance that corresponds to the maimum height. This will be the (or d) coordinate of the point you found in part (a). The horizontal distance corresponding to the maimum height is m. (c) Using the TRACE command, you find that the hose nozzle is at a height of 1 m. NSSAL Draft 009 C. D. Pilmer

11 (d) Using the TRACE command, you find that the water will hit the ground at approimately 8.5 m. (e) If the water pressure is increased the water stream would go higher and a further horizontal distance. The only thing that wouldn t change is the initial height of the water (i.e. the height of the hose nozzle). The y- (or h-) intercept would remain the same. Questions: 1. A baseball is thrown vertically into the air. It s height with respect to time can be described by the quadratic function h = 16t -.9t, where h is the height in metres and t is the time in seconds. (a) Graph the equation on a graphics calculator using the following Window settings. Sketch the graph in the space provided. Parts (b) to (l) are multiple choice questions. (b) The vertical ais (dependent variable) represents (i) the height of the ball off the ground. (ii) the speed of the ball. (iii) the time. (iv) the spot where the ball hits the ground. (v) the distance the ball travels horizontally. (c) The horizontal ais (independent variable) represents (i) the height of the ball off the ground. (ii) the speed of the ball. (iii) the time. (iv) the spot where the ball hits the ground. (v) the distance the ball travels horizontally. (d) The graph of this equation is a (i) straight line formed by a linear function. (ii) straight line formed by a quadratic function. (iii) parabola formed by a linear function. (iv) parabola formed by a quadratic function. (v) none of these NSSAL 5 Draft 009 C. D. Pilmer

12 (e) What is the initial height of the ball before it was thrown? (i) 0 m (ii) m (iii). m (iv) m (v) 6 m (f) Approimate the maimum height reached by the ball. (i) 0 m (ii) 1.6 m (iii) 10 m (iv) 15 m (v) 18 m (g) When does the ball reach its maimum height? (i) 0 s (ii) 1. s (iii) 1.6 s (iv) 15 s (v) 0 s (h) Approimate the height of the ball 1 second after it is thrown. (i) 1 m (ii) 1 m (iii) 1 m (iv) 15 m (v) 18 m (i) Approimate the time when the ball reaches a height of 10 m. (i) 0.6 s (ii).6 s (iii) 0.8 s (iv) both 0.6 s and.6 s (j) When does the ball strike the ground? (i) 1.6 s (ii).7 s (iii).1 s (iv). s (v).7 s (k) Determine the domain and range. R 0 t. (i) Domain: { tε } Range: { hε R 0 h 15 } (ii) Domain: { hε R 0 h 15 } Range: { tε R 0 t. } (iii) Domain: { tε R 0 t 1.6 } Range: { hε R h 15 } (iv) Domain: { hε R h 15 } Range: { tε R 0 t 1.6 } (l) Another individual throws a rock and the equation which describes its height with respect to time is h = 1t -.9t. How are these two situations (ball thrown/rock thrown) similar? (i) The ball and rock hit the ground at the same time. (ii) The ball and rock reached the same maimum height. (iii) The ball and rock were thrown from the same height. (iv) The ball and rock were thrown at the same speed. (v) none of these. The distance a car travels after the driver decides to slam on the brakes must consider two factors: the distance the car travels as the driver reacts to the situation (no brakes applied) and the distance the car travels when the brakes have been applied. The quadratic equation d = 0.007s 0. s describes the stopping distance, d, in terms of the initial speed, s, of the car. The distance is measured in metres and the speed is measured in kilometers per hour. Use graphing technology to answer the following questions. Use the following WINDOW settings. (Note: You will only be viewing half of the parabola.) NSSAL 6 Draft 009 C. D. Pilmer

13 (a) What is the y- (or d-) intercept and what does it represent in this situation? (b) If the stopping distance was 5 m, how fast was the driver going when he/she decided to slam on the brakes? (c) If the car was initially traveling at 100 km/h before the driver decides to hit the brakes, how far will the car travel before it comes to a stop? (d) If the car was initially traveling at 10 km/h before the driver decides to hit the brakes, how far will the car travel before it comes to a stop? (e) How do you feel about how the answers to (c) and (d) compare?. When a car is driven, the amount of gas consumed per kilometer changes based on the speed of the car. The quadratic equation c = s 0.116s 11.1 describes cost of gas per kilometre in terms of the speed of the vehicle. The cost is measured in cents per kilometre and the speed is measured in km/h. Use graphing technology to answer the following questions. Use the following WINDOW settings. (a) What speed is most cost-efficient? (b) At what speeds are you getting 9 cents per kilometre? (c) What is the cost per kilometre for a speed of 90 km/h? Think About Up to this point we have used graphs or graphing technology to understand situations that can be modeled using quadratic functions. Obviously there are many algebraic skills that we will have to learn so that we don t have to rely only on graphs and graphing technology. Over the net few weeks you will learn how to use paper and pencil techniques to: 1. graph quadratic functions. determine the equations of quadratic functions. solve quadratic equations associated with quadratic functions NSSAL 7 Draft 009 C. D. Pilmer

14 The Most Basic Quadratic Function The most basic quadratic function is y =. It has not undergone any transformations (i.e. reflections, stretches or translations). We will graph this function by generating a table of values using -values from - to, plotting the points, and connecting them with a smooth curve. y = y Please note that this type of curve is called a parabola. 1. State the domain.. State the range.. The verte of a parabola is either the highest or lowest point on the curve depending of the orientation of the curve. State the coordinates of the verte. Verte:. The ais of symmetry for a quadratic function is the vertical line that cuts the curve into two identical halves. State the equation of the ais of symmetry. Ais of Symmetry: 5. Is the function odd, even, or neither? 6. Why is this type of curve classified as a function? NSSAL 8 Draft 009 C. D. Pilmer

15 Quadratic Functions and Transformations In the last section we graphed the most basic quadratic function y = using a table of values. The resulting cupshaped curve is called a parabola. It has it s verte at ( 0,0) and the equation of the ais of symmetry is = 0. The curve is concave upwards. Name: Let s make this parabola and its corresponding table of values on a graphics calculator. Press Y =, enter the function ( y = ), and set the ZOOM feature to ZStandard. To generate the table of values, press TBLSET, set TblStart = - and set Tbl = 1. Now press TABLE. Complete the chart below. Our Basic Quadratic Function y = Enter: Y1=X^ Table of Values TblStart = - Tbl = 1 y Sketch of Graph Investigation: In each part of this investigation you are going to alter the equation of the quadratic function y = and see what effect this has on the graph and table of values. You will also identify the type of transformation (vertical stretch, horizontal stretch, vertical translation, horizontal translation or reflection in -ais) that has occurred. It is important to note that in each part of this investigation, you will be comparing your transformed quadratic function to the quadratic function y =. You will also have to determine the mapping rule. The mapping rule eplains how the ordered pairs in our original function, y =, have been changed to the NSSAL 9 Draft 009 C. D. Pilmer

16 ordered pairs in our new function. Graph each function and generate its table of values using technology. Note that the TblStart values change in parts (f) and (g) of this investigation. Complete the following chart. For the first few parts, the chart is partially completed. New Quadratic Table of Function Values (a) y = TblStart = - Tbl = 1 y Sketch of Graph Mapping Rule and Transformation (, y) (, y) We re dealing with a reflection in the -ais because the parabola is now upside down (i.e. concave downwards). (b) y = TblStart = - Tbl = 1 y (, y) (, y) (c) 1 y = TblStart = - Tbl = 1 y - (, y) (, ) We re dealing with a vertical stretch of 1 because the parabola appears wider. NSSAL 10 Draft 009 C. D. Pilmer

17 New Quadratic Table of Function Values (d) y = TblStart = - Tbl = 1 Sketch of Graph Mapping Rule and Transformation (, y) (, ) y (e) y = 5 TblStart = - Tbl = 1 (, y) (, ) y (f) y = ( ) TblStart = -6 Tbl = 1 (, y) (, ) y NSSAL 11 Draft 009 C. D. Pilmer

18 New Quadratic Function Table of Values (g) y = ( ) TblStart = 1 Tbl = 1 y 1 9 Sketch of Graph Mapping Rule and Transformation (, y) (, ) Summarize Your Findings: Equation Transformation (a) y = reflection in the -ais (b) y = (c) 1 1 y = vertical stretch of (d) y = (e) y = 5 (f) y = ( ) (g) y = ( ) Conclusions: If a quadratic function is of the form y = k( c) d, then: (i) the negative sign in front of the k indicates that a has occurred. (ii) the k indicates that a has occurred. (iii) the d indicates that a has occurred. (iv) the c indicates that a has occurred. The form y = k( c) d is referred to as the transformational form of the equation. If k is negative, then we know that we are also dealing with a reflection in the -ais. NSSAL 1 Draft 009 C. D. Pilmer

19 Questions 1. What transformation affects the -coordinate of the verte?. What transformation affects the y-coordinate of the verte?. Which transformation determines whether the parabola is concave upwards or concave downwards? Eamples For each of the following, state the transformations, state the coordinates of the verte, and determine whether the parabola is concave upwards or concave downwards. (a) y = 6( 7) 1 1 (b) y = ( 9) Answers: (a) y = 6( 7) 1 Reflection in the -ais, VS = 6, VT = 1, HT = 7 The verte is found by looking at the horizontal translation and vertical translation. The coordinates of the verte are (7, 1). The parabola is concave downwards because the quadratic function has undergone a reflection in the -ais. 1 y = 9 (b) ( ) VS = 1, HT = -9 The verte is found by looking at the horizontal translation and vertical translation. The coordinates of the verte are (-9, 0). The parabola is concave upwards because the quadratic function has not undergone a reflection in the -ais. NSSAL 1 Draft 009 C. D. Pilmer

20 State the Transformations: Quadratic Functions 1. In the previous section, you learned how to identify the transformations that occurred to y = when a quadratic function is presented in the form y = k( c) d. In the questions that follow, you will be given the equation of a quadratic function and be asked identify the transformations. Two eamples have been completed for you. Function e.g. ( ) Horizontal Translation Vertical Translation Reflection in the -ais Vertical Stretch y = none no e.g. ( ) ( ) 7 h = yes none (a) y = ( 7) 6 (b) y = g ( ) = 8 (c) y = ( ) (d) ( ) 5 (e) ( ) y = 7 (f) 5( ) 1 y = (g) ( ) 6( ) 1 h = (h) f ( ) = 8. Fill in the blanks. You may choose from the following terms. (Vertical Stretch, Vertical Translation, Horizontal Translation, Reflection in the -ais) (a) A will change a parabola from right side up (i.e. concave upwards) to upside down (i.e. concave downwards). (b) The -coordinate of the verte of a quadratic function can be determined by looking at the. (c) The y-coordinate of the verte of a quadratic function can be determined by looking at the. (d) The determines whether the graph is wider or narrower than the graph of y =. NSSAL 1 Draft 009 C. D. Pilmer

21 Visualizing Quadratic Functions We know that the graph of the quadratic function y = forms a cupshaped curve called a parabola. This function can be graphed on a graphing calculator. The resulting curve in ZStandard mode is displayed on the right. Most quadratic functions have undergone one or more transformations. The power of transformations is that we can visualize the resulting graph without having to do any formal work. For eample, if we have the quadratic function y = ( 5), we know the following. The parabola is concave downwards, due to the reflection in the -ais. The verte is located at (-5, ) due to the horizontal and vertical translations. The shape of the curve is narrower than that of y = due to the vertical stretch of. If you were asked to match this equation to one of the four graphs shown below, you would choose Graph B because it has all three features listed above. Graph A Graph B Graph C Graph D Correct Match Questions: 1. Match the equation to the appropriate graph. The scales on both the -ais and y-ais go from -10 to 10 (ZStandard mode). (a) y = ( ) Graph (i) Graph (ii) Graph (iii) Graph (iv) (b) y = ( 6) Graph (i) Graph (ii) Graph (iii) Graph (iv) NSSAL 15 Draft 009 C. D. Pilmer

22 (c) y = ( ) Graph (i) Graph (ii) Graph (iii) Graph (iv) (d) y = ( 5) Graph (i) Graph (ii) Graph (iii) Graph (iv) 1 (e) y = Graph (i) Graph (ii) Graph (iii) Graph (iv) (f) y = 1 ( ) Graph (i) Graph (ii) Graph (iii) Graph (iv) (g) y = ( ) 9 Graph (i) Graph (ii) Graph (iii) Graph (iv) NSSAL 16 Draft 009 C. D. Pilmer

23 (h) y = Graph (i) Graph (ii) Graph (iii) Graph (iv) (i) y = ( 6) Graph (i) Graph (ii) Graph (iii) Graph (iv) 1 (j) y = ( ) 5 Graph (i) Graph (ii) Graph (iii) Graph (iv). For each of the functions below, answer the following. Determine the coordinates of the verte. Describe the concavity (upwards or downwards) Describe the shape of the curve compared to that for the function narrower, wider, or no change. Two sample questions have been completed. y =. Use the terms Function Verte Concavity Shape e.g. ( ) y = (, 0) upwards narrower e.g. ( ) ( ) 7 f = 5 (-5, 7) downwards no change (a) y = ( ) 6 1 (b) y = ( ) 9 (c) y = 8 10 (d) g ( ) = ( ) 5 NSSAL 17 Draft 009 C. D. Pilmer

24 Graphing Quadratic Functions Using Transformations Before you can graph any quadratic function using transformation, you must understand what the table of values and graph for the function y = look like. The function y = is the most basic quadratic function; it hasn t undergone any transformations. y = y Notice the following for the graph. The verte is at (0, 0). The curve is concave upwards. Domain { ε R} yε R y 0 Range { } Using transformations, mapping rules, and your understanding of the basic quadratic function y =, you can graph any quadratic function of the form y = k( c) d. This can be accomplished using the following procedure. 1. State the transformations. Separate the transformations that affect the -values (horizontal translations) from those that affect the y-values (reflections in the -ais, vertical stretches, and vertical translations).. Construct the mapping rule.. Using the mapping rule, create the table of values for the desired quadratic function by altering the -values and y-values for y =.. Graph the points and draw the curve. 5. Check to see if the graph looks reasonable. Is it parabolic in shape? Does the location of the verte correspond to the horizontal and vertical translations? If the function has undergone a reflection in the -ais, is the curve concave downwards? If the function has undergone a vertical sketch, does the graph look correspondingly narrower or wider? Eample 1: Graph the function = ( ) 8 y using transformations. Answer: - Horizontal Translation of - Reflection in the -ais - Vertical Stretch of - Vertical Translation of 8 NSSAL 18 Draft 009 C. D. Pilmer

25 The mapping rule is created by eamining the transformations that affect the -values in the table, and eamining the transformations that affect the y-values. The -values are only affected by horizontal translations. In this case, the function has undergone a horizontal translation of such that the -values increase by. That is why the mapping rule shows that the -values change to. The y-values are affected by reflections in the -ais, vertical stretches, and vertical translations. In this case, the function has undergone a reflection in the -ais, a vertical stretch of, and a vertical translation of 8. That is why the mapping rule shows that the y-values change to y 8., y, y 8 Mapping Rule: ( ) ( ) Old Table ( y = ) New Table ( y = ( ) 8) y y - 9 ( ) = = 9 - ( ) = = -1 1 ( ) = = ( ) = = ( ) = = 1 ( ) = = = 9 9 ( ) = 0-10 ( ) ( ) ( ) ( ) ( ) ( ) ( ) The graph looks reasonable for the following reasons. The graph is concave downwards due to the reflection in the -ais. The coordinates of the verte, (, 8), correspond to the horizontal and vertical translations. The graph looks narrower than the graph of y = due to the vertical stretch of. Eample : Graph the function ( ) 1 y = using transformations. Answer: 1 - Horizontal Translation of - - Vertical Stretch of or 0.5 NSSAL 19 Draft 009 C. D. Pilmer

26 Mapping Rule: (, y) (, 0.5y) or (, y), y y y - 9 ( ) = -7.5 = 0.5( 9) - ( ) = -6 = 0.5( ) -1 1 ( 1 ) = = 0.5( 1) 0 0 ( 0 ) = - 0 = 0.5( 0) 1 1 ( 1 ) = = 0.5( 1) ( ) = - = 0.5( ) 9 ( ) = -1.5 = 0.5( 9) The graph looks reasonable for the following reasons. The graph is concave upwards because we do not have a reflection in the -ais. The coordinates of the verte, (-, 0), correspond to the horizontal and vertical translations. The graph looks wider than the graph of y = due to the vertical stretch of 0.5. Eample : Graph the function ( 5) 1 y = using transformations. Answer: - Horizontal Translation of 5 - Reflection in the -ais - Vertical Translation of -1 Mapping Rule: (, y) ( 5, y 1) y y ( ) 5 = 1 1 ( ) 5 = ( ) 5 = - 9 ( ) = -10 = ( 9) 1 - ( ) = -5 = ( ) ( 1 ) = - = ( 1) = ( 0) = ( 1) = ( ) 1 9 ( ) 5 = 8-10 = ( 9) 1 NSSAL 0 Draft 009 C. D. Pilmer

27 The graph looks reasonable for the following reasons. The graph is concave downwards because we have a reflection in the -ais. The coordinates of the verte, (5, -1), correspond to the horizontal and vertical translations. The graph does not look wider or narrower than the graph of y = because we do not have a vertical stretch. Questions: 1. For each quadratic function, state the transformations, and construct the mapping rule. The first question is partially completed. Please note that in the first question, the transformations that affect the -values are on the left opposed to the transformations that affect the y-values which are found on the right. If we do this, people generally find it easier to construct the mapping rule. (a) y = 5( 7) 1 Transformations: - Horizontal Translation of 7 - Reflection in the -ais Mapping Rule: (, y) ( 7, ) - Vertical Stretch of - Vertical Translation of (b) ( 8) 5 y = Transformations: Mapping Rule: (, y) (, ) NSSAL 1 Draft 009 C. D. Pilmer

28 1 y = 5 Transformations: (c) ( ) Mapping Rule: (, y) (, ) 1 (d) y = 1 5 Transformations: Mapping Rule: (, y) (, ). This question is partially completed. Complete the question. Graph ( ) 9 y = using transformations. - Horizontal Translation of - - Vertical Stretch of - Vertical Translation of -9 Mapping Rule (, y) (, y 9) y y - 9 ( ) = - ( ) = -1 1 ( ) = = ( 9) 9 = ( ) 9 1 = ( 1) The graph looks reasonable for the following reasons. The graph is concave because we do not have a reflection in the -ais. The coordinates of the verte, (, ), correspond to the horizontal and vertical translations. The graph looks than the graph of y = because we have a vertical stretch of. NSSAL Draft 009 C. D. Pilmer

29 . This question is partially completed. Complete the question. y = 1 using transformations. Graph ( ) - Horizontal Translation of - Reflection in the -ais - Vertical Stretch of 1 Mapping Rule (, y) (, ) y y ( ) = - ( ) = Vertical Translation of The graph looks reasonable for the following reasons. The graph is concave because we have a reflection in the -ais. The coordinates of the verte, (, ), correspond to the horizontal and vertical translations. The graph looks than the graph of y = because we have a vertical stretch of.. Graph each of the following quadratic functions using transformations. (a) ( ) 5 y = Transformations: Mapping Rule: y NSSAL Draft 009 C. D. Pilmer

30 (b) y = 10 Transformations: Mapping Rule: y Transformations: (c) y = ( ) 1 Mapping Rule: y For each of these quadratic functions, determine the coordinates of the verte whether the function is concave upwards or concave downwards whether the function is narrower, wider or the same as the function y =. (a) y = ( 7) 1 (b) y = ( 5) 10 (c) ( ) y = 1 (d) y = 11 Verte Concavity Narrower/Wider/Same NSSAL Draft 009 C. D. Pilmer

31 Verte Concavity Narrower/Wider/Same 1 g = 9 7 h = (e) y = ( 8) 6 (f) ( ) ( ) (g) ( ) 8 y = 1 using transformations. State the domain, range, coordinates of the verte and the equation of the ais of symmetry. 6. Graph ( 6) The height, h, of a projectile relative to the distance, d, it travels horizontally can be described by the equation h = 1 ( d ) 5. (a) Graph the function using transformations. (b) Use the graph to determine the maimum height reached by the projectile. (c) Determine the distance the projectile will travel horizontally when it reaches its maimum height. (d) Determine the initial height of the projectile. (e) State the equation of the ais of symmetry NSSAL 5 Draft 009 C. D. Pilmer

32 Using Finite Differences to Identify Quadratic Functions Previously we learned how to determine if a table of values could be represented by a linear function (i.e. y = m b). We discovered that if the -values were changing by the same increment and a common difference occurred between successive y-values, then the table of values was generated from a linear function. Eample 1: Determine if the following table of values can be modeled using a linear function. y Answer: y - 5 y In this particular eample, the -values are changing by increments of and the successive y- values display a common difference of. This table of values was generated by a linear function, specifically y = 8. If we can identify linear functions by looking for patterns in the table of values, can we do the same for quadratic functions? The answer is yes. Linear functions display a common difference at the D1 Level. Quadratic functions display a common difference at the D level. Consider the following table of values that was generated by the quadratic function y =. y y D ( 1) ( ) = D ( 1) = 6 6 = = = = = = = NSSAL 6 Draft 009 C. D. Pilmer

33 To the left we ve rewritten the table and started eamining it for patterns. As the -values change by the same increment (1), the y-values do not display a common difference. Since there is no common difference at the D1 level, we know that this table of values is not generated by a linear function. In the last column, however, when we take the differences of the differences, we do see a common difference. Common differences for quadratic functions occur at the D level. Check another quadratic function to see if this pattern holds up. Here is the table of values for the quadratic function y = 5. Analyze the table and determine if a common difference occurs at the D level? y y D D So how do we know that this common difference occurs at the D level for all quadratic functions? Consider the analysis of the function y = a b c; this equation can be used to represent all quadratic functions. A table of values using -values from 1 to 5 has been generated for this function. y = a b c 1 a b c a b c 9a b c 16a b c 5 5a 5b c Now we can analyze this table using finite differences. y = a b c 1 a b c D1 1 a b c a b D 1 9a b c 5a b a 1 16a b c 7a b a 1 5 5a 5b c 9a b a Notice that as the -values changed by the same increment (1), a common difference of a occurred at the D level. Using the general equation y = a b c, we have shown that all tables of values for quadratic functions will have a common difference at the D level when the -values are changing by the same increment. NSSAL 7 Draft 009 C. D. Pilmer

34 Eample : Determine if the following table of values can be modeled using a linear function, a quadratic function or neither. (a) (b) (c) y y y Answer: (a) y - - D ( 8) ( ) = D 1 ( 8) = = - -1 ( 1) = -1 ( 1) 10 = ( 50) ( 1) = -8 ( 8) ( 1) = ( 6) ( 8) = ( ) ( ) = Since the -values change by the same increment () and there is a common difference of - at the D level, then this table of values can be modeled using a quadratic function. (b) y D D Although there is a common difference at the D level, the -values are not changing by the same increment. This table does not appear to be generated by a linear or quadratic function. (c) y D D 1 9 NA NA NA NA NSSAL 8 Draft 009 C. D. Pilmer

35 Since the -values change by the same increment () and there is a common difference of 9 at the D1 level, then we know that this table of values can be modeled using a linear function. Questions: 1. Generate the table of values for y = 7 using -values from -1 to, and determine the common difference at the D1 level. y -1 D Generate the table of values for y = 1 using -values from - to, and determine the common difference at the D level. y - D1-1 D 0 1. Determine if the following table of values can be modeled using a linear function, a quadratic function or neither. (a) (b) y y 1 D1 1-1 D D 9 D NSSAL 9 Draft 009 C. D. Pilmer

36 (c) (d) y y - 5 D1 1 9 D1-8 D 7 D (e) (f) y y - 16 D D D - -0 D (g) (h) y y - -1 D D D.5 D NSSAL 0 Draft 009 C. D. Pilmer

37 Using Finite Differences to Determine the Equation We can identify quadratic functions using finite differences, but we can also use this technique to determine the equation of the quadratic function. The equation that we will generate will be in the form y = a b c. This is called the standard form of the equation. The procedure involves comparing the table of values for the function y = a b c to the table of values for our unknown quadratic function. Eample 1: Determine the equation of the function that is represented by the following table of values. y Answer: Start by analyzing the table using finite differences to see whether we are dealing with a quadratic function. y 1 9 D D We are dealing with a quadratic function because we have a common difference at the D level. Now we will generate the table of values for y = a b c using the same -values (1,,,, 5, 6). y = a b c 1 a b c D1 1 a b c a b D 1 9a b c 5a b a 1 16a b c 7a b a 1 5 5a 5b c 9a b a 1 6 6a 6b c 11a b a We can now compare the tables. Specific elements of one table are equal to specific elements in the other table. For eample, the a b c in the second table is equal to the 9 in the first table. Knowing these relationships we can generate the following three equations. a = a b = 6 a b c = 9 NSSAL 1 Draft 009 C. D. Pilmer

38 Now we can use these three equations to solve for a, b, and c. a = a = a =1 a b = 6 ( 1) b = 6 b = 6 b = a b c = 9 ( 1) ( ) c = 9 c = 9 c = 5 The equation of the quadratic function is y = 1 5 or y = 5. Eample : Determine the equation of the function that is represented by the following table of values. y Answer: Start by analyzing the table using finite differences to see whether we are dealing with a quadratic function. y -1 1 D D We are dealing with a quadratic function because we have a common difference at the D level. Now we will generate the table of values for y = a b c using the same -values (-1, 1,, 5, 7). y = a b c -1 a b c D1 1 a b c b D 9a b c 8a b 8a 5 5a 5b c 16a b 8a 7 9a 7b c a b 8a We can now compare the tables. Specific elements of one table are equal to specific elements in the other table. Knowing this we can generate the following three equations. 8a = 16 b = 8 a b c =1 NSSAL Draft 009 C. D. Pilmer

39 Now we can use these three equations to solve for a, b, and c. 8a = 16 b = 8 a b c =1 8a 8 = 16 b 8 = 8 ( ) ( ) c =1 6 c =1 a = b = c = 7 The equation of the quadratic function is y = 7. Eample : Determine the equation of the function that is represented by the following table of values. y Answer: y - - D1 0 0 D y = a b c - 9 a b c D1 0 c 9 a b D 9a b c 9 a b 18 a 6 6 a 6b c 7 a b 18 a 9 81 a 9b c 5 a b 18 a 18a = 18 18a 18 = a = 1 9a b = 9( 1) b = 9 b = b 15 = b = 5 9a b c = 9 ( 1) ( 5) c = 9 15 c = c = 9 15 c = 0 The equation of the quadratic function is y = or y = 5. NSSAL Draft 009 C. D. Pilmer

40 Eample : The main support cables on a particular suspension bridge form a parabolic curve when viewed from the side. The cables are suspended from two support towers 0 metres apart. The following table shows the height in metres of the cable above the roadway relative to the distance in metres from the vertical support tower on the left hand side of the bridge. (a) Determine the equation that describes the height, h, of the main support cables with respect to distance, d, from the left support tower. (b) Use the equation to determine the height of the main support cables when we are 7 metres from the left support tower. (c) Use the equation and graphing technology to determine the minimum height of the cable. (d) State the domain and range. Answers: (a) d h d 0 8 D D d h = ad bd c d 0 c D1 a b c a b D 16 a b c 1 a b 8 a 6 6 a 6b c 0 a b 8 a 8 6 a 8b c 8 a b 8 a Horizontal Distance (in metres) Height (in metres) a = 0.1 8a 0.1 = 8 8 a = a b = 1.1 c = 8 ( 0.015) b = b = 1.1 Therefore: b 1.0 = h = 0.015d 0.6d 8 b = 0.6 (b) h = 0.015d h = d 8 ( 7) 0.6( 7) h = h = The main support cable is 5.15 m above the roadway. NSSAL Draft 009 C. D. Pilmer

41 (c) Use the TRACE feature on a graphing calculator. To set the WINDOW setting, look at the information supplied in the question and the table of values. The minimum height of the main support cables is metres. (d) Domain: { dε R 0 d 0} Range: { hε R h 8} Note: Suspended cables form curves called hyperbolas; however, when subjected to a uniform load, as is the case with a suspension bridge, the cables deform and approach the shape of a parabola. Using a quadratic function to model the shape of these cables is acceptable, although not perfect. Questions: 1. Determine the equation of the function that is represented by the following table of values. The question has been partially completed. y Your Answer: y 1 - D D y = a b c 1 a b c D1 1 a b c a b D 1 9a b c 5 NSSAL 5 Draft 009 C. D. Pilmer

42 . Determine the equation of the function that is represented by the following table of values. y Determine the equation of the function that is represented by the following table of values. y NSSAL 6 Draft 009 C. D. Pilmer

43 . Sapphire and Manish are organizing a provincial softball tournament. It is a round-robin tournament where each team must play every other team eactly once. It s pretty easy to figure out how many games must be scheduled if only a few teams enter the tournament but what happens if many teams decide to participate. Sapphire and Manish want to see if they can use their knowledge of mathematics to address this issue. Number of Teams Number of Games Played (a) Determine the equation that describes the number of games, g, played in terms of the number of teams, t, signed up for the tournament. (b) Using the equation, determine number of games that must be played if 9 teams participate in the tournament. (c) Using the equation and graphing technology, determine how many teams signed up for the tournament if 66 games are played. (Hint: Remember to adjust the calculator s WINDOW settings. Please note that the -values on the calculator represent the number of teams, and the y-values represent the number of games played. Think about this when deciding upon Xmin, Xma, Ymin, and Yma values for the WINDOW settings. You will likely have to eperiment a little with these values until you find ones that give a good view of the function.) NSSAL 7 Draft 009 C. D. Pilmer

44 5. A projectile is fired vertically into the air. Eventually the projectile will reach its maimum height and fall back to the ground. Its height in metres is recorded at specific times. The time is measured in seconds. The following data was collected. t h (a) Determine the equation of the function that describes the height of the projectile with respect to time. (b) Using your equation from (a) and a graphing calculator, determine each of the following. maimum height reached by the projectile the time when the projectile reached its maimum height the initial height of the projectile the time when the projectile strikes the ground (c) State the domain and range. NSSAL 8 Draft 009 C. D. Pilmer

45 6. In 1971, Apollo 1 astronaut, Alan Shepard, hit a golf ball while on the surface of the moon. He made this shot using only one hand and while encumbered in a spacesuit. Although the eact trajectory of the ball was not recorded, we have taken liberal license and created our own data. Horizontal Distance Traveled (in metres) Height (in metres) (a) Determine the equation of the function that describes the height, h, of the golf ball with respect to horizontal distance, d, traveled. (b) Using the equation, determine the height of the ball after it traveled horizontally 10 m. (c) Using your equation and a graphing calculator, determine each of the following. maimum height reached by the golf ball the position where the ball strikes the moon s surface (d) State the domain and range. NSSAL 9 Draft 009 C. D. Pilmer

46 Find the Equation Given Three Points Up to this point we have learned how do determine the equation of a quadratic function using finite differences. In this section, we will learn how to find the equation given three points on the quadratic function. With this type of question, we must be told eplicitly that we are dealing with a quadratic function. Otherwise we would not be able to determine if these three points belonged to a quadratic function or another nonlinear function. The process of finding the equation given three points is a two step process. We first use the three points to generate three equations. We then solve the by system of equations using a technique we learned in the Systems of Equation Unit that we completed a few weeks ago. This is best illustrated using eamples. Eample 1: Determine the equation of the quadratic function that passes through the points (1, 7), (, 15), and (7, 79). Answer: As with the technique of finite differences, we will find the equation in its standard form ( y = a b c ). We need to find the values of a, b, and c if we want to know the equation of this quadratic function. If we look at the first point, (1, 7), we have been supplied with an -coordinate (1) and a y- coordinate (7). We can substitute these values of and y into the equation y = a b c. This gives us one equation with three unknowns, a, b, and c. y = a b c 1, 7 ( 1) b( 1) 7 = a 7 = a b c c ( ) Now do the same thing with the other two points that were supplied. y = a b c,15 y = a b c ( ) ( 7, 79) ( ) b( ) c 79 = a( 7) b( 7) c 15 = a 15 = 9a b c 79 = 9a 7b c We now have three equations and three unknowns. a b c = 7 9 a b c = 15 9 a 7b c = 79 We need to solve for a, b, and c. This can be done using a TI-8 or TI-8 calculator or by doing by elimination. We learned how to use the calculator to accomplish this back in the Systems of Equation Unit. We create two matrices on the calculator. Matri A is the matri of coefficients. Matri B is the matri of constants [ A ] = 9 1 [ B ] 7 7 = NSSAL 0 Draft 009 C. D. Pilmer

47 After the two matrices have been entered in the calculator, take the inverse of matri A (i.e. [ A ] 1 ) and multiply it by matri B. A complete description of this procedure can be found in the Systems of Equation Unit in the section titled Checking Answers Using a TI-8 or TI-8. Since a =, b = -, and c = 9, we can now state that the equation is y = 9. Online System of Equations Calculator (Optional) If you do not have a TI-8/8, you can use the following site to solve for the three unknowns. (or Google Search: Solve Systems of Equations Calculator) Eample : Determine the equation of the quadratic function that passes through the points (-, -6.6), (, -5.), and (1, -8.6). Answer: Create the three equations using the three points. y = a b c -, y = a ( ) b c (, - 5.) a( ) b( ) c 5. = a( ) b( ) c 6.6 = 6.6 = a b c y = a b c ( 1, - 8.6) ( 1) b( 1) c 8.6 = a 8.6 = 1a 1b c 5. = 16a b c Now use the TI-8 or TI-8 calculator to solve for a, b, and c. Equation: y = 0. You may have realized that the method of finite differences can be used on these types of questions, and vise versa. NSSAL 1 Draft 009 C. D. Pilmer

48 Questions: With these questions, make sure you show clearly how you generated your three equations. Obviously the work on the calculator does not have to be shown. 1. In each case, three points on the graph of a quadratic function have been supplied. Determine the equation of the quadratic function. You may use graphing technology to complete the by elimination. (a) (1, -5), (6, -0), (8, -5) (b) (-, ), (1, 5), (, 1) (c) (-1, -9), (, -9), (6, -7) (d) (-, ), ( -, ), (10, ) NSSAL Draft 009 C. D. Pilmer

49 . The amount of lift generated from a wing is proportional to the speed of the air flowing across the wing. Consider the following data that was collected for one of the Wright Brothers early planes. Speed of Air (in miles per hour) Lift (in Pounds) (a) Determine the equation of the quadratic function that describes the lift, l, of the plane measured in pounds in terms of the speed, s, of the air flowing across the wing in miles per hour. (b) How much lift is generated by this plane if the air flows across the wing at 0 miles per hour? (c) Assuming that the top speed the air can flow across this wing is 0 miles per hour, determine the range and domain of the quadratic function that models this situation.. Angela is interested in purchasing a new flat screen LCD HD television. She is told the price of the television varies quadratically with the screen size. She has found the following information for one brand of televisions. Screen Size (Diagonal Measure in Inches) Price (in Dollars) (a) Determine the equation that describes the price, p, in terms of screen size, s. (b) Use the equation to determine the cost of a LCD television with a screen size of 5 inches. NSSAL Draft 009 C. D. Pilmer

50 Find the Equation Given the Verte and a Point Up to this point we have learned how to determine the equation of a quadratic function: using finite differences where a table of values is supplied using by elimination when three points on the function are supplied In both cases, the equation that we determined was in the standard form, y = a b In this section, we will determine the equation given the verte and one point on the quadratic function. In this case, the equation will be written in the transformational form, y = k( c) d. The reason for this form stems from the fact that the coordinates of the verte supply one with the horizontal and vertical translations. Eample 1: Determine the equation of the quadratic function that passes through the point (7, -1) and whose verte is located at (, 5). Answer: y = k( c) d In the equation y = k( c) d y = k( ) 5 y = k( ) 5 ( 7, 1) 1 = k( 7 ) = k( ) 18 = 9k = k Therefore : y = ( ) 5 NSSAL Draft 009 C. D. Pilmer c. we need to solve for k, c, and d to complete the equation. Since the verte is at (, 5), we know that the horizontal translation is and the vertical translation is 5. In our equation c is the horizontal translation and d is the vertical translation. Substitute the values and 5 into the equation. Now we only need to solve for k. To accomplish this substitute the and y values from point, (7, -1), that was supplied. Now the only unknown in the equation is k. Solve for k. Once this is done, the complete equation can be written. Eample : Determine the equation of the quadratic function that passes through the point (5, 19) and whose verte is located at (-1, 7). Answer: y = k( c) d Verte: (-1, 7) HT (or c) equals -1. VT (or d) equals 7. y = k( 1) 7 y = k( 1) 7 ( 5,19) Substitute the point (5, 19) into the equation and solve for k. 19 = k = 6k 1 = k ( ) = k( 6) y = 1 Therefore: ( 1) 7

51 Eample : A projectile is fired from a height of meters. After it travels horizontally 5 metres, it reaches its maimum height of 8 metres. (a) Determine the equation that describes the height, H, of the projectile in terms of the horizontal distance, D, traveled. (b) Using the equation, determine the height of the projectile after it has traveled 7 metres horizontally. (c) Using graphing technology and the equation, approimate the distance the projectile travels horizontally when it hits the ground. (d) State the domain and range. Answer: H = k (a) ( D c ) d H = k( D 5) 8 H = k( D 5) 8 ( 0,) = k( 0 5) 8 8 = k( 5) (b) 6 = 5k H H H H H 6 5 = k 6 = 5 6 = ( 7 5) 5 6 = ( ) 8 5 = = 7 5 Verte: (5, 8) HT (or c) equals 5. VT (or d) equals 8. ( D 5) 8 8 Substitute the point (0, ) into the equation and solve for k. 6 5 Therefore: H = ( D 5) 8 The projectile is at a height of 1 7 m. 5 (c) Use the TRACE feature on the graphing technology. Think about the information supplied in the question when playing with the WINDOW settings. The projectile travels horizontally approimately 10.8 m before it hits the ground. (d) Domain: { Dε R 0 D 10.8} Range: { Hε R 0 H 8} NSSAL 5 Draft 009 C. D. Pilmer

52 Questions: 1. Determine the equation of each quadratic function given the following point and verte. (a) Point: (7, 1) Verte: (5. ) (b) Point: (1, -7) Verte: (-, -1) (c) Point: (7, -1) Verte: (, -) (d) Point: (, 11) Verte: (-1, 5) NSSAL 6 Draft 009 C. D. Pilmer

53 (e) Point: (8, -0) Verte: (, 0). A ball is thrown from an initial height of 1 metre. Its trajectory is parabolic. Seven metres from where it was thrown, it reaches its maimum height of 5 metres. (a) Determine the equation that describes the height, H, of the projectile in terms of the horizontal distance, D, traveled. (b) Using the equation, determine the height of the projectile after it has traveled metres horizontally. (c) Using graphing technology and the equation approimate the distance the projectile travels horizontally before it hits the ground. (d) State the domain and range. NSSAL 7 Draft 009 C. D. Pilmer

54 . The Gateway Arch in St. Louis is the tallest national monument in the United States. The arch is parabolic in shape. The arch is 60 feet wide at the base. It reaches a maimum height of 60 feet in the middle. (a) If we set up a Cartesian coordinate system with one end of the arch at the origin, determine the equation of equation that describes the height, H, of the arch in terms of the horizontal distance, D, from one end of the arch. (b) Using the equation, determine the height of the arch when the horizontal distance is 100 feet. (c) Using the equation and graphing technology approimate the horizontal distances that correspond to a height of 00 feet. (d) Determine the domain and range of the function. NSSAL 8 Draft 009 C. D. Pilmer

55 . The Parkes radio telescope, located in Australia, is comprised of a large parabolic dish with a diameter of 10 feet and a depth of ft. If the dish is pointed straight up into the sky, its cross-section can be modeled by a quadratic function. (a) If we orient the parabola dish such that its verte is at the origin, determine the equation that describes its cross-section. (b) If you moved 10 feet to the right or left of the verte, how much would height change by? (c) Determine the domain and range of the function. Did You Notice? Although the verte was supplied in questions and, two other points could be determined based on the written descriptions. For eample, in question you could determine that we are dealing with the verte (15, 60) and the points (0, 0) and (60, 0). This means that one could determine the equation using the verte and one point method, or the three point method. Either is perfectly acceptable. With some of the questions you encounter, you will find that there is more than one acceptable method. NSSAL 9 Draft 009 C. D. Pilmer

56 Putting It Together, Part I Up to this point we have learned the following. Interpretation of real world applications that can be modeled using quadratic functions. Graphing of quadratic functions of the form y = k( c) d using transformations. Determining the equation of a quadratic function using finite differences when a table of values is supplied. Determining the equation of a quadratic function using by elimination when three points are supplied Determining the equation of a quadratic function in its transformational form when a point and the verte are supplied. Although it may be difficult to believe, we still have more material to learn regarding quadratic functions. Instead of pursuing new material, it is probably a good time to do a few review questions to solidify some of the concepts you have already learned. Questions: 1. For each of the functions below, answer the following. State the transformations. Determine the coordinates of the verte. Describe the concavity (upwards or downwards) Describe the shape of the curve compared to that for the function narrower, wider, or no change. y =. Use the terms Function Transformations Verte Concavity Shape (a) ( 5) 1 y = (b) y = ( 6) (c) y = ( 8) 1 (d) y = 9. Provide an equation based on the description. There is more than one acceptable answer. A quadratic function has its verte at (, -7), is concave downwards, and is narrower than the curve for the function y =. NSSAL 50 Draft 009 C. D. Pilmer

57 . (a) Graph = ( 1) 18 y using transformations. Show all your work (i.e. state transformations, show mapping rule, ). (b) State the domain, range, coordinates of the verte and the equation of the ais of symmetry. (c) Classify as odd, even or neither. (d) If the equation represented the flight path of a projectile, the y-intercept was the launch point for the projectile, and the -ais represented ground level, what would be the domain and range of the function modeling this situation? In your own words, eplain how you can distinguish between the table of values for a linear function and the table of values for a quadratic function assuming that the -values, in both cases, are changing by the same increment. NSSAL 51 Draft 009 C. D. Pilmer

58 5. Massato s company presently sells 800 Blu-Ray Disks a month, each for $0. The revenue from Blu-Ray sales is $ a month. After conducting some market research, he realizes that every time he drops the price by $1, the sales increase by 100 disks per month and in turn the revenue increases. The following table shows the relationship between the number of times the price is dropped by $1 and the monthly revenue. Number of Times the Price is Dropped (n) Monthly Blu-Ray Revenue (r) (a) Determine the equation that describes the monthly Blu-Ray revenue in terms of the number of times the price is dropped by $1. (b) What will the monthly revenue be if the price is dropped 8 times? (c) Using the equation and graphing technology, determine how many times the price should be dropped to maimize the revenue. What is that maimum revenue be? NSSAL 5 Draft 009 C. D. Pilmer

59 6. A projectile is fired, travels along a parabolic path, reaches a maimum height of 10 m after it has traveled horizontally 0 m, and finally strikes a target 70 m away at a height of m. (a) Determine the equation that describes the height, h, of the projectile in terms of the horizontal distance, d, the projectile travels. (b) Was the projectile fired from ground level or from the edge of a cliff? Eplain. (c) What is the height of the projectile after it has traveled horizontally 50 m? (d) State the domain and range. 7. Mrs. Leck gives her academic learners a math puzzle. She tells them that there specific numbers called triangular numbers. The second triangular number is. The fourth triangular number is 10. The seventh triangular number is 8. (a) Determine the equation of the quadratic function that describes the triangular number, n, in terms its ordinal, o (i.e. the number indicating the position in the sequence. e.g. 1 for first, for second, ) (b) Determine the fifteenth triangular number. NSSAL 5 Draft 009 C. D. Pilmer

60 Multiplying Polynomials There seems to be two glaring omissions so far in this unit. 1. We know how to graph quadratic functions when they are in the transformational form ( y = k( c) d ) but often quadratic functions are written in the standard form ( y = a b c ). How do we change a quadratic function from its standard form to its transformational form (and vise versa)? This seems like an important type of question that we should learn to solve.. When we worked with linear functions ( y = m b), we learned how to solve for algebraically, given a value for y. Consider the eample below. Eample: Given that y = 7, find when y equals 1. Answer: y = 7 1 = = = = What happens, however, when we are dealing with a similar question involving a quadratic function? For eample, if you were given the function y = 5 and told to find when y equals 1, then how would you accomplish this without using technology. This seems like another important type of question that we should learn to solve. Both of these critical types of questions require one to understand two concepts: multiplying polynomials and factoring. We will take several days to practice and master these two skills. What are Polynomials? The first thing we must do is define polynomial. A polynomial is an algebraic epression of one or more terms. The three types of polynomials (monomials, binomials, and trinomials) that we will encounter in this course are listed in the following chart. Name Number of Terms With One Variable With Two Variables With Three Variables Monomial one y 5 1a bc Binomial two 1 r t 5a b 7 c Trinomial three 10 6 a 7ab b r st 5s t NSSAL 5 Draft 009 C. D. Pilmer

61 Notice that the eponents on each of the variables are whole numbers (0, 1,,, ). The eponents must be whole number in order for the epressions to be classified as polynomial. 1 Not Polynomial: 5 1 y 5 y Multiplying Monomials You are probably very familiar with the topic of multiplying monomials. It is just a matter of m n m n applying the power rule of eponents which states =. (a) = 5 7 (b) a = 7 a a (c) 6 r r = r 7 (d) b = b 5 b 8 (e) 5y = y 8 9 6y 0 (f) 5 9 7g g = 7g (g) a b 6 a c = a b c (h) y = y 5 y 1 (i) 7 9g h g h = 9g h Multiplying Monomials and Binomials To understand how to multiply a monomial and a binomial, we first have to learn about the distributive property. Consider the following questions. There are two ways to solve these problems. 7 ( ) ( ) Answers: Method 1: Order of Operations ( ) ( 7 ) = ( 6) = ( ) = 18 = 8 Method : Distributive Property ( ) ( 7 ) = ( ) ( ) = 7 = 1 6 = 1 6 = 18 = 8 ( ) ( ) According to the order of operations, we complete the operations within the parentheses before attempting the multiplication. With the distributive property, we distribute the multiplication to both of the terms within the parentheses first and then do the addition or subtraction. Notice that the same final answer was obtained regardless of the method. We can now apply this distributive property when multiplying monomials and binomials. 7 ( ) = 7( ) 7( ) = ( 7) = 5( ) 5( 7) = 0 5 NSSAL 55 Draft 009 C. D. Pilmer

62 (a) ( ) = 6 (b) y( y 5) = y 10y (c) ( 7 ) = 7 (d) ( ) 5a 8 a = 0a 10a (e) ( y) = 1 8y (f) cd( c d ) = c d 6cd (g) r s( rt s) = r st r s (h) pq( 1 6 p) = pg 18 p q Multiplying Two Binomials Understanding the distributive property is critical when learning how to multiply binomials. Consider the problem ( )( 5 1). There are two ways to solve this problem. Answer: Method 1: Order of Operations ( )( 5 1) According to the order of operations, we complete the = ( 5)( 6) operations within the parentheses before attempting the multiplication. = 0 Method : Distributive Property ( )( 5 1) With the distributive property, we distribute the = ( 5) ( 1) ( 5) ( 1) multiplication from both of the terms in the first parentheses to both of the terms in the second = parentheses and then do the addition or subtraction. = 0 This means the is multiplied by both the 5 and 1, and the is multiplied by both the 5 and 1. Notice that the same final answer was obtained regardless of the method. We can now apply this distributive property when multiplying two binomials. ( 1)( ) = ( ) ( ) 1( ) 1( ) = 6 = One way to remember how to use the distributive property when multiplying binomials is to use the acronym FOIL. If we consider the eample above, we start by multiplying the first terms, and, in both binomials. We then multiply the outside terms and. The inside terms, 1 and are then multiplied. And finally the last terms 1 and are multiplied. FOIL: First Terms Outside Terms Inside Terms Last Terms NSSAL 56 Draft 009 C. D. Pilmer

63 ( 1)( 5 ) = ( 5) ( ) 1( 5) 1( ) = 10 = (a) ( )( 5) = 5 15 = 8 15 (b) ( a )( a 6) = a 6a a = a a (c) ( p 1)( p 5) = 6 p 15 p p 5 = 6 p 17 p 5 (d) ( 1)( 1) = 1 = 1 (e) ( )( ) 5 d d = 15 0d 6d 8d = 15 6d 8d (f) ( )( ) a 7b a b = a 1ab 7ab 8b = a 5ab 8b (g) ( )( ) y 5 6y = 0 y 10 y 1y = 0 1 y 1y It should be mentioned that FOIL is not a mathematical term. It is an acronym that helps learners remember how to multiply two binomials. The distributive property is the mathematical term that truly describes what we are doing. We are distributing the multiplication across all the terms. Visualizing FOIL We just learned how to use the distributive property to multiply two binomials. This is done using the acronym FOIL (first, inside, outside, last). Eample: ( )( ) = 6 = 5 6 Some learners struggle to understand why this technique works. In an attempt to address this issue a visual model is going to be used to try to eplain the rationale of FOIL. NSSAL 57 Draft 009 C. D. Pilmer

64 Step 1: Suppose we have a rectangle that measures m by m. Using the formula A = l w, we can determine the area of the rectangle. Its area is 6 m. Step : We decide to add to both the length and width of this rectangle. We add the same amount,, to both of these dimensions. This means that the new length of the rectangle will be, and the new width of the rectangle will be. Now we will again work out the area of this rectangle using the formula A = l w. If so then the area will be equal to the following. A = l w A = ( )( ) Step : To find the total area, we will divide the rectangle into four smaller rectangles. The rectangle in the upper left measures by, and therefore has an area of. The upper right rectangle measures by and has an area of 6. The lower left rectangle measures by and has an area of. The lower right rectangle measures by and has an area of. To find the total area, we add the four areas. A = 6 Area = Area = 6 Area = Area = Step : Using the last lines of steps and, we can conclude: = Yes! We see the distributive property at work. ( )( ) 6 You do not need to know this page. It was only supplied for learners who were not sure why the FOIL technique worked. Using the Distributive Property with Other Polynomials We use the distributive property to multiply a variety of polynomials. 1. Multiplying Monomials and Trinomials (a) 6( 7 ) = 18 1 (b) ( 5 6) = 5 6 (c) p( p 6 p 5) = 1 p 18 p 15 p NSSAL 58 Draft 009 C. D. Pilmer

65 NSSAL 59 Draft 009 C. D. Pilmer (d) ( ) 6 8 cd d c d c d cd c cd = (e) ( ) 5 y y y y y y =. Squaring Binomials (a) ( ) ( )( ) = = = (b) ( ) ( )( ) = = = c c c c c c c c (c) ( ) ( )( ) = = = (d) ( ) ( )( ) p np n p np np n p n p n p n = = =. Multiplying Binomials and Trinomials (a) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) = = = (b) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) = = = g g g g g g g g g g g g g g g g g g (c) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) = = = (d) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) b ab b a a b ab b a ab b a a b b ab b a b b a ab a a a b ab a b a = = =

66 These rules for multiplying polynomials can be used to change a quadratic function from its transformational form to its standard form. Eample: Change the following quadratic functions from their transformational forms to their standard forms. (a) y = ( ) 1 y = 1 (b) ( 6) 5 Answers: (a) ( ) y = 1 y = ( )( ) y = y = y = y = ( 16) ( 8 16) (b) 1 y = 1 y = 1 y = 1 y = 1 y = 1 y = ( 6) 5 ( 6)( 6) ( 6 6 6) ( 1 6) Questions: 1. Multiply the following monomials. (a) 5 = (b) m 6 m = (c) y 7 y = (d) p 7 p = 6 (e) 5g g = (f) 5t t = 6 (g) a b a b = (h) pq p q = (i) s t s t = (j) 6e f d e =. Multiply the following monomials and binomials. (a) 5( 6) = (b) ( y 7) = (c) ( ) = (d) ( 8) = (e) 5 p ( p ) = (f) 7 c( 6c) = NSSAL 60 Draft 009 C. D. Pilmer

67 (g) ( y) = (h) 6 cd( 1 c) = (i) a b( a b) = (j) y ( 7 y ) =. Multiply the following binomials. (a) ( )( 6) (b) ( )( 5) (c) ( p 6 )( p 1) (d) ( 7 )( ) (e) ( y 7)( y ) (f) ( 5)( 5) (g) ( 6 )( ) (h) ( 9 h )( 1 h) (i) ( 1)( ) (j) ( )( ) (k) ( )( 5) (l) ( 5a 1)( a ) (m) ( d 5)( d 1) (n) ( 1)( 1) (o) ( 7 1)( 1) (p) ( )( 5) (q) ( 7y)( y) (r) ( a b)( a b) (s) ( y)( y) (t) ( d e)( d e) NSSAL 61 Draft 009 C. D. Pilmer

68 (u) ( 9g h)( g h) (v) ( r s)( r s). Complete the indicated operation (a) ( ) (b) d ( d 5d ) (c) ( 6 1) (d) 7g ( g g 5) (e) p( 7 6 p p ) (f) y ( 6y y 9) (g) y( 6y y ) (h) 5ab ( a ab b ) (i) pq ( 5p 7 pq q ) (j) g ( g g 5) (k) 6 y ( 5y y ) (l) 5 5 m n( m mn n ) (m) ( 7) (n) ( n 9) (o) ( 5 6) (p) ( p 5) (q) ( 5y) (r) ( p q) (s) ( )( 5 1) (t) ( d )( d 5d 1) NSSAL 6 Draft 009 C. D. Pilmer

69 (u) ( y 1)( y 5y ) (v) ( h )( h h 5) 5. Change each of the quadratic functions from their transformational form to their standard form. (Other resources will refer to this as changing from the standard form to the general form.) (a) y = ( 5) (b) y = ( 1) 1 y = 1 (c) y = ( ) 1 (d) ( ) 1 6. Complete the indicated operation. 5 7 (a) ( )( ) (b) ( p ) (c) ( y y 1) y (d) ( 1 5d )( d ) (e) ( )( 5 ) (f) y ( y 1) NSSAL 6 Draft 009 C. D. Pilmer

70 (g) ( d 9) (h) ( k 7)( k 7) (i) ( e )( e 5e 1) (j) ab ( a ab b ) (k) ( ) c (l) r s( 6r 5s) 5d (m) ( y)( 5 y) (n) 7g h ( g g h 5h ) 7. Graph the three functions = ( 1 )( ) y =, and ( ) y, y = 1 simultaneously on a graphing calculator. What do you notice? Why is this so? Challenge Yourself (a) ( 1) (b) ( 5) ( 1)( ) NSSAL 6 Draft 009 C. D. Pilmer

71 If you feel that you need additional practice with these types of questions, you can complete the optional questions titled Additional Practice: Multiplying Polynomials found in the appendi of this unit. For Your Information (Optional) There is an interesting and alternate approach for multiplying polynomials. It involves using rectangular arrays. The rows of the array are defined by the terms in one of the polynomials and the columns are defined by the terms in the other polynomial. Eample 1: ( )( 5) Eample : ( 1)( 1) 5 1 Multiply the terms ( )( 5) = 6 = Multiply the terms. ( 1)( 1) = = NSSAL 65 Draft 009 C. D. Pilmer

72 Factoring Part 1 Factoring Whole Numbers Back in grade school, we learned how to multiply whole numbers. In each case we were given two numbers and asked to find the product. Eample: When we multiply and 7, we get 1. A few weeks later, we learned about factors. We were trying to find pairs of numbers that when multiplied generated that product. Eample: Two of the factors of 1 are and 7 (There are two other factors; 1 and 1.) To understand factors, we had to understand multiplication. Factoring Polynomial Epressions In this section, we are going to learn how to factor polynomials, specifically binomials and trinomials. In order to understand the factoring of polynomials, we first had to understand the multiplication of polynomials. That is why the last section of material was dedicated to this topic. In the last section, we were given ( )( ) In this section, we will be given 5 6 Multiplying Polynomials and generated the answer 5 6. and generate the answer ( )( ). ( )( ) = 5 6 To factor a polynomial epression is to epress it as a product. Here are some additional eamples. Factor 1 Factor 8 Factor 10 5 Factor 81 Factor 6 5 Factoring Polynomials. Answer: ( ). Answer: ( 7 )( ). Answer: ( 5)( 5) or ( 5) Answer: ( 9)( 9) Answer: ( 1)( ) We are not epected to know how to generate these types of answers just yet. There are actually five different types of factoring techniques used in the eamples above. We will study each technique separately over the net three sections of this unit. This will give us time to master one factoring technique before proceeding onto a new one. NSSAL 66 Draft 009 C. D. Pilmer

73 Common Factoring Common factoring is the easiest type of factoring and, for some odd reason, the most commonly overlooked technique. We are trying to find the greatest common factor of all of the terms in the polynomial. For the first eight eamples we will concentrate on questions where the greatest common factor is a monomial. This monomial will be one of the factors. The other factor is obtained by dividing the original polynomial by the monomial we factored out. Eamples: (a) Factor: 6 9 Answer: 6 9 = ( ) The great common factor (GCF) of 6 and 9 is. That means that one of the factors is. To find the other factor, divide 6 9 by. This gives us the binomial. (b) Factor: 10 (c) Answer: 10 = 5 1 Factor: Answer: 7 p 5p 7 p 5p = p (d) Factor: p 16 Answer: p 16 = 8 p ( ) ( 7 p 5) ( ) (e) Factor: 1t 6t t Answer: 1t 6t t = t 7t t ( ) The great common factor (GCF) of 10 and is. That means that one of the factors is. To find the other factor, divide 10 by. This gives us the binomial 5 1. The GCF is p. To find the other factor, divide 7 p p. The GCF is 8. p 5 by To find the other factor, divide p 16 by 8. The GCF is t. To find the other factor, divide 1t 6t t by t. (f) (g) 7 5 Factor: 10d 5d 5d Answer: 7 10d 5d 5 5d = 5d Factor: 1 y 1 y ( d d 1) Answer: 1 y 1 y = 7 y y ( ) The GCF is 5d. To find the other factor, divide d 5d 5d by 5d. The GCF is 7 y. To find the other factor, divide 1 y 1 y by 7 y. NSSAL 67 Draft 009 C. D. Pilmer

74 (h) Factor: 9ab 7a b The GCF is 9ab. Answer: 9ab 7a b = 9ab 1 a ( ) To find the other factor, divide by 9ab. 9ab 7a b Sometimes the greatest common factor is a binomial, rather than a monomial. (i) Factor: ( ) 5( ) Answer: 5 = 5 ( ) ( ) ( )( ) (j) Factor: 7 ( ) ( ) Answer: 7 = 7 ( ) ( ) ( )( ) (k) Factor: p( p) 7( p) Answer: p p 7 p = p p 7 ( ) ( ) ( )( ) (l) Factor: d ( d 5) ( d 5) Answer: d d 5 d 5 = d 5 d 1 ( ) ( ) ( )( ) The GCF is. Factoring out leaves us with the binomial 5. This is our other factor. The GCF is. Factoring out leaves us with the binomial 7. This is our other factor. The GCF is p. Factoring out p leaves us with the binomial p 7. This is our other factor. The GCF is d 5. Factoring out d 5 leaves us with the binomial d 1. This is our other factor. When factoring polynomial epressions, you always attempt to find a common factor first. Factoring Difference of Squares In the last section, Multiplying Polynomials, we noticed that in most cases when two binomials were multiplied, the resulting product was a trinomial. Eamples: ( )( 5) = 5 15 p 1 p 5 = 6 p 15 p p = 8 15 = 6 p 17 p 5 ( )( ) 5 There were, however, a few isolated cases where the product of two binomials was another binomial. Eamples: ( 5)( 5) = p 1 p 1 = p 1p 1p = 5 = p 1 ( )( ) 1 NSSAL 68 Draft 009 C. D. Pilmer

75 This occurred when the binomials were of the form a b and a b. When two binomials only differ in terms of operation of addition and subtraction, they are called conjugates. Conjugates: 5 and 5 p 1 and p 1 8 and 8 1 7y and 1 7y 5 a 6b and 5a 6b g h and g h The product of two conjugates is always a binomial where two perfect squares are separated by the operation of subtraction. These products are often referred to as the difference of squares. The Products of Two Conjugates: 5 p y 5a 6b g 9h Now when we see a difference of squares, we know that its two factors must be conjugates. The two terms in each conjugate will be found by taking the square root of each perfect square in the original binomial. Factor 9 5. Square Root Square Root 5 Make your conjugates ( 5)( 5) Eamples: (a) Factor: 16 Answer: 16 = ( )( ) (b) Factor: 5 9 Answer: 5 9 = 5 5 ( )( ) (c) Factor: 9 16t (d) Factor: 1 6y Answer: 9 16t = 7 t 7 t ( )( ) Answer: 1 6y = 1 6y 1 6y ( )( ) (e) Factor: a 81b (f) Factor: 11r 9s t Answer: a 81b = a 9b a 9b ( )( ) Answer: 11r 9s t = 11r 7st 11r 7st ( )( ) Factoring by Inspection This type of factoring is most commonly used when we have a trinomial of the form a b c when a is equal to 1 (e.g ). NSSAL 69 Draft 009 C. D. Pilmer

76 Let s consider the trinomial If this trinomial is a product of two binomials, then the first terms in each binomial must be. That is the only way we can produce a trinomial where the first term is = ( )( ) The last term in our trinomial is 15. The last terms in our two binomials must multiply to give this number. That means that our possible combinations are and 5, 1 and 15, - and -5, and finally -1 and = ( )( 5) ( 1)( 15) ( )( 5) ( 1)( 15) We have considered the first and last terms in our original trinomial but we have not looked at the middle term. The middle term will tell us which one of the four combinations is correct. Using the distributive property the middle term generated by: and 5 will be 8 desired middle term 1 and 15 will be 16 and 5 will be -8 1 and 15 will be -16 Therefore: 8 15 = ( )( 5) Which one of these four combinations is right? We can create a rule so that one can quickly factor a trinomial of the form a b c where a = 1. The two binomial factors first terms will be. The second terms in each factor must multiply to give the last term in the original trinomial and add to give the middle term in the original trinomial. Eamples: (a) Factor: 6 8 Answer: 6 8 = (b) Factor: 5 ( )( ) Answer: 5 = 5 1 (c) Factor: 5 6 Answer: 5 6 = ( )( ) ( )( ) Start with: ( )( ) We now need two numbers that multiply to give us 8 and add to give us 6. The only numbers that work are and. Start with: ( )( ) We now need two numbers that multiply to give us -5 and add to give us. The only numbers that work are 5 and -1. Start with: ( )( ) We now need two numbers that multiply to give us 6 and add to give us -5. The only numbers that work are - and -. NSSAL 70 Draft 009 C. D. Pilmer

77 (d) Factor: p p 10 Answer: p p 10 = (e) Factor: h 7h 8 ( p )( p 5) Answer: h 7h 8 = h 8 h 1 (f) Factor: a ab ( )( ) b Answer: a ab b = a b a b (g) Factor: y ( )( ) 10 y Answer: 10y y = y 6y ( )( ) Start with: ( p )( p ) We now need two numbers that multiply to give us -10 and add to give us -. The only numbers that work are and -5. Start with: ( h )( h ) We now need two numbers that multiply to give us -8 and add to give us 7. The only numbers that work are 8 and -1. a b a b Start with: ( )( ) We now need two numbers that multiply to give us and add to give us. The only numbers that work are and 1. y y Start with: ( )( ) We now need two numbers that multiply to give us and add to give us -10. The only numbers that work are - and -6. Up to This point, we have learned how to: common factor polynomials factor difference of squares polynomials factor trinomials of the form a b c where a = 1 using inspection. There are still two other types of factoring that we need to learn. Rather than doing this, it is probably better to practice the three factoring techniques that we have just learned before learning any more. We will leave the two other types for the net section in this unit. Questions 1. Common factor the following polynomials. (a) 10 (b) 15n 5n (c) 8 (d) 9 6 (e) 1y 7 y (f) 8d 1d NSSAL 71 Draft 009 C. D. Pilmer

78 (g) 8p 16 p p (h) 1 y 6 y 18y (i) 5 ( 7) 6( 7) (j) 7h ( h ) ( h ) (k) d( 8 d ) 6( 8 d ) (l) 5g( 1 g) ( 1 g). Factor the following difference of squares polynomials. (a) 6 (b) 9g 5 (c) 1 (d) 5 16 (e) 81 d (f) 1 6t (g) 16 y 9 (h) 9a 6b. Factor the following trinomials by inspection. (a) 8 1 (b) k 10k 16 (c) t 10t 1 (d) 0 (e) p 5 p (f) y 1 7y (g) r 1rs 5s (h) a 5ab 6b. Factor each of the following. (a) 1 (b) 5p NSSAL 7 Draft 009 C. D. Pilmer

79 (c) 7n 18n (d) t 6t 0 (e) a 7a (f) 16g 1 (g) g 11gh 0h (h) 15d 5d (i) 5 (j) 6 ( 8) 5( 8) 81 h (k) 9 q p 6 (l) y 1 y 5 (m) p q 1p q (n) a ab 18b (o) 5v( 10v) ( 10v) (p) 5s 9t (q) 7y 1y (r) c ( c 5) ( c 5) NSSAL 7 Draft 009 C. D. Pilmer

80 Factoring Part Factoring By Decomposition In the last section we learned how to factor trinomials of the form y = a b c where a = 1 using inspection. What happens when the leading numerical coefficient, a, is not equal to 1. Consider the trinomial where a equals 6. If this trinomial is the product of two binomials and the leading term in the trinomial is 6, then the first terms in the two binomials could be and, or 6 and. ( )( ) or (6 )( ) The last term in our trinomial is -1. The last terms in our two binomials must multiply to give this number. That means that our possible combinations are 1 and -1, -1 and 1, 7 and -, and finally -7 and. ( 1)( 1) ( )( ) 1 1 ( 1)( 1) ( 1)( 1) ( 7)( ) ( )( 7) ( 7)( ) ( )( 7) This is enough to drive = a person crazy! ( )( ) ( 6 1)( 1) ( 6 1)( 1) ( 6 1)( 1) ( 6 7)( ) ( 6 )( 7) ( 6 7)( ) ( 6 )( 7) If we consider the middle term, -17, in our original trinomial, then we can conclude the following = 7 desired factors ( )( ) Which one of these siteen combinations is right? The technique that we used above is very time-consuming and seems very impractical. There is a better way to deal with trinomials of this form. NSSAL 7 Draft 009 C. D. Pilmer

81 Let s tackle the trinomial again. Follow this five step process. Step 1: Multiply the coefficient of the first term by the last term. 6 1 = ( ) 8 Step : Find two numbers that multiply to give you the product you obtained in step 1, and those same two numbers must add to give us the coefficient of the second term. = -8 = -17 Answer: -1 and Step : Split the middle term apart using the two numbers from step = Step : Group the first two terms together and the last two terms together. Common factor from each of those groups separately = 6 1 = ( 6 1) ( 1) = 7 7 ( ) ( ) Step 5: Common factor out the binomial = 6 1 = ( 6 1) ( 1) = ( 7) ( 7) = ( 7) desired factors ( ) Eample 1 Factor Answer: 11 1 () (8) = () (8) = 11 = 8 1 = = ( ) ( 8 1) ( ) ( ) ( )( ) = Step 1: Multiply and 1. Step : Find two numbers whose product is and whose sum is 11. Step : Split the middle term, 11, apart. Step : Group first two terms and last two terms together. Common factor from each group separately. Step 5: Common factor out the binomial. If we are unsure whether we have factored the trinomial correctly, we can check our answer by multiplying the two binomials and seeing if you obtain the original trinomial. ( )( ) = 8 1 = 11 1 Correct! NSSAL 75 Draft 009 C. D. Pilmer

82 Eample Factor Answer: 7 15 (1) (-5) = -60 (1) (-5) = 7 = ( 1) ( 5 15) = = ( ) 5( ) ( )( 5) = Step 1: Multiply and -15. Step : Find two numbers whose product is -60 and whose sum is 7. Step : Split the middle term, 7, apart. Step : Group first two terms and last two terms together. Common factor from each group separately. Step 5: Common factor out the binomial. Eample Factor 8y 6y 15. Answer: 8y 6y 15 (-6) (-0) = 10 (-6) (-0) = -6 = 8y 6y 0y 15 = = ( y 6y) ( 0y 15) 8 y( y ) 5( y ) ( y )( 5) = y Step 1: Multiply 8 and 15. Step : Find two numbers whose product is 10 and whose sum is -6. Step : Split the middle term, -6y, apart. Step : Group first two terms and last two terms together. Common factor from each group separately. Step 5: Common factor out the binomial y -. Eample Factor 10g g. Answer: 10g g (5) (-6) = -0 (5) (-6) = -1 = 10g 5g 6g = = ( g 5g) ( 6g ) 10 5g( g 1) ( g 1) ( g 1)( 5 ) = g Step 1: Multiply 10 and -. Step : Find two numbers whose product is -0 and whose sum is -1. Step : Split the middle term, -g, apart. Step : Group first two terms and last two terms together. Common factor from each group separately. Step 5: Common factor out the binomial g 1. NSSAL 76 Draft 009 C. D. Pilmer

83 Questions: 1. Fill in the blanks. (a) 6 11 ( ) ( ) = ( )( ) = 11 = 6 8 = = ( 8) ( ) 6 ( ) 1( ) = ( )( ) (b) ( ) ( ) = ( )( ) = = ( 10) ( 8) = 5 = 5 ( ) ( ) = ( )( 5 ) (c) 1d 17d 6 ( ) ( ) = 7 ( )( ) = -17 = 1d 9d 8d 6 = = = ( d 9d ) ( 8d 6) 1 d( ) ( ) ( )( ) Step 1: Multiply and. Step : Find two numbers whose product is and whose sum is 11. Step : Split the middle term, 11, apart. Step : Group first two terms and last two terms together. Common factor from each group. Step 5: Common factor out the binomial. Step 1: Multiply 5 and -8. Step : Find two numbers whose product is and whose sum is. Step : Split the middle term,, apart. Step : Group first two terms and last two terms together. Common factor from each group. Step 5: Common factor out the binomial -. Step 1: Multiply and. Step : Find two numbers whose product is and whose sum is. Step : Split the middle term,, apart. Step : Group first two terms and last two terms together. Common factor from each group. Step 5: Common factor out the binomial.. Factor using decomposition. Show all your work. (a) 1 15 (b) NSSAL 77 Draft 009 C. D. Pilmer

84 (c) (d) 6 p 17 p 7 (e) c 11c (f) 8 1 (g) 8a 6a 9 (h) 6y 11y 10 (i) (j) d 16d 15 (k) (l) 8 p 1 p. Factor completely. You will encounter all four types of factoring questions that you have learned up to this point. (a) 81 5 (b) 0 NSSAL 78 Draft 009 C. D. Pilmer

85 (c) 1 18 (d) 15c 5c (e) 1 (f) d( 1 d ) 7( 1 d ) (g) 9 9 (h) 6 11y (i) 10ab b (j) 15 p 11p a (k) 9 d 9 (l) 6 p 10 p q p NSSAL 79 Draft 009 C. D. Pilmer

86 Alternate Forms of Decomposition (Optional) In the last section, we learned a five step process for decomposition. There is an alternate form of decomposition which is slightly easier, but many students struggle a little more to understand why this process works. Eample 1 Factor Alternate Method #1 Since the leading numerical coefficient is, then we will create two binomials whose leading numerical coefficients are. ( )( ) We can see that we already have a problem. The product of and is, rather than the found in the original trinomial that we are attempting to factor. In order to deal with this problem, we will divide by. ( )( ) Reason: = Now we will follow the first two steps that were used in the five step process for decomposition learned in the previous section. First we multiply the coefficient of the first term () by the last term (1). Net we find two numbers that multiply to give us the product we just obtained (), and those same two numbers must add to give us the coefficient of the second term (11). 1 = = = 11 Answer: 8 and Substitute the two values from the previous step into the epression from the second step. ( 8)( ) If the number in the denominator divides evenly into one of the binomials in the numerator, complete the division. Such is the case with this eample. The binomial 8 is divisible of. ( 8)( ) = ( )( ) final answer If the number in the denominator does not divide evenly into either of the binomials in the numerator, factor the number such that each of the resulting factors divides into one of the binomials. Alternate Method # Create a two-by-two grid, putting the first term in the upper left-hand corner and the last term in the lower right-hand corner. 1 NSSAL 80 Draft 009 C. D. Pilmer

87 Multiply the coefficient of the first term () by the last term (1). Net we find two numbers that multiply to give us the product we just obtained (), and those same two numbers must add to give us the coefficient of the second term (11). 1 = = = 11 Answer: 8 and Take these two numbers, complete with their signs and variables, and insert them in the two missing parts of our two-by-two grid. 8 1 Now we will common factor from each row and column. For eample, column one is comprised of and. The common factor in this case is. This factor is written above the column. Column two is comprised of 8 and 1. The factor in this case is. This factor is written above column two. The same thing is done for the two rows. The common factor for row one is. The common factor for row two is. 8 1 The answer can now be read from across the top and along the side of the grid = ( )( ) Eample Factor Alternate Method #1 Since the leading numerical coefficient is 6, then we will create two binomials whose leading numerical coefficients are 6. (6 )(6 ) The product of 6 and 6 is 6, rather than the 6 found in the original trinomial that we are attempting to factor. In order to deal with this problem, we divide by 6. ( 6 )(6 ) 6 Reason: = We start by multiplying the coefficient of the first term (6) by the last term (-1). Net we find two numbers that multiply to give you the product we just got (-8), and those same two numbers must add to give us the coefficient of the second term (-17). 6 ( 1) = 8 = -8 = -17 Answer: -1 and Substitute the two values into the epression. NSSAL 81 Draft 009 C. D. Pilmer

88 ( 6 1)( 6 ) 6 Since the number in the denominator (6) does not divide evenly into either of the binomials in the numerator, then factor the number such that each of the resulting factors divide into one of the binomials. ( 6 1)( 6 ) Notice that divides evenly into the binomial 6 1, and divides evenly into 6. Now complete the division so that we can obtain the desired factors. 7 ( )( ) Alternate Method # Create a two-by-two grid, putting the first term in the upper left-hand corner and the last term in the lower right-hand corner. 6 1 Multiply the coefficient of the first term (6) by the last term (-1). Net we find two numbers that multiply to give us the product we just obtained (-8), and those same two numbers must add to give us the coefficient of the second term (-17). 6 ( 1) = 8 = -8 = -17 Answer: -1 and Take these two numbers, complete with their signs and variables, and insert them in the two missing parts of our two-by-two grid Now we will common factor from each row and column. For eample, column one is comprised of 6 and. The common factor in this case is. This factor is written above the column. Column two is comprised of 1 and 1. The factor in this case is 7. This factor is written above column two. The same thing is done for the two rows. The common factor for row one is. The common factor for row two is The answer can now be read from across the top and along the side of the grid = 7 ( )( ) NSSAL 8 Draft 009 C. D. Pilmer

89 Eample Factor Answer: Alternate Method #1 ( )( ) (1) (-5) = -60 (1) (-5) = 7 ( 1)( 5) ( )( 5) Alternate Method # (1) (-5) = -60 (1) (-5) = ( )( 5) 7 15 = Eample Factor 8y 6y 15. Answer: Alternate Method #1 ( 8y )(8y ) 8 (-6) (-0) = 10 (-6) (-0) = -6 ( 8y 6)( 8y 0) 8 ( 8y 6)( 8y 0) ( y )( y 5) Alternate Method # 8y (-6) (-0) = 10 (-6) (-0) = -6 8y y 8y 5 0y 15 6y 0y 15 y 6y 15 ( y )( y 5) 8y 6y 15 = If you find either of these alternate forms of decomposition easier to remember and complete successfully, feel free to use in remaining sections of this unit. NSSAL 8 Draft 009 C. D. Pilmer

90 Questions: 1. Factor each of the following. (a) 5 1 (b) 7 1 (c) 6a 17a 5 (d) 9 p 9 p (e) y 1y 0 (f) (g) 1d d 10 (h) 6h 9h 0 NSSAL 8 Draft 009 C. D. Pilmer

91 Factoring Part Factoring Perfect Square Trinomials When we talk about perfect squares, we think of numbers like 1,, 9, 16, 5, and 6. These types of numbers can be obtained by squaring a whole number. 1 = 1 = = 9 = 16 5 = 5 6 = 6 When we talk about perfect square trinomials, we are dealing with trinomials that are obtained by squaring binomials. ( 5) = 10 5 ( p 7) = p 1 p 9 ( h ) = h 1h 9 ( 5 6 ) = y = 9 16y ( g 9h) = g 6gh 81h ( ) Note: The sign of the second term in the binomial is always the same as the sign of the second term in the perfect square trinomial. Factoring perfect square trinomials is easy as long as we recognize that we are dealing with such a trinomial. Perfect square trinomials have three specific traits. 1. We can take the square root of the first term.. We can take the square root of the third term.. When we multiply the square root of the first term by the square root of the second term, then double that product, we will obtain the middle term (ignore the sign of the term) p 1 p 9 h 1h 9 Square Root Double Square Root Square Root Double Square Root Square Root Double Square Root 5 Multiply p 7 Multiply h Multiply y 16y g 6gh 81h Square Root Double Square Root Square Root Double Square Root Square Root Double Square Root 5 6 Multiply y Multiply g 9 h Multiply 1 6 is not a perfect square trinomial because we cannot take the square root of is not a perfect square trinomial because the middle term is not 1 or -1. NSSAL 85 Draft 009 C. D. Pilmer

92 If you know you are dealing with a perfect square trinomial, then its factored form is comprised of a squared binomial where the first term is the square root of first term in the trinomial, the second term is square root of the third term in the trinomial, and the sign of the second term in the binomial is the same as the sign of the second term in the trinomial. Eample 1: Factor each of following perfect square trinomials. (a) 16 6 (b) 9d 0d 5 (c) 9 8y y Answer: Answer: Answer: d 0d 5 9 8y y Square Root Take the sign. Square Root Square Root Take the sign. Square Root Square Root Take the sign. Square Root 8 Therefore: 16 6 = ( ) 8 d - 5 Therefore: 9d 0d 5 = d ( ) y Therefore: 9 8y y = 7 y ( ) Some people wonder why we learn this type of factoring because we can use inspection or decomposition to factor these types of trinomials successfully. Unfortunately inspection and decomposition are not always the most efficient techniques. We must also learn how to recognize and manipulate perfect square trinomials so that we can change quadratic functions from their standard form to their transformational form. One of the critical aspects of this is learning how to take a binomial of the form b and adding a number to it such that it becomes a perfect square trinomial. For eample if we take the binomial 6 and add 9 to it, it becomes the perfect square trinomial 6 9. This process is referred to as completing the square. How can we quickly complete the square for any binomial of the form b? All we have to do is take half of the coefficient of the second term, square that value, and add that value to our epression. Eample : Complete the square to change each of the following binomials to a perfect square trinomial. (a) 16 (b) d 1 d (c) y y Answer: Half of 16 is 8 Square the Answer: Half of -1 is -7 Square the -7 d 1d 9 Answer: Half of - is -1 Square the -1 y y 1 NSSAL 86 Draft 009 C. D. Pilmer

93 Using Two Methods of Factoring Up to this point we have learned how to: Common Factor Factor Difference of Squares Factor by Inspection Factor by Decomposition Factor Perfect Square Trinomials Now we will look at polynomial epressions which are factored using two methods for factoring. Specifically we will start by common factoring and then use one of the four remaining types of factoring to complete the question. Eample : Factor each of the following completely. (a) 0 5 (b) cd 1cd 0c (c) y y 6 Answer: ( 9) ( )( ) Answer: cd 1cd 0c c c ( d 7d 10) ( d )( d 5) Answer: y y 6 ( y 6y 9) ( y ) Questions: 1. Which of the following are perfect square trinomials? (a) 11 (b) y 1y 6 (c) g 6g 9 (d) (e) 16d 18d 9 (f) 6 p 60 p 5 (g) 6a b 16ab (h) 9 1y 81y. Factor the following perfect square trinomials. (a) 1 (b) t 0t 100 (c) g 8g 16 (d) 5 10 (e) 9k 6k 1 (f) 9 p 70 p 5 (g) 16d 7d 81 (h) 9a 11ab 6b NSSAL 87 Draft 009 C. D. Pilmer

94 . Part of these perfect square trinomials is missing. Fill in the blank to complete the trinomial and then factor the trinomial. (a) 00 (b) h h 900 (c) 11p p (d) 81y y 100 (e) 16 y y 11 (f) 6a ab 9b. Complete the perfect square trinomials by adding a specific number to the epression. (i.e. Complete the square.) (a) (c) (e) (g) 1 (b) 10 t 8t (d) w 0w 6 (f) p 60 p d d (h) 0 5. Factor each of the following completely. All of these require that you use two types of factoring. (a) 75 (b) 18 8 (c) y y 6 (d) t t 0t (e) (f) 0 p 5p NSSAL 88 Draft 009 C. D. Pilmer

95 (g) (h) 9 18 (i) 18 (j) 16 c 8c c (k) 1ab 75a (l) 8 pq 8 pq 6 p 6. Factor each of the following. Identify the type or types of factoring you used. (a) 1 y 8 y (b) 5p 9 (c) t 1 t (d) d ( d ) 7( d ) (e) 10 5 (f) 6a 5a NSSAL 89 Draft 009 C. D. Pilmer

96 (g) (h) 9 16h (i) 16 9 (j) y 1 y (k) ( p 1) ( p 1) p (l) (m) 18c 50c (n) 6 6 (o) 9 d d 9 (p) 5p q 1 p q 7 pq (q) 70 (r) 7c 8c 8c If you feel that you need additional practice with these types of questions, you can complete the optional questions titled Additional Practice: Factoring Polynomials found in the appendi of this unit. NSSAL 90 Draft 009 C. D. Pilmer

97 Standard Form to Transformational Form As previously mentioned, one of the reasons we spent so much time learning how to factor polynomials was so that we can change quadratic functions from their standard form ( y = a b c ) to their transformational form y = k( c) d. (Remember that many resources will refer to this as changing from the general form to standard form.) To accomplish this we must be good at common factoring and factoring perfect square trinomials. Eample 1 Change y = 1 into its transformational form. Answer: y = 1 y = 1 ( ) ( 6 ) y = y = y = y = ( 6 9) ( 9) ( 6 18 ( ) Eample Change y = into its transformational form. Group the first two terms together. Look at the coefficient of the. Common factor that value from the two terms grouped in the parentheses. In this case, we common factor out a. We want to take the binomial 6 and change it to a perfect square trinomial. This is accomplished by adding 9 (completing the square: half of 6, then squared). We cannot simply add a number to one part of the equation without throwing the balance of the equation off. To compensate we have to subtract an equivalent value from the same side of the equation. Since the 9 that we added was multiplied by, then we must also subtract 9. Factor the perfect square trinomial. The function is now in its transformational form. Answer: y = Group the first two terms together. y = ( 5 0) 79 y = 5 8 Common factor out the -5 from the grouped terms. y = 5 y = 5 ( ) 79 ( 8 16) 79 ( 5 16) ( ( ) 1 Change the binomial 8 to a perfect square trinomial by adding 16. Since the 16 that we added was multiplied by -5, then we must also subtract 5 16 to ensure that the equation remains balanced. y = 5 Factor the perfect square trinomial. NSSAL 91 Draft 009 C. D. Pilmer

98 Eample Change y = 87 into its transformational form and state the transformations. Answer: y = 87 Group the first two terms together. y = 87 Common factor out the from the grouped terms. ( 16 ) 87 y = This is accomplished by dividing both terms by (equivalent to multiplying both terms by ). y = y = ( 16 6) 87 6 ( 16 6) y = ( 8) 9 VS =, VT = -9, HT = 8 Change the binomial 16 to a perfect square trinomial by adding 6. Since the 6 that we added was multiplied by, then we must also subtract 6 to ensure that the equation remains balanced. Factor the perfect square trinomial and state the transformations. Eample Change y = 1 11 to its transformational form. Answer: y = 1 11 y = y = y = y = y = ( 1) 11 ( ) 11 ) 11 ( ) ( 111 ( ) 1 Eample 5 Change y = 8 to its transformational form. Answer: y = 8 y = y = y = y = y = 1 ( ) 8 1( ) 8 1) 8 ( 1 1) 1( 8 1 ( 1) 7 NSSAL 9 Draft 009 C. D. Pilmer

99 Eample 6 Change 1 y = 8 to its transformational form and state the transformations. Answer: 1 y = 8 1 y = 8 1 y = ( 8) y = ( 8 16) y = ( 8 16) y = ( ) Eample 7 Change y = 8 9 to its transformational form and state the transformations. Answer: y = 8 9 y = 8 9 y = ( 1) 9 y = ( 1 6) 9 y = ( 1 6) 9 y = ( 6) 5 6 VS = 1, HT = - Reflection in -ais, VS =, VT = -5, HT = 6 Questions: 1. Fill in the blanks to complete the answer. (a) Change y = 1 to its transformational form and state the transformations. Answer: y = 1 y = y = y = y = y = ( ) 1 ( ) 1 ( ) 1 ( 9) ( ) 1 ( ) 5 VS =, VT =, HT = NSSAL 9 Draft 009 C. D. Pilmer

100 (b) Change y = 6 7 to its transformational form and state the transformations. Answer: y = 6 7 y = y = y = y = y = ( 6) 7 ( ) 7 ( 1) 7 ( ) ( 7 ( ) 10 Reflection in -ais, VS =, VT =, HT = (c) Change 5 y = 10 7 to its transformational form, state the transformations and describe the resulting parabola. Answer: 5 y = y = y = ( ) 7 5 y = 5 y = 5 y = ( ) ( ) ( ) VS =, VT =, HT = The graph is concave (downwards or upwards) since there is no reflection in the -ais. The coordinates of the verte should be (, ). The graph looks (narrower or wider) than the graph of y = due to the vertical stretch. NSSAL 9 Draft 009 C. D. Pilmer

101 . Change each of the following into their transformational forms. Show all your work. (a) y = 1 1 (b) y = (c) y = 10 1 (d) y = 6 (e) y = 5 (f) y = 6 NSSAL 95 Draft 009 C. D. Pilmer

102 (g) y = 8 7 (h) y = (i) y = 1 7 (j) y = 0 (k) y = (l) y = 7 NSSAL 96 Draft 009 C. D. Pilmer

103 (m) 11 y = (n) y = 1 6. Graph y = 0 1 using transformations. If you cannot remember how this is done, look back at the section titled Graphing Quadratic Functions Using Transformations. Also state the domain, range, coordinates of the verte and the equation of the ais of symmetry If you feel that you need additional practice with these types of questions, you can complete the optional questions titled Additional Practice: Standard Form to Transformational Form found in the appendi of this unit. NSSAL 97 Draft 009 C. D. Pilmer

104 Is There a Formula? After spending the last few days mastering how to change a quadratic function from standard form to transformational form, you may be wondering if there is an easier way to accomplish this. The answer is yes if one is prepared to derive the necessary formula. y = a b c y = ( a b) c b y = a c a b b b y = a c a a a a b b y = a c a a We usually refer to our transformational form as y = k( c) d but for this eplanation we will use the form y = k( d ) e. The reason is that we do not want to confuse the variable c in the equation y = a b c with the variable c in the equation y = k( c) d. They are not equivalent. Using the form y = k( d ) e, we can conclude the following. k = a b d = a b e = c a Eample Change y = 1 into its transformational form. Answer: k = a k = Therefore: y = ( ) b d = a 1 d = d = b e = c a 1 e = e = Most people do not bother with these formulas because they find the formulas too difficult to remember. NSSAL 98 Draft 009 C. D. Pilmer

105 Solving Quadratic Equations by Factoring In this section we will learn how to solve factorable quadratic equations of the form a b c = 0. These quadratic equations often arise when we are given a quadratic function ( y = a b c ) and asked to solve for when provided with a specific value of y. Before we can learn how to solve quadratic equations by factoring, we must first talk about the zero-factor property. Zero-Factor Property If a and b are real numbers and a b = 0, then a = 0 or b = 0. Translated this means that if the product of two numbers is 0, then at least one of the numbers must be 0. One number must be zero, but it is possible that both numbers are 0. We will use this property once we have the quadratic equation in its factored form. Eample 1 Given 0 =, find. Answer: 0 = 8 = 0 ( 7 )( ) = 0 7 = 0 or = 0 = 7 = Set the equation equal to zero by adding to both sides. We have to do this because the zero-factor property only applies when the product is 0. Factor the quadratic equation completely. In this case we factored by inspection. We now use the zero-factor property. If the product of two binomials is 0, then one of the binomials must be equal to 0. We can now solve the two resulting equations. We end up with two real roots (i.e. two answers for ). Eample Given 6a 7 = 15a 7, find a. Answer: 6a 7 = 15a 7 Set the equation equal to zero by subtracting 15a and 7 6a 15a = 0 from both sides. Factor the quadratic equation completely. In this case we a ( a 5) = 0 common factored. a = 0 or a 5 = 0 We now use the zero-factor property. If the product of a monomial and a binomial is 0, then either the monomial or a 0 a = 5 = the binomial is be equal to 0. We can now solve the two 5 a = resulting equations. We end up with two real roots (i.e. a = 0 two answers for a). NSSAL 99 Draft 009 C. D. Pilmer

106 Eample Given the function y = 9 0 7, solve for when y equals -. Answer: y = 9 0 = 9 0 = 9 0 = = ( 5) Therefore: 5 = 0 = 5 = Between the third and fourth steps we multiplied both sides of the equation by -1. By doing this we change the leading numerical coefficient to a positive number. Most people find it easier to factor polynomial epressions when the leading numerical coefficient is positive. In this case we only end up with one real root (i.e. one answer for ). Eample A metal rod is attached to the top of two poles and shaped in a manner that can be described by the equation h = d 1d 1 where h is the height in metres and d is the horizontal distance in metres. Determine the distances that correspond to a rod height of 18 metres. Answer: h = d 1d 1 18 = d 1d 1 0 = d 1d 5 0 = d 1d 5 Method I 0 = d 10d d 5 0 = d 10d d 5 0 = d( d 5) 1( d 5) 0 = d 5 d 1 ( ) ( ) ( )( ) (Multiply by -1to make the leading numerical coefficient positive.) (Decomposition) Method II d 0 = 0 = 0 = 0 = Therfore: d 5 = 0 or d 1 = 0 d = 5 d = 1 d = 5 metres d = ( )( d ) ( d 10)( d ) ( d 10)( d ) ( d 5)( d 1) metres NSSAL 100 Draft 009 C. D. Pilmer 1

107 Eample 5 f Given ( ) = 5 (a) when f ( ) = 1. (b) f ( ). Answers: f = (a) ( ) 5 1 = 0 = 0 = 0 = ( )( ) = or =, find: = 0 or = 0 (b) f ( ) = 5 f ( ) = ( ) ( ) 5 f ( ) = ( 16) ( ) 5 f ( ) = 15 In this case we have been supplied with the dependent variable and been asked to find the independent variable,. In this case we have been supplied with the independent variable,, and been asked to find the dependent variable. Questions: 1. Complete the following solutions by filling in the blanks. (a) 0 = 8 (b) 16 = = 0 ( 6 )( 8) = 0 6 = 0 or 8 = 0 = or = (c) = 1 (d) 11 = ( 5 9) 0 ( 11) = 0 7 = 7 = 0 or 5 9 = 0 NSSAL 101 Draft 009 C. D. Pilmer

108 . Solve each of the following equations. (a) 8 15 = (b) 10 5 = (c) 9 5 = 6 (d) 6 = 1 17 (e) d d 0 = 0 (f) 16t 5 = 0t (g) h = 10h (h) y = 1y 0 (i) 15p p = p (j) 5 9 = 0 (k) 9c 1 = c 8 (l) 5y 15y 7 = y NSSAL 10 Draft 009 C. D. Pilmer

109 (m) 1 = 0 (n) 5n = 1 10n. Given the function y = 8 15, solve for when y equals 6.. Given the function p = q 17q 80 and that p equals -8, solve for q. 5. A steel ball is rolled down a long ramp. It is initially given a gentle push. The equation d = t t describes the distance, d, the ball travels in metres in terms of the time, t, measured in seconds. (a) How far does the ball travel in 1.5 seconds? (b) How long will it take the steel ball to travel 15 metres down the ramp? NSSAL 10 Draft 009 C. D. Pilmer

110 6. A ball is thrown from a raised patio. Its height, h, in feet with respect to time, t, in seconds can be represented by the equation h = 16t t 7. (a) How high is the ball at 1. seconds? (b) What is the initial height of the ball? (c) When is the ball at a height of 15 feet? (d) When does the ball hit the ground? (e) Given that the maimum height reached by the ball is 16 feet, state the domain and range of the quadratic function. 7. If the points P ( 5, y) and Q (,1) missing coordinates. lie on the quadratic function y = 6 9, determine the 8. Given g ( ) = 0 0 (a) when g ( ) = 5. (b) when g ( ) = 0. (c) g ( )., find: NSSAL 10 Draft 009 C. D. Pilmer

111 Solving Quadratic Equations Using the Quadratic Formula In the last section, we learned how to solve quadratic equations by factoring. The reality is that most quadratic equations can not be solved using this method. Such is the case with the three equations listed below. None of these can be factored. 9 7 = 0 5 = 0 7 = 0 We need another approach that will work with any quadratic equation. The quadratic formula is that alternate approach. Before we start using this formula we should take a few minutes to see where it comes from. Read over the following derivation. The Derivation of the Quadratic Formula a b c = 0 a b = c a b c = a a a b c = a a Subtract c from both sides of the equation. Divide everything on both sides of the equation by a. Complete the square by taking half of a b, squaring b b b c = b that value. This new value,, must be added to a a a a a both sides of the equation to ensure that the equation remains balanced. b b ac = The left hand side of the equation was a perfect square a a a trinomial which is then factored. On the right hand b b ac side of the equation, we will subtract the two = epressions by creating the common denominator a. a a b b ac We want to solve for so we must get rid of the = ± a a squaring by square rooting both sides of the equation. b We can subtract from both sides of the equation b b ac a = ± a a so that we can isolate. We will also take the square b ± = b ac a root of a. The two terms on the right side of the equation can be combined because they have a common denominator. The resulting formula is called the quadratic formula. NSSAL 105 Draft 009 C. D. Pilmer

112 Eample 1 Solve for given that 9 = 0 Answer: b ± b ac = where a =, b = 9, and c = a 9 ± = 9 ± = 9 ( ) 81 ( )( ) = or = = 0.6 or =.17 (two real roots) Eample For the function f ( ) = 1, find when ( ) f = 5. Answer: Substitute 5 into the equation. 5 = 1 Set the equation equal to 0 by subtracting 5 from both sides of the equation. The quadratic formula only works when the equation is set equal to 0. 0 = b ± b ac = where a =, b = -, and c = - a ( ) ± ( ) ( )( ) = ( ) ± 8 = = or = 6 6 = 1.55 or = (two real roots) Eample A projectile is fired vertically into the air. The height of the projectile with respect to time is represented by the equation h =.9t 15t 1. The height, h, is measured in metres. The time, t, is measured in seconds. (a) At what time(s) is the projectile at a height of 10 m? (b) At what time does the projectile strike the ground? (c) At what time(s) is the projectile at a height of 17 m? NSSAL 106 Draft 009 C. D. Pilmer

113 Answer: (a) 10 =.9t 15t 1 0 =.9t 15t 9 15 ± t = 15 ± t = ( 15) (.9)( 9) (.9) t = or t = t = or t =. The projectile reaches a height of 10 metres at approimately 0.8 seconds and. seconds. One would epect two times because the projectile reaches 10 metres as it travels up and as it falls back to the ground. (b) The height of the projectile when it strikes the ground is 0 metres. Enter this value into the equation for h. 0 =.9 t 15t 1 ( ) 15 ± 15 (.9 )( 1 t = ).9 ( ) 15 ± t = t = or t = 9.8 t = or t = The first time does not make sense in this contet. How can you have negative time? The only answer that is reasonable for this situation is the second value for t. The projectile strikes the ground at approimately.1 seconds. (c) 17 =.9t 15t 1 0 =.9t 15t ± t = ( 15) (.9)( 16) (.9) 15 ± t = t = or t = 9.8 We cannot take the square root of a negative number therefore there is no real solution for this quadratic equation (i.e. no real roots). Based on this we can conclude that the projectile never reaches a height of 17 metres. NSSAL 107 Draft 009 C. D. Pilmer

114 Eample For the function y = 6 determine: (a) the y-intercept. (b) the -intercept(s). Answers: (a) The y-intercept for any type of function occurs when equals 0. y = 6 y = y = ( 0) 6( 0) (b) The -intercept(s) for any type of function occurs when y equals 0. y = 6 b ± b ac = 0 = 6 a = 6 ± = ( 6) ± ( 6) ( )( ) ( ) = =.66 or 6 1 or = = 0.6 Questions: 1. Solve each of the following equations. Show your work. (a) 9 5 = 0 (b) 7 6 = 0 NSSAL 108 Draft 009 C. D. Pilmer

115 (c) 1 = (d).1a = 7.a 9 (e) 0.9 p 8p. = (f) 1.75t =.8t (g).5a = 6a NSSAL 109 Draft 009 C. D. Pilmer

116 (h) = 0. Solve the quadratic equation 5 = 0 by factoring and using the quadratic formula to prove that we will obtain the same answer using either method.. Given h ( ) = 1 (a) h ( 1). (b) when h ( ) = 10., find:. For the function y = 8 1 determine: (a) the y-intercept. (b) the -intercept(s). NSSAL 110 Draft 009 C. D. Pilmer

117 5. The distance a car travels after the driver decides to slam on the brakes must consider two factors: the distance the car travels as the driver reacts to the situation (no brakes applied) and the distance the car travels when the brakes have been applied. The quadratic equation d = 0.007s 0. s describes the stopping distance, d, in terms of the initial speed, s, of the car. The distance is measured in metres and the speed is measured in kilometers per hour. (a) How fast was the car traveling at the point when the driver saw the braking situation if the stopping distance was 95 metres? (b) How far did the car travel after the driver initially saw the braking situation if the car was traveling at 60 km/h? 6. How many real roots (i.e. real solutions) can a quadratic equation have? (a) no solution (b) one solution (c) two solutions (d) either a, b, or c NSSAL 111 Draft 009 C. D. Pilmer

118 Programming the Quadratic Formula into a TI-8 or TI-8 The quadratic formula is a powerful tool for solving quadratic equations. Eample: Given = 0, solve for. Solution: b ± = 1 ± = b ac a 1 ( 8)( 15) ( 8) 1 ± = = or = = 0.75 or = Using this formula can be quite time-consuming. This issue can be rectified if we program our graphics calculators to solve quadratic equations using the quadratic formula. (a) Creating a New Program PRGM > NEW > select Create New > enter the name of the program > ENTER (the ALPHA key will automatically be activated so that you can type in letters. (b) Entering the Program The quadratic formula program is listed below. Screen shots from the calculator weren t supplied because they wouldn t allow you to see the full program. Look at the net page to see how you can access the Prompt and Disp commands found in this program. Prompt A Prompt B Prompt C (-B (B - AC))/(A) D (-B - (B - AC))/(A) E Disp D Disp E NSSAL 11 Draft 009 C. D. Pilmer

119 The commands Prompt and Disp are found by pressing PRGM and moving over to I/O. The arrow used in the fourth and fifth lines of the program is obtained by pressing the STO key. Once all seven lines are correctly entered, press QUIT. (c) Running the QUAD Program PRGM > EXEC > QUAD > enter the values of A, B, and C We may want to test our program by using the quadratic equation = 0. The answers should be 0.75 and -.5. If the program does not generate these values, then press PRGM, move to EDIT, and select the program. This will put us back into the program and we can now make the necessary changes. Optional (Transferring the Program) We may wish to transfer this program to other calculators in the classroom. This can be done using the link cable and a few simple key strokes on the two calculators. (a) Insert the link cable into the port of each calculator. (b) Set One Calculator to Receive LINK > RECEIVE > select Receive (c) Set the Other Calculator to Transmit LINK > SEND > Prgm > SELECT > select the program > TRANSMIT > select Transmit Online Quadratic Formula Calculator (Optional) (or Google Search: Math Warehouse Quadratic Formula Calculator & Solver) NSSAL 11 Draft 009 C. D. Pilmer

120 Word Problems Involving Quadratic Equations For the questions in this section, feel free to use the QUAD program you have entered in the TI-8 or TI-8 graphing calculator. Eample 1 Find the two consecutive positive even numbers whose sum of their squares is 580. Answer: If you think about any two consecutive even numbers (e.g. and 6, 8 and 0, 7 and 7, ), the second number is always two more than the first number. Based on this, we can let represent our first even number, and represent the second even number. - first even number - second even number Now we create an equation based on the statement whose sum of their squares is 580. We need to square both of the even numbers separately and then add those values together. ( ) 580 = We will now epand the equation and set it equal to 0. = = 0 We can now use the QUAD program on the graphing calculator. = 16 or -18 (Ignore the -18 because the question asked for positive numbers.) = 16 = 18 The two consecutive positive even numbers are 16 and 18. Eample A rectangular building measuring 0 m by 5 m is going to have its area increased by 00 m by adding a strip of uniform width to all four sides. Determine the width of the strip. Answer: For the diagram of this situation, we have a smaller rectangle (the original building) surrounded by a larger rectangle (the new building). We will let represent the unknown width of the strip. If the dimensions of the original building were 0 m by 5 m, then the dimensions of the new building will be 0 by 5 If the area of the original building was 750 m, then the area of the new building will be 950 m. 0 5 NSSAL 11 Draft 009 C. D. Pilmer

121 Now we look at the area of the new building. A = l w 950 = l w 950 = ( 0 )( 5 ) We will now epand the equation and set it equal to = = We can now use the QUAD program on the graphing calculator. = 1.7 or -9. (Ignore the -9. because the width cannot be a negative number.) The width of the strip is 1.7 m. Eample The hypotenuse of a right angle triangle is cm longer than one of the legs, and 6 cm longer than the other leg. Find the lengths of the three sides of the triangle. Answer: If we let represent the length of the hypotenuse, then the longer leg would be, and the shorter leg would be 6. The relationship between the three sides of a right-angle triangle is described by the Pythagorean Theorem. a b = c ( ) ( 6) = = 18 5 = 0 = or 15 (Ignore the because it will generate leg lengths of 0 cm and - centimetres.) - 6 The hypotenuse is 15 cm. The longer leg is 1 cm. The shorter leg is 9 cm. - Questions: 1. Find the two consecutive positive even numbers whose sum of their squares is NSSAL 115 Draft 009 C. D. Pilmer

122 . The square of the sum of two consecutive positive even numbers is 916. Find the numbers.. The sum of the squares of three consecutive positive odd numbers is 71. Find the numbers.. The sum of the squares of three consecutive numbers is 590. Find the numbers. 5. There are three consecutive positive numbers. The sum of the squares of the first two numbers eceeds the square of the third number by 896. Find the three numbers. NSSAL 116 Draft 009 C. D. Pilmer

123 6. There are three consecutive positive even numbers. The sum of the squares of the first two numbers eceeds the square of the third number by 8. Find the three numbers. 7. There are three consecutive positive odd numbers. The sum of the squares of the first and third numbers eceeds the square of the second number by 9. Find the three numbers. 8. There are two consecutive positive numbers. If the square of the first number is decreased by double the second number, then the result is 118. Find the two numbers. 9. There are two consecutive positive even numbers. If the square of the second number is increased by triple the first number, then the result is 5. Find the two numbers. NSSAL 117 Draft 009 C. D. Pilmer

124 10. A building measuring 50 metres by 0 metres is going to have its area double by adding a strip of uniform length to all four sides. Find the width of the uniform strip. 11. A building measuring 60 metres by 0 metres is going to have its area increased by 00 m by adding a strip of uniform length to two adjoining sides. Find the width of the uniform strip. 1. A rectangular building is built on a lot measuring 70 m by 50 m. The building will be surrounded by a strip of lawn of uniform width. If the building takes by 60% of the lot, determine the width of the uniform strip. 1. A rectangular building is built on a lot measuring 80 m by 60 m. The building will be tucked in the corner of the lot such that a strip of lawn of uniform width will be along two adjoining sides of the building. If the building takes up 70% of the lot, determine the width of the uniform strip. NSSAL 118 Draft 009 C. D. Pilmer

125 1. The hypotenuse of a right angle triangle is cm longer than one of the legs, and 9 cm longer than the other leg. Find the lengths of the three sides of the triangle. 15. The longer leg of a right angle triangle is 1 cm shorter than the hypotenuse and 17 cm longer than the shorter leg. Find the lengths of the three sides. 16. If the length of the shorter leg of a right-angle triangle is doubled and then increased by cm, one obtains the length of the other leg of the triangle. If the length of the shorter leg is tripled and then decreased by cm, one obtains the length of the hypotenuse. Find the side lengths. 17. If the length of the shorter leg of a right-angle triangle is tripled and then decreased by 1 cm, one obtains the length of the other leg of the triangle. If the length of the shorter leg is doubled and then increased by 1 cm, one obtains the length of the hypotenuse. Find the side lengths. NSSAL 119 Draft 009 C. D. Pilmer

126 18. The height of a projectile with respect to time is described by the equation h =.9t t. The height, h, is measured in metres. The time, t, is measured in seconds. Determine the times when the projectile is 5 metres off the ground. 19. There are three consecutive positive odd numbers. The sum of the squares of the second and third numbers eceeds the square of the first number by 85. Find the three numbers. 0. A building measuring 70 metres by 0 metres is going to have its area increased by 60% by adding a strip of uniform length to two adjoining sides. Find the width of the uniform strip. 1. If the length of the shorter leg of a right-angle triangle is increased by cm, one obtains the length of the other leg of the triangle. If the length of the shorter leg is doubled and then decreased by 1 cm, one obtains the length of the hypotenuse. Find the side lengths. NSSAL 10 Draft 009 C. D. Pilmer

127 NSSAL 11 Draft 009 C. D. Pilmer Finding the Verte The verte of a parabola either represents the highest point on the curve (i.e. the maimum) or the lowest point on the curve (i.e. the minimum) depending on whether the quadratic function has undergone a reflection in the -ais. When dealing with real life applications that can be modeled using quadratic functions, it is sometimes necessary to find that maimum or minimum point on the curve. For eample, if we were given the equation that described the flight path of a projectile, then the verte could be used to determine the maimum height reached by the projectile. To find the coordinates of the verte, we could change the equation of the quadratic function into its transformational form and use the horizontal translation as the -coordinate and the vertical translation as the y-coordinate of the verte. Changing equations into their transformational form can be tricky particularly if the coefficients are fractions or decimals. It would be easier if we had formulas for determining the coordinates of the verte. Derivation of the Verte Formulas ( ) a ac b a b a y a b a ac a b a y a b c a b a y a b a c a b a b a y c a b a y c b a y c b a y = = = = = = = The horizontal translation (HT) = a b The vertical translation (VT) = a ac b Therefore the coordinates of the verte are a ac b a b, Note: Most adult learners do not have a problem identifying minimum/maimum word problems. The issues usually lie in terms of deciding which one of the two formulas should be used to answer the specific question. - - (HT, VT)

128 Eample 1 For the quadratic function y = 16 5, determine the coordinates of the verte. State whether the verte is a minimum or a maimum. Answer: b b ac, a a ( 16) ( ), 56 00, 8 (,7) ( 16) ( )( 5) ( ) The verte, (-, 7), will be a maimum because the quadratic function has undergone a reflection in the -ais. We can tell because the leading numerical coefficient, a, is a negative number (-). The reflection indicates that the curve is concave downwards and the verte is therefore the highest point on the curve. Eample The height of a projectile with respect to time can be described by the equation h =.9t 18t. The height, h, is measured in metres and the time, t, is measured in seconds. (a) What is the maimum height reached by the projectile? (b) When does the projectile reach its maimum height? (c) At what time(s) is the projectile at a height of 7 metres? (d) What is the initial height of the projectile? Answers: (a) Since we are dealing with the maimum height, we know that at least one of the verte formulas must be used. In this situation the dependent variable is h and the independent variable is t. This means that any point on this curve, including the verte, would have coordinates written in the form (t, h). Since we are asked to find the height (dependent variable), then the second verte formula must be used. b ac h = a ( 18) (.9)( ) h = (.9) 58.8 h = 19.6 h = The projectile reaches a maimum height of approimately 19.5 metres. NSSAL 1 Draft 009 C. D. Pilmer

129 (b) In this case we must find the time (independent variable) when the maimum height occurs. We need to use the first verte formula. b t = a 18 t =.9 ( ) t = 1.87 The projectile reaches its maimum height at approimately 1.8 seconds. (c) This question has nothing to do with the verte. Simply find t given h. h =.9t 18t 7 =.9t 18t 0 =.9t 18t Use the QUAD program. The projectile reaches the height of 7 m at approimately 0. seconds and. seconds. (d) This question has nothing to do with verte. The initial height would occur when t = 0. h =.9t 18t h =.9 h = ( 0) 18( 0) Eample The fuel economy of a vehicle is measured in litres per 100 km (L /100 km). A vehicle with a fuel economy of 8. L/100 km uses less fuel than a vehicle with a fuel economy of 9.8 L/100 km. The fuel economy is affected by the speed of the vehicle. If a car goes too slow or too fast, the fuel economy is sacrificed. The equation E = 0.00s 0.8s describes the fuel economy, E, of a particular European subcompact car in terms of the car s speed, s. (a) At what speed would this particular car travel in order to obtain a fuel economy of 9. L/100 km? (b) At what speed should this car travel in order to obtain the best fuel economy? (c) What is the best fuel economy this car can obtain? Answer: (a) This question has nothing to do with the verte. We have been given E and been asked to solve for s. E = 0.00s 0.8s 9. = 0.00s 0.8s 0 = 0.00s 0.8s 1.7 QUAD program: s =.9 and 96. The car can obtain a fuel economy of 9. L/100 km if it cruises at.9 km/h or 96. km/h. NSSAL 1 Draft 009 C. D. Pilmer

130 (b) The best fuel economy occurs when we have the lowest number of litres per 100 kilometres. For this reason, we need to look at the verte. In this situation the dependent variable is E and the independent variable is s. This means that any point on this curve, including the verte, would have coordinates written in the form (s, E). Since we are asked to find the speed (independent variable), then the first verte formula must be used. b s = a 0.8 s = s = 70.6 ( 0.00) The best fuel economy is obtained when the car travels at 70.6 km/h. (c) Since we are asked to find the best fuel economy (dependent variable), then the second verte formula must be used. b ac E = a E = E = 7.06 ( 0.8) ( 0.00)( ) ( 0.00) The best fuel economy for this car is 7.06 L/100km. Important Note: With the remaining four eamples, the equation of the quadratic function has not been supplied. We will be required to use the information supplied to generate the equation. Eample Two numbers are related by the equation y =. Find the values of and y so that their product is a maimum. Answer: The two unknown numbers that we have been asked to find are represented by the variables and y. We have been given the equation y = but it is linear (not quadratic). We have been told that their product is a maimum. We can use this to generate the following equation where P represents the product. P = y This equation is not quadratic. Even worse, it has three variables (P,, and y) in it, rather than two that we would epect for a quadratic function. We will now take the equation that was supplied and the equation we generated, and substitute one into the other so that we can end up with a quadratic function. NSSAL 1 Draft 009 C. D. Pilmer

131 y = P = y y = ( ) P = P = P = We now have our quadratic function that has the variables P (the product) and (one of the two numbers). Any point on this curve, including the verte, would have coordinates written in the form (, P). Since the question asks us to find and y, we will use the first verte formula to find (independent variable). We will not use the second verte formula in this question because we were not asked to determine the product, P. b = a = = ( ) We now need to find the other number y. y = y = y = 16 ( ) The number represented by the variable is. The number represented by the variable y is 16. Eample 5 Two numbers differ by. Find their values if the sum of their squares is a minimum. Answer: We can create two equations based on each of the sentences in the question. Neither of the equations will be quadratic functions. y = s = y For the equations above, represents the larger number, y represents the smaller number, and s represents the sum of the squares. We will now take the two equations, substitute one into the other such that we end up with a quadratic function. y = = y s = y NSSAL 15 Draft 009 C. D. Pilmer

132 s = s = y s = y ( y ) 8y 576 y y 8y 576 We now have our quadratic function that has the variables s and y. Any point on this curve, including the verte, would have coordinates written in the form (y, s). Since the question asks us to find and y, we will use the first verte formula to find y (independent variable). We will not use the second verte formula in this question because we were not asked to determine the sum of their squares, s. b y = a 8 y = y = 1 ( ) = y = 1 = 1 The larger number is 1 and the smaller number is -1. Eample 6 Angela wants to create two identical and adjacent pens (as illustrated) using 50 m of fencing. What are the dimensions of the entire enclosure if the area must be a maimum? Answer: Let l represent the length of the entire enclosure, w represent the width and A represent the area. Create two equations (neither will be quadratic). The first equation is created by looking at the 50 m of fencing that is used. The other equation represents the area of the entire enclosure. l w = 50 A = lw Substitute one equation into the other to create a quadratic function. l w = 50 A = lw l = 50 w l = 5 1.5w ( ) A = w w A = 5w 1.5w A = 1.5w 5w Now use the first verte formula to find the width. Once we have the width, w, then we can solve for the length, l. NSSAL 16 Draft 009 C. D. Pilmer

133 b w = a 5 w = w = 1 8 ( 1.5) The entire enclosure measures l = 5 1.5w 1 l = l = l = metres by 1 8 metres. Eample 7 A lifeguard wants to create a rectangular supervised swimming zone along a popular beach. She only has 00 m of rope with marker buoys to enclose three sides of the rectangular zone (see illustration). If she wishes to maimize the area of the zone, what should the dimensions be? Answer: Let l represent the length of the supervised zone, w represent the width and A represent the area. Create two equations (neither will be quadratic). The first equation is created by looking at the 00 m of rope that is used to enclose three sides of the supervised zone. The other equation represents the area of the rectangular zone. l w = 00 A = lw Substitute one equation into the other to create a quadratic function. l w = 00 A = lw l = 00 w ( 00 ) Supervised Zone Rope with Marker Buoys Beach A = w w A = w 00w Now use the first verte formula to find the width. Once we have the width, w, then we can solve for the length, l. b w = a 00 l = 00 w w = ( ) l = 00 ( 100) w = 100 l = 00 The rectangular zone measures 00 m by 100 m. NSSAL 17 Draft 009 C. D. Pilmer

134 Questions: 1. Determine the coordinates of the following quadratic functions without using graphing technology or by changing the equation into its transformational form. In each case, state whether the verte is a minimum or a maimum. (a) y = (b) y = 1 7. The height of a projectile (in feet) with respect to time (in seconds) can be represented by the equation h = 16t t. (a) How high is the projectile at 1. seconds? (b) What is the maimum height reached by the projectile? (c) When does the projectile reach its maimum height?. The sum of two numbers is -8. Find the two numbers if their product is a maimum. NSSAL 18 Draft 009 C. D. Pilmer

135 . A rectangular dog kennel is being constructed along one side of a house. Fencing will be placed along three sides of the kennel. If one must use all 1 metres for the three sides, what should the dimensions of the kennel be to maimize the area of the kennel? House Kennel 5. Twice the first number increased by the second number is 16. Find the two numbers if the sum of their squares is a minimum. 6. In 1971, Apollo 1 astronaut, Alan Shepard, hit a golf ball while on the surface of the moon. Although the eact trajectory of the ball was not recorded, we have taken liberal license and created our own equation, h = 0.00d 0. 6d, that describes the height, h, of the ball in metres in terms of horizontal distance, d, the ball travels in metres. (a) What is the maimum height reached by the golf ball? (b) At what horizontal distances will the ball be at a height of 15 m? NSSAL 19 Draft 009 C. D. Pilmer

136 7. Montez decided to build five adjoining rectangular dog kennels using 90 m of fencing. The fencing not only goes around the perimeter but also separates the individual kennels. What should the dimensions of the entire structure be if he wishes to maimize the area covered? 8. Triple the first number increased by double the second number is 60. Find the two numbers if their product is a maimum. 9. Three times the first number decreased by the second number is 1. Find the numbers if the square of their sum (not sum of their squares) is a minimum. NSSAL 10 Draft 009 C. D. Pilmer

137 10. The height of the main support cables on a suspension bridge relative to the water can be described by the equation h = 0.015d d 8. The height, h, is measured in metres. The horizontal distance, d, from one of the support towers is also measured in metres. (a) What is the height of the support cable relative to the water when one is 0 metres from the support tower? (b) How far does one travel horizontally to reach the minimum height of support cable? (c) What is the minimum height of the cable? (d) At what horizontal distances is the support cable at a height of 6 m? 11. An enclosure of the following configuration is to be constructed using 10 m of fencing. (a) If one wishes to maimize the area, determine that area. (b) What are the values of and y that maimize the area? y y NSSAL 11 Draft 009 C. D. Pilmer

138 Application Questions In this section we will focus on multiple step application questions that require us to use a variety of concepts that we learned in this unit. Eample: A projectile is fired vertically into the air from a raised platform. The height of the projectile at specific times is recorded in the table below. The height is measured in metres and the time is recorded in seconds. time height (a) Determine the equation that describes the height, h, of the projectile in terms of time, t. (b) What is the maimum height reached by the projectile? (c) When does the projectile reach its maimum height? (d) For this question we will use function notation and epress the equation we found in the first question as h ( t). If so, determine h ( 1.5) and eplain what the answer represents in terms of this real world situation. (e) At what times is the projectile at a height of 6 m? What does one of these values represent? (f) When will the projectile strike the ground? (g) What is the equation of the ais of symmetry? (h) State the domain and range. Answer: (a) We are going to use finite differences because we were supplied with a table of values. t h t D D It is quadratic because we have a common difference at the D level. t h = at bt c t 1 a b c D1 1 a b c a b D 1 9a b c 5a b a 1 16a b c 7a b a 1 5 5a 5b c 9a b a 1 6 6a 6b c 11a b a a 7b c 1 a b a We can now compare the tables. Specific elements of one table are equal to specific elements in the other table. Knowing this we can generate the following three equations. NSSAL 1 Draft 009 C. D. Pilmer

139 a = 9.8 a b = 0. a b c = 6. 1 Now we can solve for a, b, and c. a = 9.8 a =.9 a b = 0. (.9) 1.7 b = 0. b = b = 5 Our equation is h =.9t 5t 6. b = 0. a b c = c = c = 6.1 c = c = 6 Remember we have two other techniques for determining the equation of a quadratic function. In one case we are supplied with three points, we create three equations, and then solve the by system. In the other case we are supplied with a verte and a point, and must find the equation in its transformational form. In the end, we have to decide which technique is most appropriate for the information supplied in the question. (b) We need to find the maimum height so we are going to use one of the verte formulas. Since the height is the dependent variable (i.e. the variable on the vertical ais), then we will use the second formula. b ac a 5 (.9)( 6) = (.9) = 19.6 = 68.5 The maimum height reached by the projectile is 68.5 m. (c) We are going to other verte formula to find the time when projectile reaches its maimum height. b a 5 = (.9) =.6 The maimum height is reached in.6 seconds. (d) h ( t) =.9t 5t 6 h( 1.5) =.9( 1.5) 5( 1.5) h( 1.5) = h(1.5) = 7.75 Eplanation: The height of the projectile at t = 1.5 seconds is approimately 7.5 m. NSSAL 1 Draft 009 C. D. Pilmer

140 (e) h =.9t 5t 6 6 =.9t 5t 6 Set the equation equal to zero. 0 =.9t 5t Solve by factoring, specifically common factoring 0 = t (.9t 5) t = 0 or.9. t 5 = 0.9t = 5 t = 7.1 The projectile is at a height of 6 m at t = 0 seconds and t = 7.1 seconds. The significance of the first value is that we know that the initial height of the projectile was 6 m (i.e. height of the raised platform). (f) h =.9t 5t 6 0 =.9t 5t 6 This equation can not be factored so we have to use the quadratic formula or the QUAD program on our graphing calculators. Quadratic Formula: b ± b ac t = a 5 ± 5.9 t = (.9) 5 ± 1.6 t = 9.8 t = - 0. or 7. ( ) ( )( 6) QUAD Program: Using either method we can see that he projectile strikes the ground at t = 7. seconds. (g) The ais of symmetry for a quadratic function is the vertical line that passes through the verte and cuts the parabola in half. In this case the equation of the ais of symmetry is t =.6. (h) We know that the projectile reaches a maimum height of 68.5 metres and strikes the ground at t = 7. seconds. These two pieces of information are critical when determining the domain and range of this function. R 0 t 7. hε R 0 h 68.5 Domain: { tε } Range: { } NSSAL 1 Draft 009 C. D. Pilmer

141 Questions: 1. The cables on a particular suspension bridge are subjected to a uniform load such that they essentially form a parabolic shape that can be modeled using a quadratic function. An engineer has three measurements for the height of the cable relative to the water below. When one is 50 metres to the right of one of the support towers, the cable is 6.5 metres above the water. When one is 100 metres to the right of the same tower, the cable is 50 metres above the water. And finally when one is 150 metres to the right of this tower, the cable height is.5 metres. (a) Determine the equation that describes the height, h, of the cable with respect to the distance, d, from the support tower. (b) How high is the cable above the water when one is 0 metres to the right of the support tower? (c) What is the h-intercept and what does it represent in this situation? (d) What is the minimum height of the cable relative to the water? (e) What is the equation of the ais of symmetry? (f) For this question we will use function notation and epress the equation we found in the first question as h ( d ). If so, determine d when h ( d ) = 60 and eplain what the answer represents in terms of this real world situation. (g) The other support tower is 00 metres away. That means that the height of the cable relative to the water can no longer be modeled with our quadratic function beyond this point. Based on this, determine the domain and range of our quadratic function. NSSAL 15 Draft 009 C. D. Pilmer

142 . Shima is constructing different figures by adding new layers of squares. She thinks that there is a relationship between the number of squares in a figure and its figure number. For eample the second figure has a figure number of and is comprised of 8 squares. Figure 1 Figure Figure Figure Figure 5 (a) Determine the equation of the function that describes the number of squares, s, in a figure in terms if its figure number, n. (b) How many squares are required to construct the tenth figure? (c) Which figure is comprised of 80 squares? (d) If we do not continue beyond the tenth figure, then which two of these three answers for the domain of this function would be correct? nε R 1 n 10 (i) { } (ii) { nε I 1 n 10 } (iii) {1,,,, 5, 6, 7, 8, 9, 10} NSSAL 16 Draft 009 C. D. Pilmer

143 . A small object is dropped into a tank of still water. A ripple forms and moves away in all directions from the point of impact. The area covered by this epanding ripple changes with respect to time. This relationship between the ripple area and time can be modeled using a quadratic function. At t = 0 seconds the ripple area is logically 0 cm. At t = 0.5 seconds the ripple area is 80 cm. At t = 1 seconds, the ripple area is 150 cm. (a) Determine the equation of the quadratic function that describes the ripple area, A, in terms of time, t. (b) If the ripple disappears after seconds, determine the domain and range of our function. (c) For this question we will use function notation and epress the equation we found in the first question as A ( t). If so, determine t when A ( t) = 10 and eplain what it represents in terms of this real world situation. (d) Using the same function as used in the previous question, determine A ( ) and eplain what it represents in terms of this real world situation. (e) What are the coordinates of the verte of this quadratic function? Does this answer make sense in terms of this real world application? Eplain. NSSAL 17 Draft 009 C. D. Pilmer

144 . A parabolic archway is constructed. At its highest point the archway is.5 metres off the ground. At the base the archway is metres wide. (a) If we set a coordinate system over the archway such that the lower left hand corner of the archway is positioned at the origin (see diagram), determine the equation that describes the height, h, of the archway in terms of the horizontal distance, d, from the lower left hand corner. (b) What horizontal distances correspond to an archway height of metres? (c) What height corresponds to a horizontal distance of 0.7 metre? (d) Determine the domain and range. (e) What is the equation of the ais of symmetry? (f) Can a vehicle that is. metres wide and.5 metres high (see diagram) fit through the archway?.5 m (0, 0). m NSSAL 18 Draft 009 C. D. Pilmer

145 Putting It Together Part You have now completed all of the new material for the Quadratic Functions Unit, the longest unit in the Level IV Academic Math program. In this section we will review the concepts covered in this unit. To recap, you have learned to do the following. 1. Interpret real life situations modeled by quadratic functions given their graphs or using graphing technology (pages 1 to 7).. Graph quadratic functions using transformations (pages 1 to 5).. Determine the equation of a quadratic function given: - a table of values (pages 1 to 9) - three points (pages 0 to ) - a verte and a point (pages to 9). Multiply and factor polynomial epressions (pages 5 to 90) 5. Change a quadratic function from its standard form to its transformational form (pages 91 to 98) 6. Solve quadratic equations using factoring, the quadratic formula, and technology (i.e. QUAD program) (pages 99 to 10) 7. Find the verte of a quadratic function to solve a variety of maimum/minimum problems (pages 11 to 11) 8. Multiple Step/Multiple Concept Application Questions (pages 1 to 18) Questions: The questions in this section do not appear in an order that corresponds to the order presented in the unit. In addition, several application word problems require that we use multiple concepts to complete them. For eample in one question with multiple parts we may need to find the equation of function given three points, determine the coordinates of the verte, and solve a quadratic equation using the QUAD program. 1. For each of the functions below, answer the following. Determine the coordinates of the verte. Describe the concavity (upwards or downwards) Describe the shape of the curve compared to that for the function narrower, wider, or no change. y =. Use the terms Function Verte Concavity Shape (a) y = ( ) 6 (b) y = ( ) 7 (c) y = 1 5 (d) ( ) y = 6 NSSAL 19 Draft 009 C. D. Pilmer

146 . (a) Graph ( ) = ( ) 6 (b) Find g ( ). g using transformations. (c) State the domain and range. (d) State the equation of the ais of symmetry.. Kendrick has correctly changed a quadratic function from its standard form to its transformational form. He showed all his work. His classmate, Meera, has provided written eplanations corresponding to each of Kendrick s steps. Match each of Meera s eplanations with the appropriate steps that Kendrick has shown. Kendrick s Work: y = Eplanation y = ( 5 0) 7 y = 5( 6) 7 y = 5( 6 9) 7 ( 5 9) y = 5 ( ) Meera s Eplanations: A Factor the perfect square trinomial and complete the multiplication and subtraction. C Group the first two terms of the trinomial together. B D Common factor from the two terms that you grouped together earlier. Complete the square and compensate by subtracting an equivalent amount from the same side of the equation. NSSAL 10 Draft 009 C. D. Pilmer

147 . Change the following quadratic functions from their standard forms to their transformational forms and state the transformations. (a) y = 6 19 (b) y = 8 95 (c) 1 y = 16 (d) y = Change the function 5( ) 9 y = from its transformational form to its standard form. 6. If you do not have access to graphing technology, is it easier to determine the domain and range of the quadratic function if its equation is in the standard form or transformational form? Eplain. 7. Find the coordinates of the verte for the function y = without changing the function to its transformational form. Is the verte a minimum or maimum? NSSAL 11 Draft 009 C. D. Pilmer

148 8. Pat wants to create three identical and adjacent rectangular pens (as illustrated) using 10 m of fencing. What are the dimensions of the entire enclosure if the area must be a maimum? 9. Triple the first number increased by si times the second number is 7. Find the two numbers if their product is a maimum. 10. Which one of these tables represents a quadratic function? Eplain how you arrived at this answer. (a) y (b) y (c) y (d) y NSSAL 1 Draft 009 C. D. Pilmer

149 11. Factor each of the following. (a) 0 0 (b) p 9 (c) ( ) 7( ) (d) 10 1 (e) h h 11 (f) a 1a 1 (g) 16 d 5 (h) 0 (i) p 16 (j) b ( b 1) 5( b 1) 8 p (k) 9g 8g 6 (l) (m) 50 8 (n) Solve the following without using the quadratic formula or the QUAD program on a graphing calculator. Show all your work. (a) Given 17 7 =, find. NSSAL 1 Draft 009 C. D. Pilmer

150 (b) Given f ( ) = 16 1, find when ( ) = f. 1. Solve the following without using the QUAD program on a graphing calculator. Show all your work. (a) Given 9 7 = 10, find. (b) Given = 0., find. (c) Given h ( ) = 1.1, find when ( ) =. 7 h. NSSAL 1 Draft 009 C. D. Pilmer

151 1. There are three consecutive positive even numbers. The sum of their squares is 596. Find the three numbers. 15. A building measuring 80 metres by 50 metres is going to have its area increased by 1000 m by adding a strip of uniform length to three sides as shown in the illustration. Notice that the strip is along one length and the two widths of the original building. Find the width of the uniform strip. 16. The longer leg of a right angle triangle is 18 cm shorter than the hypotenuse and 7 cm longer than the shorter leg. Find the lengths of the three sides. 17. Determine the equation of the quadratic function that passes through the point (-, -0) and has its verte at (-5, 7) NSSAL 15 Draft 009 C. D. Pilmer

152 For the remaining questions, you are not permitted to use the graphing features on your calculator. However, you are permitted to use the matri features and the QUAD program. 18. The trajectory of a projectile is recorded in a table. The height, h, and the horizontal distance, d, are both measured in metres. d h (a) Determine the equation that describes the height of the projectile in terms of horizontal distance. (b) What is the initial height of the projectile? (c) How far does the projectile travel horizontally when it is at a height of 15 m? (d) What is the maimum height reached by the projectile? (e) How far does the projectile travel horizontally to reach its maimum height? (f) How high is the projectile when it has traveled a horizontal distance of 8 m? (g) How far does the projectile travel horizontally when it strikes the ground? (h) State the domain and range. (i) State the equation of the ais of symmetry. NSSAL 16 Draft 009 C. D. Pilmer

153 19. Travis is constructing steps using cubes. If he only wants one step, he uses two cubes. If he wants two steps, he uses 6 cubes. If he wants steps, he uses 1 cubes. His instructor informs him that the relationship between the number of steps and the number of cubes can be described using a quadratic function. One Step Two Steps Three Steps (a) Determine the equation that epresses the number of cubes, c, in terms of the number of steps, s. (b) What is the c-intercept of this function and what does it represent in this situation? (c) How many cubes are needed to construct seven steps? (d) How many steps can Travis construct using 1 cubes? NSSAL 17 Draft 009 C. D. Pilmer

154 0. Angela s company produces toasters. She knows that there is a specific price per toaster that will maimize the company s monthly profit. If her toasters sell each for $1, then the company achieves their maimum monthly profit of $6 thousand. However, if she increases the price to $15, the monthly profits drop to $ thousand. The price increase reduces the number of sales and in turn reduces the profits. She also knows that if she drops the price, profits will also suffer. She assumes that this relationship between the price per toaster (i.e. unit price) and the monthly profits can be modeled using a quadratic function. (a) Determine the equation of the quadratic function that describes monthly profits, P, in terms of the unit price, u. (b) What will be the monthly profit if the unit price is $1.50? (c) What prices would result in no monthly profits? (d) State the equation of the ais of symmetry. (e) Is this function odd even, or neither? NSSAL 18 Draft 009 C. D. Pilmer

155 Post-Unit Reflections What is the most valuable or important thing you learned in this unit? What part did you find most interesting or enjoyable? What was the most challenging part, and how did you respond to this challenge? How did you feel about this math topic when you started this unit? How do you feel about this math topic now? Of the skills you used in this unit, which is your strongest skill? What skill(s) do you feel you need to improve, and how will you improve them? How does what you learned in this unit fit with your personal goals? NSSAL 19 Draft 009 C. D. Pilmer

156 Additional Practice: Multiplying Polynomials 1. Complete the indicated operation. (a) 6 pq p q (b) 7 s t s t (c) 6cd( 1 c) (d) a b( a b) (e) ( )( ) (f) ( )( 5) (g) ( 9g h)( g h) (h) ( r s)( r s) (i) y ( 6y y 9) (j) y( 6y y ) (k) ( p q) (l) ( ) (m) ( )( 5 1) (n) ( d )( d d 1). Change each of the quadratic functions from their transformational form to their standard form. (a) y = ( 6) 7 (b) y = 1 ( ) NSSAL 150 Draft 009 C. D. Pilmer

157 Additional Practice: Factoring Polynomials 1. Factor each of the following completely. (a) 0 (b) 9d 16 (c) 6a 15a (d) p 1 p 5 (e) 16 6 (f) 6 f ( f 1) 5( f 1) (g) 10 c d 0c d 5c d (h) p pq 1 q (i) y 5y 1y (j) 8n 1n (k) 7( 1 p) p( 1 p) (l) 18d 50d (m) 8y 9y (n) 6g 6g h 1gh NSSAL 151 Draft 009 C. D. Pilmer

158 Additional Practice: Standard Form to Transformational Form Change each of the following from standard form to transformational form. (a) y = 18 5 (b) y = 0 7 (c) y = 1 1 (d) y = (e) 1 y = (f) y = NSSAL 15 Draft 009 C. D. Pilmer

159 Additional Practice: Word Problems Involving Quadratic Equations You may use the QUAD program on the graphing calculator to solve these questions. 1. There are three consecutive positive numbers. The square of the second number, it is 6 more than eleven times the sum of the first and third numbers. Find the three numbers.. There are three consecutive odd numbers. The square of the third number is 1575 less than the square of the sum of the first and second numbers. Find the three numbers.. There are three consecutive even numbers. The sum of the squares of the first two numbers eceeds ten times the third number by 50. Find the three numbers.. A rectangular building measuring 0 m by 6 m is going to have its area increased by 70% by adding a strip of uniform width to two adjoining sides. Determine the width of the uniform strip. NSSAL 15 Draft 009 C. D. Pilmer

160 5. A rectangular building measuring 50 m by m is going to have its area increased by 850 m by adding a strip of uniform width to three of the four sides as illustrated in the diagram. Determine the width of the uniform strip. 6. The hypotenuse of a right angle triangle is 5 cm longer than one of the legs, and 18 cm longer than the other leg. Find the lengths of the three sides of the triangle. 7. If the length of the shorter leg of a right-angle triangle is doubled and then decreased by cm, one obtains the length of the hypotenuse. The length of the shorter leg is cm less than the length of the longer leg. Find the side lengths. NSSAL 15 Draft 009 C. D. Pilmer

161 Review Sheet: Types of Factoring For this course, there are only five types of factoring we need to master. 1. Common Factoring We always try common factoring first. We need to find the greatest common factor of all of the terms in the polynomial. (a) 6 8 (b) 10 = ( ) = 5 1 ( ). Factoring Difference of Squares This type of factoring is used when we are dealing with a binomial where two perfect squares are separated by the operation of subtraction. (a) 6 (b) 6 5 = 8 8 = ( )( ) ( )( ). Inspection This type of factoring is used when we have a trinomial of the form equal to 1 (e.g ). a b c where a is (a) 0 (b) 1 (-5) (8) = -0 (-5) (8) = ( 5 )( 8) = (-1) (-) = (-1) (-) = -1 ( 1)( ) =. Decomposition This technique is used to factor trinomials of the form to 1. y = a b c, where a is not equal (a) 1 0 (b) (5) (8) = 0 (5) (8) = 1 (-1) (-) = 8 (-1) (-) = -5 = = = ( 5) ( 5) = 5 ( ) ( ) ( )( ) = = = ( 7) ( 7) = 7 ( ) ( ) ( )( ) NSSAL 155 Draft 009 C. D. Pilmer

162 5. Factoring Perfect Square Trinomials Perfect square trinomials have three specific traits. We can take the square root of the first term. We can take the square root of the third term. When we multiply the square root of the first term by the square root of the second term, then double that product, we will obtain the middle term (ignore the sign of the term). (a) (b) 9d 56d d 56d 16 Square Root Take the sign. Square Root Square Root Take the sign. Square Root 9 Therefore: = ( ) 9 7 d - Therefore: 9d 56d 16 = 7d ( ) Note: Do not forget that some questions require that you use more than one type of factoring. NSSAL 156 Draft 009 C. D. Pilmer

163 Factoring Flow Chart Always try to common factor first. Look for a term that divides into all of the terms in the epression. 6 = 1 e.g. ( ) If you have two perfect squares separated by a subtraction sign, then you are dealing with a difference of squares question. 9 = 7 7 e.g. ( )( ) Trinomial (Three Terms) If the leading numerical coefficient is 1, try factoring by inspection. 10 = 5 e.g. ( )( ) Leading numerical coefficient is not equal to 1. If the first and last terms are perfect squares, determine if you are dealing with a perfect square trinomial. e.g. 9 1 = ( ) If it is not, then you will have to factor using decomposition e.g (-1) (-) = 8 (-1) (-) = -5 = = = ( 7) ( 7) = 7 ( ) ( ) ( )( ) NSSAL 157 Draft 009 C. D. Pilmer

164 Terminology In this unit, you were eposed to the following terminology. These terms are presented in the order they appear in this resource. Quadratic Function Parabola Verte Ais of Symmetry Transformational Form Finite Differences D Level Standard Form Polynomial Monomial Binomial Trinomial Distributive Property Factoring Common Factoring Factoring Differences of Squares Factoring by Inspection Factoring by Decomposition Factoring Perfect Square Trinomials Completing the Square Zero-Factor Property Roots b ± b ac Quadratic Formula = a b b ac Verte Formulas, a a NSSAL 158 Draft 009 C. D. Pilmer

165 Answers Introduction to Quadratic Functions (pages 1 to ) 1. (a) 8. m (b).7 m (c) 5 m (d) m (e) 1 m (f) 0.7 m and 7. m R 0 d 8. hε R 0 h 5 (g) { dε } (h) { }. (a) 8000 alternators (b) $ (c) 000 or alternators (d) $ nε R 0 n 16 (e) 000 or alternators (f) { } (g) { pε R -1 p 18 }. (a) 05 ft (b) 167 ft (c) 0 ft and 90 ft (d) 0 R 0 h 05 dε R -167 d 0 (e) { hε } (f) { } (g) neither Using a Graphing Calculator to Interpret Quadratic Functions (pages to 7) 1. (a) (b) i (d) iv (f) iv (h) ii (j) iv (l) iii (c) iii (e) ii (g) iii (i) iv (k) i. (a) The d-intercept equals 0. It represents the stopping distance if the car was initially traveling at 0 km/h. (b) 55 km/h (c) 9 m (d) 10 m. (a) 70 km/h (b) 1 km/h and 119 km/h (c) 7.6 cents per kilometer NSSAL 159 Draft 009 C. D. Pilmer

166 The Most Basic Quadratic Function (page 8) 1. Domain: { R } ε. Range: { yε R y 0 }. Verte: (0, 0). Ais of Symmetry: = 0 5. Even 6. It is a function because for each member of the first set, there is only one corresponding value in the second set. Quadratic Functions and Transformations (pages 9 to 1) Our Basic Quadratic Function y = Enter: Y1=X^ Table of Values TblStart = - Tbl = 1 y Sketch of Graph New Quadratic Table of Function Values (a) y = TblStart = - Tbl = 1 y Sketch of Graph Mapping Rule and Transformation (, y) (, y) We re dealing with a reflection in the -ais because the parabola is now upside down (i.e. concave downwards). NSSAL 160 Draft 009 C. D. Pilmer

167 New Quadratic Table of Function Values (b) y = TblStart = - Tbl = 1 y Sketch of Graph Mapping Rule and Transformation (, y) (, y) We re dealing with a vertical stretch of because the parabola appears narrower. (c) 1 y = TblStart = - Tbl = 1 y (, y), y 1 We re dealing with a vertical stretch of 1 because the parabola appears wider. (d) y = TblStart = - Tbl = 1 y (, y) (, y ) We re dealing with a vertical translation of because the parabola appears to have been slid up four units. NSSAL 161 Draft 009 C. D. Pilmer

168 New Quadratic Table of Function Values (e) y = 5 TblStart = - Tbl = 1 y (f) y = ( ) TblStart = -6 Tbl = 1 y (g) y = ( ) TblStart = 1 Tbl = 1 y Sketch of Graph Mapping Rule and Transformation (, y) (, y 5) We re dealing with a vertical translation of -5 because the parabola appears to have been slid down five units. (, y) (, y) We re dealing with a horizontal translation of - because the parabola appears to have been slid to the left three units. (, y) (, y) We re dealing with a horizontal translation of because the parabola appears to have been slid to the right four units. NSSAL 16 Draft 009 C. D. Pilmer

169 Summarize Your Findings: Equation Transformation (a) y = reflection in the -ais (b) y = vertical stretch of (c) 1 1 y = vertical stretch of (d) y = vertical translation of (e) y = 5 vertical translation of -5 (f) y = ( ) horizontal translation of - (g) y = ( ) horizontal translation of Conclusions: If a quadratic function is of the form y = k( c) d, then: (i) the negative sign in front of the k indicates that a reflection in the -ais has occurred. (ii) the k indicates that a vertical stretch of k has occurred. (iii) the d indicates that a vertical translation of d has occurred. (iv) the c indicates that a horizontal translation of c has occurred. 1. horizontal translation. vertical translation. reflection in the -ais NSSAL 16 Draft 009 C. D. Pilmer

170 State the Transformations: Quadratic Functions (page 1) 1. Function Horizontal Translation Vertical Translation Reflection in the -ais Vertical Stretch (a) y = ( 7) no none (b) y = 1 none 1 yes 1 y - none yes 1 g ( ) = no (c) = ( ) (d) ( ) 5 (e) ( ) y = yes none (f) 5( ) 1 y = - 1 yes 5 (g) ( ) = 6( ) 1 h -1 no 6 (h) f ( ) = 8 none 8 yes. (a) A reflection in the -ais will change a parabola from right side up (i.e. concave upwards) to upside down (i.e. concave downwards). (b) The -coordinate of the verte of a quadratic function can be determined by looking at the horizontal translation. (c) The y-coordinate of the verte of a quadratic function can be determined by looking at the vertical translation. (d) The vertical stretch determines whether the graph is wider or narrower than the graph of y =. 1 1 Visualizing Quadratic Functions (pages 15 to 17) 1. (a) Graph (iii) (b) Graph (i) (c) Graph (iv) (d) Graph (i) (e) Graph (ii) (f) Graph (i) (g) Graph (iv) (h) Graph (i) (i) Graph (iv) (j) Graph (iii) NSSAL 16 Draft 009 C. D. Pilmer

171 . Function Verte Concavity Shape (a) y = ( ) 6 (, -6) upwards no change y 1 (-, -9) downwards wider (b) = ( ) 9 (c) y = 8 (0, 8) downwards narrower g 10 (, -5) upwards wider (d) ( ) = ( ) 5 Graphing Quadratic Functions Using Transformations (pages 18 to 5) 1. (a) HT = 7, reflection in -ais, VS = 5, VT = 1, (, y) ( 7, 5y 1) (b) HT = -8, VS =, VT = -5, (, y) ( 8,y 5) 1 5, y 1 (c) HT = 5, reflection in -ais, VS =, ( ) 1, y, y 1 5, y 1 (d) VS =, VT = 1, ( ) 5. (, y) (, y 9) y concave upwards verte (-, -9) narrower vertical stretch of. 1 (, y), y y concave downwards verte (, ) wider 1 vertical stretch of NSSAL 165 Draft 009 C. D. Pilmer

172 . (a) HT = -, Reflection in -ais, VT = 5 (, y) (, y 5) y (b) Reflection in -ais, VS =, VT = 10 (, y) (, y 10) y (c) HT = -, VS = 1, VT = -1 (, y), y y Verte Concavity Narrower/Wider/Same (a) y = ( 7) 1 (7, 1) upwards same y = (-5, -10) upwards narrower (b) ( 5) 10 (c) ( ) y = 1 (1, 0) downwards narrower (d) y = 11 (0, 11) downwards same (e) y = 1 ( 8) 6 (-8, 6) upwards wider (f) g ( ) = ( ) 9 (, 0) downwards wider 7 (g) h ( ) = 8 (0, 8) upwards narrower NSSAL 166 Draft 009 C. D. Pilmer

173 6. 1 HT = 6, reflection in -ais, VS =, VT = 1 (, y) 6, - y ε R Domain: { } Range: { yε R y } Verte: (6, ) Ais of Symmetry: = 6 y (, y), - y 5 1 (b) Ma Height: 5 m (c) Distance: m (d) Initial Height: 0.5 m (e) d = y Using Finite Differences to Identify Quadratic Functions (pages 6 to 0) 1. y -1 9 D NSSAL 167 Draft 009 C. D. Pilmer

174 . y - -1 D D (a) linear (b) quadratic (c) quadratic (d) neither (e) linear (f) quadratic (g) neither (h) quadratic Using Finite Differences to Determine the Equation (pages 1 to 9) 1. y = 5 1. y = 8. y = 7 1. (a) g = 1 t t (b) 6 games (c) 1 teams (Xmin = 1, Xma = 15, Ymin = 0, Yma = 100) 5. (a) h = 5t 0t 1 (b) maimum height: 81 m (Xmin = 0, Xma = 10, Ymin = 0, Yma = 100) time when projectile reaches maimum height: s initial height: 1 m time when projectile strikes ground: 8 s (approimately) R 0 t 8 hε R 0 h 81 (c) Domain: { tε } Range: { } 6. (a) h = 0.00d 0. 6d (b) 7.8 m (c) maimum height: 5 m (Xmin = 0, Xma = 00, Ymin = 0, Yma = 100) horizontal distance traveled when the ball strikes the moon s surface: 00 m R 0 d 00 hε R 0 h 5 (d) Domain: { dε } Range: { } NSSAL 168 Draft 009 C. D. Pilmer

175 Find the Equation Given Three Points (pages 0 to ) 1. (a) y = 6 (b) y = 7 (c) y = 6 (d) y =.5 8. (a) l = 0.7s 1.s 19 (b) 1091 pounds R 0 s 0 (c) Domain: { sε } Range: { lε R 0 l 1091}. (a) p = 1.5s 50s 600 (b) $056 Find the Equation Given the Verte and a Point (pages to 9) 1. (a) y = ( 5) (b) y = ( ) 1 1 (c) y = ( ) (d) y = ( 1) 5 5 y = (e) ( ) H = D 9 (b) m or appro..7 m 9 (c) appro. 1.8 m. (a) ( 7) 5 (d) Domain: { Dε R 0 D 1.8} Range: { Hε R 0 H 5} 15 (b) appro. 6.5 m (c) appro. 55 m and 575 m R 0 D 60. (a) H = ( D 15) 60 (d) Domain: { Dε } Range: { Hε R 0 H 60}. (a) H = D 1105 (b) appro. 0.9 m NSSAL 169 Draft 009 C. D. Pilmer

176 (c) Domain: { Dε R 105 D 105} Range: { Hε R 0 H } Putting It Together, Part 1 (pages 50 to 5) 1. Transformations Verte Concavity Shape (a) Reflection in -ais, VS =, HT = 5, VT = 1 (5, 1) downwards narrower (b) VS =, HT = -6 VT = 9 (-6, 9) upwards narrower (c) Reflection in -ais, VS = 7 5, HT = 8 (8, 0) downwards wider (d) VS = 9 1, VT = - (0, -) upwards wider. Answers will vary. Here are a few acceptable answers. 1 y = ( ) 7 y =.5( ) 7 y = ( ) 7. (a) Mapping Rule: (, y) ( 1, y 18) y (b) Domain: { R} Ais of Symmetry: = 1 (c) Neither R 0 ε Range: { R y 18} yε Verte: (1, 18) (d) Domain: { ε } Range: { yε R 0 y 18}. When using finite differences: Linear functions have a common difference at the D1 level. Quadratic functions have a common difference at the D level. 5. (a) r = 100n 100n (b) $19 00 NSSAL 170 Draft 009 C. D. Pilmer

177 (c) 6 times, maimum revenue: $ h = 1 00 (b) edge of a cliff, reason: h-intercept is 5.5 (c) 8 m R 0 d 70 hε R 6. (a) ( 0) 10 (d) Domain: { dε } Range: { d 10} 7. (a) n = 0.5o 0. 5o (b) 10 Multiplying Polynomials (pages 5 to 65) 1. (a) (c) (e) (g) (i) y g (b) (d) (f) a 5 b 5 (h) 7 8 1s t (j) NSSAL 171 Draft 009 C. D. Pilmer 10 m 1p 10t 6 p q d e f. (a) 5 0 (b) 1y 1 (c) (d) (e) 10 p 0 p (f) 8c c (g) 8 1y (h) 6cd 18c d (i) a b 9a b (j) 1 y 8y. (a) 8 1 (b) 0 (c) p 5p 6 (d) 9 1 (e) y 10y 1 (f) 5 (g) 10 (h) 9 8h h (i) 9 (j) 10 8 (k) 1 10 (l) 5a 1a (m) 6d 1d 5 (n) 1 (o) (p) (q) y 1y (r) a 7ab b (s) 9 y (t) 6d 5de e (u) 9g 19gh h (v) r 11rs s. (a) (b) d 5d d (c) 1 (d) 8g 1g 5g

178 (e) (g) (i) (k) 8p p 1 p (f) 18y y 7y 6 y 1 y y (h) 5a b 10a b 0ab 5 5p q 7 p q pq (j) 8g 1g 0g y 0 y 6 y (l) n 1m n m n (m) 1 9 (n) n 18n 81 (o) (p) 9 p 0 p 5 (q) y 0y 5 (r) m 16 p 8pq q (s) 7 11 (t) d d 16d (u) 6y 7y y (v) 1h 5h 17h (a) y = 0 7 (b) y = 6 1 (c) 7 y = (d) y = 6. (a) 1 5 (b) 16 p p 9 (c) 1y 8y y (d) 11d 5d (e) 11 1 (f) 1y y (g) d 6d 81 (h) k 9 (i) 8e e 17e (j) 1a b a b 6ab (k) 9c 0cd 5d (l) 5 r s 0r s (m) 15 11y y 7 5 (n) 7g h 1g h 5g h 7. The same graph is produced. They are all equivalent. Challenge Yourself (a) (b) Factoring Part 1 (pages 66 to 7) 1. (a) ( 5) (b) 5n ( n 1) (c) ( 8) (d) ( ) (e) 7 y ( y 1) (f) d ( d ) (g) p ( p p 1) (h) 6y ( ) (i) ( 7 )( 5 6) (j) ( h )( 7h ) (k) ( 8 d )( d 6) (l) ( 1 g )( 5g 1). (a) ( 8)( 8) (b) ( 7 g 5)( 7g 5) (c) ( 1)( 1) (d) ( 5 )( 5 ) NSSAL 17 Draft 009 C. D. Pilmer

179 (e) ( 9 d )( 9 d ) (f) ( 1 6t)( 1 6t) (g) ( 7y)( 7y) (h) ( a 8b)( a 8b). (a) ( )( 6) (b) ( k )( k 8) (c) ( t )( t 7) (d) ( 5 )( ) (e) ( p 8)( p ) (f) ( y)( 9y) (g) ( r 5s)( r 7s) (h) ( a 9 b)( a b). (a) ( 6)( 7) (b) ( 5 p )( 5p ) (c) 9 n ( n ) (d) ( t 10)( t ) (e) 7a ( 6a 1) (f) ( g 1)( g 1) (g) ( g 6 h)( g 5h) (h) 5d ( d 1) (i) ( 9 5h)( 9 5h) (j) ( 8)( 6 5) (k) ( p 8q)( p 8q) (l) ( 11y)( y) (m) p q ( p 7q) (n) ( a 6 b)( a b) (o) ( 10v )( 5v ) (p) ( 5 s 7t)( 5s 7t) (q) ( y)( y) (r) ( c 5)( c 1) Factoring Part (pages 7 to 79) 1. (a) 6 11 (8) () = (8)() = 11 = 6 8 = = ( 8) ( ) 6 ( ) 1( ) ( )( 1) = (b) (-10) () = -0 (-10)() = -6 = = = ( 5 10) ( 8) 5( ) ( ) ( )( 5 ) = Step 1: Multiply 6 and. Step : Find two numbers whose product is and whose sum is 11. Step : Split the middle term, 11, apart. Step : Group first two terms and last two terms together. Common factor from each group. Step 5: Common factor out the binomial. Step 1: Multiply 5 and -8. Step : Find two numbers whose product is -0 and whose sum is -6. Step : Split the middle term, -6, apart. Step : Group first two terms and last two terms together. Common factor from each group. Step 5: Common factor out the binomial -. NSSAL 17 Draft 009 C. D. Pilmer

180 (c) 1d 17d 6 (-9) (-8) = 7 (-9)(-8) = -17 = 1d 9d 8d 6 = ( 1d 9d ) ( 8d 6) = d = ( d ) ( d ) ( d )( d ) Step 1: Multiply 1 and 6. Step : Find two numbers whose product is 7 and whose sum is -17. Step : Split the middle term, -17d, apart. Step : Group first two terms and last two terms together. Common factor from each group. Step 5: Common factor out the binomial d -.. (a) ( )( 5) (b) ( 5)( ) (c) ( )( 5) (d) ( p 1)( p 7) (e) ( c 1)( c ) (f) ( 7)( ) (g) ( a )( a ) (h) ( y )( y 5) (i) ( 6 5)( ) (j) ( d 5)( d ) (k) ( 5 )( ) (l) ( p )( p 1). (a) ( 9 5)( 9 5) (b) ( 8)( 5) (c) ( 9)( ) (d) 5c ( c 7) (e) ( 7)( 6) (f) ( 1 d )( d 7) (g) ( 1)( ) (h) ( 6 11y)( 6 11y) (i) ( a b)( a 6b) (j) ( 5p )( p 1) (k) ( 7d )( 7d ) (l) p ( p 5q 1) Alternate Forms of Decomposition (Optional) (pages 80 to 8) 1. (a) ( 5 )( ) (b) ( 7 )( ) (c) ( a 1)( a 5) (d) ( p 1)( p ) (e) ( y 5)( y ) (f) ( )( ) (g) ( d )( d 5) (h) ( 6h 5)( h ) Factoring Part (pages 85 to 90) 1. (a), (d), (f), and (g). (a) ( 1) (b) ( t 10) (c) ( g ) (d) ( 5 ) or ( 5) (e) ( k 1) (f) ( 7 p 5) NSSAL 17 Draft 009 C. D. Pilmer

181 (g) ( d 9) (h) ( 7a 8b) (a) 0; (c) ; (e) 88; ( 0) (b) 60; ( 11 p ) (d) 180; ( 11y) (f) 8;. (a) 6 (b) 5 (c) 16 (d) 100 (e) 169 (f) 900 (g) 11 (h) 5 ( h 0) ( 9y 10) ( 8a b) 5. (a) ( 5)( 5) (b) ( )( 7) (c) ( y ) (d) t ( t )( t 5) (e) 6 ( 1) (f) 5 p ( p )( p ) (g) ( ) (h) ( 6)( ) (i) ( )( ) (j) c ( c 1) (k) a ( b 5)( b 5) (l) p ( q 1 )(q 6. (a) y( y ) Common (b) ( 5 p 7)( 5 p 7) Difference of Squares (c) ( 7)( ) Inspection (d) ( d 7)( d ) Common (e) ( 5) Perfect Square Trinomial (or Inspection) (f) ( a )( a 1) Decomposition (g) 5( )( 7) Common and Inspection (h) ( h)( h) Difference of Squares (i) ( ) Common and Perfect Square Trinomial (or Common and Decomposition) y 8y Inspection p p 1 Common 5 1 Decomposition c c 5 c 5 Common and Difference of Squares Common and Decomposition (j) ( )( ) (k) ( )( ) (l) ( )( ) (m) ( )( ) (n) ( )( ) (o) ( d 7) Perfect Square Trinomial (or Decomposition) (p) 7 ( 5 pq p q 1) pq Common 7 5 Common and Inspection 7 c c Common and Perfect Square Trinomial (or Common and Inspection) (q) ( )( ) (r) ( ) NSSAL 175 Draft 009 C. D. Pilmer

182 NSSAL 176 Draft 009 C. D. Pilmer Standard Form to Transformational Form (pages 91 to 98) 1. (a) ( ) ( ) ( ) ( ) ( ) ( ) = = = = = = y y y y y y VS =, VT = -5, HT = - (b) ( ) ( ) ( ) ( ) ( ) ( ) = = = = = = y y y y y y Reflection in -ais, VS =, VT = 10, HT = 1 (c) ( ) ( ) ( ) ( ) = = = = = = y y y y y y VS = 5, VT = -, HT = - The graph is concave upwards since there is no reflection in the -ais. The coordinates of the verte should be (-, -). The graph looks narrower than the graph of y = due to the vertical stretch.

183 . (a) y = ( 6 9) 1 ( 9) (b) y = 7( 1) 5 ( 7 1) y = ( ) 5 y = 7( 1) (c) y = ( 10 5) 1 ( 1 5) (d) y = 6( ) ( 6 ) y = ( 5) 6 y = 6( ) 9 (e) y = ( ) 1 (f) y = ( ) (g) y = ( 1) 7 ( 1) (h) y = 5( ) 1 ( 5 ) y = ( 1) y = 5( ) 6 (i) y = ( 7) 1 (j) y = ( 5) (k) y = ( 1 6) 1 6 (l) y = ( 8 16) 7 16 = 1 6 = 1 y = 1 y = 8 y ( ) ( ) 1 (m) y ( ) 5 (n) ( ). y = ( 5) 9 mapping rule: (, y) ( 5, y 9) y Domain: { R} ε Range: { R y 9} yε Verte: (-5, 9) Ais of Symmetry: = -5 1 Solving Quadratic Equations by Factoring (pages 99 to 10) 1. (a) 8 = 0, = 6 or = 8 (b) ( )( ) = 0, = 0 or = 0 9 (c) 5 6 = 0, = 0 or = 5 11 (d) 11 = 0, 11 = 0, =. (a) or 6 (b) 0 or 7 NSSAL 177 Draft 009 C. D. Pilmer

184 (c) 1 1 or (d) (e) 6 or 5 (f) (g) 0 or (h) 5 or (i) or (j) or (k) (l) or 1 (m) 1 or (n) 5 or or. 8 or 9 5. (a) 6 metres (b).5 seconds 6. (a) 1.76 feet (b) 7 feet (c) 0.5 seconds and 1 second (d) 1.75 seconds R 0 t 1.75 hε R 0 h 16 (e) Domain: { tε } Range: { } 7. Point P: ( 5, ) Point Q: two answers: (,1) or (,1) 8. (a) 5 (b) 0 or 5 (c) 6 Solving Quadratic Equations Using the Quadratic Formula (pages 105 to 111) 1. (a) or -.6 (b) 0.71 or -.1 (c) no real roots (d) or.1 (e) or (f) or 0.67 (g) (h) no real roots. -5 or.5. (a) 1 (b) or (a) 1 (b) or -.55 NSSAL 178 Draft 009 C. D. Pilmer

185 5. (a) km/h (b) 9.1 m 6. (d) either a, b, or c Word Problems Involving Quadratic Equations (pages 11 to 10) 1. ( ) = 1060, answers: and. ( ) = 916, answers: 6 and 8. ( ) ( ) = 71, answers: 9, 11 and 1. ( 1) ( ) = 590, answers: 1, 1 and 15 or -1, -1 and ( 1) = ( ) 896, answers: 1,, and 6. ( ) ( ) 8 =, answers: 1, 1, and ( ) ( ) 9 =, answers: 19, 1 and 8. ( 1) = 118, answers: 1 and 1 9. ( ) = 5, answers: 18 and ( 50 )( 0 ) = ( 60 )( 0 ) = ( 70 )( 50 ) = ( 80 )( 60 ) = ( ) ( ) 9 = 15. ( ) ( ) 17 = ( ) ( ) = 17. ( ) ( ) 1 = 1 ; answer: 7.8 metres ; answer:.9 metres ; answer: 6.5 metres ; answer: 11. metres ; answers: 8 cm, 15 cm, and 17 cm ; answers: 7 cm, cm and 5 cm ; answers: 5 cm, 1 cm and 1 cm ; answers: 1 cm, 5 cm and 7 cm NSSAL 179 Draft 009 C. D. Pilmer

186 18. 5 =.9t t ; answers: 0.1 seconds and.56 seconds 19. ( ) ( ) = 85 ; answers:, 5 and 7 0. ( 70 )( 0 ) = ( ) ( ) = 1 ; answers: 11. metres ; answer: cm, 56 cm and 65 cm Finding the Verte (pages 11 to 11) 1. (a) (1,-6), minimum (b) (-, 5), maimum. (a) 17.6 feet (b) 18 feet (c) 1 second. Hint: y = 8 and P = y Answer: -1 and -1. Hint: l w = 1 and A = l w Answer: 6 m by m 5. Hint: y = 16 and s = y Answer: 6. and. 6. (a) 5 metres (b) 7.5 m and 7.5 m 7. Hint: l 6w = 90 and A = l w Answer:.5 m and 7.5 m 8. Hint: y = 60 and P = y Answer: 10 and Hint: y = 1 and s ( y) = Answer: and (a) 19.5 m (b) 0 m (c) 18 m (d) 1.7 m and 65. m 11. Hint: y ( y) = 10 and A = ( y) ( ) (a) 76. m (b) 8.18 m and 7.8 m NSSAL 180 Draft 009 C. D. Pilmer

187 Application Questions (pages 1 to 18) 1. (a) Hint: We have been given the three points (50, 6.5), (100, 50), (150,.5). Generate three equations using the three points and then complete the by elimination using the inverse matri method on the graphing calculator. Answer: h = 0.001d 0.d 80 (b) 5. metres (c) h-intercept: 80 metres: It represents the initial height of the cable relative to the water or the height at which the cable is attached to the support tower. b ac (d) Hint: a Answer: 0 metres (e) Hint: The ais of symmetry cuts the parabola in half and passes through the verte. We b need to use the formula to find the first coordinate of the verte. a Answer: d = 00 (f) Hints: 60 = 0.001d 0.d 80 Set the equation equal to zero and solve using the quadratic formula or the QUAD program on you graphing calculator Answers: 58.6 m and 1. m R 0 d 00 hε R 0 h 80 (g) Domain: { dε }, Range: { }. (a) Hint: We can generate a table of values based on the diagrams and then use finite differences to find the equation of the quadratic function. n s Answer: s = n n (b) 10 squares (c) figure number 8 (d) Domains (ii) and (iii) are correct.. (a) A = 150t (b) Domain: { tε R 0 t }, Range: { Aε R 0 A 1680 } (c) The ripple area reaches 10 cm at 0.9 seconds. (Ignore the 0. 9; we can not have negative time) (d) At seconds the ripple area reaches 6080 cm. (e) The verte is at (0, 0). This makes sense because the minimum area of 0 cm is going to occur when the time is equal to 0 seconds (i.e. time of impact). NSSAL 181 Draft 009 C. D. Pilmer

188 . (a) Hint: This question can be done two ways and you end up with two equivalent equations. Method 1 (verte and one point): h = ( d 1.5). 5 Method (three points): h = d 6d (b) Hint: = d 6d, set equal to zero and solve by factoring, using the quadratic formula, or using the QUAD program on the graphing calculator. Answer: 1 metre and metres (c).5 m R 0 d hε R 0 h.5 (d) Domain: { dε }, Range: { } (e) d = 1. 5 (f) Hint: If the vehicle passes through the center of the archway, its outer edges would go 1.1 metres to the left and right of the center. Another way this could be stated is that the outer edges of the vehicle are 0. metres (1.5 m 1.1 m) from the outer edges of the archway. Substitute 0. into the equation of the function and find the corresponding height. If it is greater than.5, the vehicle can pass through. If it is less than.5, then the vehicle hits the archway. Answer: The vehicle cannot pass through the archway. Putting It Together, Part (pages 19 to 18) 1. Function Verte Concavity Shape (a) = ( ) 6 y (-, -6) downwards no change (b) ( ) 7 y = (, 7) upwards Narrower (c) y = 1 5 (0, -5) downwards wider (d) ( ) y = 6 (-6, 0) upwards narrower. (a) (, y) (, y 6) y (b) g ( ) = 9 (c) Domain: { R} (d) = ε Range: { yε R y 6} NSSAL 18 Draft 009 C. D. Pilmer

189 . C, B, D, A. (a) = 6( ) 5 (b) = ( 7) y ; VS = 6, VT = -5, HT = - y ; Reflection in the -ais, VS =, VT =, HT = 7 y = ; VS =, VT = 7, HT = -6 y = 5 ; Reflection in the -ais, VT = -1, HT = 5 (c) ( ) 7 (d) ( ) 1 5. y = Transformational Form (Eplanation not supplied.) 7. (6, -10), maimum 8. 5 m and 17.5 m 9. 1 and Table (d); (Eplanation: As the -values change by the same increment, there is a common difference at the D level for the y-values) 11. (a) 10 ( ) (b) ( p 7)( p 7) (c) ( )( 7) (d) ( )( 7) (e) ( h 11) (f) ( a 7)( a ) (g) ( d 5)( d 5) (h) ( 8 )( 5) (i) 8p( 1 p) (j) ( b 1)( b 5) (k) ( g 8) (l) ( 7)( 1) (m) ( 5 )( 5 ) (n) ( 1) 1 1. (a) or (b) 5 NSSAL 18 Draft 009 C. D. Pilmer 1. (a) 0.6 or (b) no real solutions (c) or The three positive even numbers are 1, 1, and The width of the uniform strip is 5.5 m cm, 55 cm, and 7 cm 17. y = ( 5) 7

190 18. (a) h = 0.05d d 1 (b) 1 m (c) 9.05 m and 0.95 m (d) 1m (e) 0 m (f).8 m (g) 0.9 m R 0 d 0.9 hε R 0 h 1 (h) Domain: { dε } Range: { } (i) d = (a) c = s s (b) c-intercept = 0 (Eplanation: If Travis does not use any cubes, he ends up with 0 steps.) (c) 56 cubes (d) 11 steps 0. (a) ( 1) 6 P = u (b) $5 thousand (c) $11 and $17 (d) u = 1 (e) neither Additional Practice: Multiplying Polynomials (pages 150) 1. (a) p q (b) 1s t (c) 6cd 18c d (d) a b 9a b (e) 10 8 (f) 1 10 (g) 9g 19gh h (h) r 11rs s (i) 18y y 7y (j) 6 y 1 y y (k) 16 p 8pq q (l) 9 1 (m) 7 11 (n) d d 7d. (a) y = 79 (b) 1 y = 11 Additional Practice: Factoring Polynomials (page 151) 1. (a) ( 5 )( 8) (b) ( 7 d )( 7d ) (c) a ( a 5) (d) ( p 1)( p 5) (e) ( 8) (f) ( 6 f 5)( f 1) (g) 5c d( c d 1) (h) ( p q)( p 8q) (i) y ( )( 1) (j) ( n 1)( n ) (k) ( 7 p)( 1 p) (l) d( 5d )( 5d ) NSSAL 18 Draft 009 C. D. Pilmer

191 (m) ( 7y) (n) 6 g( g h)( g h) Additional Practice: Standard Form to Transformational Form (page 15) (a) y = ( 6 9) 5 ( 9) (b) y = ( 10 5) 7 ( 5) y = ( ) y = ( 5) 57 (c) y = ( 1 9) 1 ( 1 9) (d) y = 6( 8 16) 10 ( 6 16) y = ( 7) 50 y = 6( ) = y = 8 (e) y = ( 1 6) 6 (f) y = ( 16 6) 6 y ( ) 10 ( ) 100 Additional Practice: Word Problems Involving Quadratic Equations (pages 15 to 15) 1. ( ) 11( ) 6 1 = Answers:,, and 1575 =. ( ) ( ) Answers:, 5, and 7. ( ) 10( ) 50 = Answers: 18, 0, and. ( 0 )( 6 ) = 8 Answer: 11.5 metres 5. ( 50 )( ) = 950 Answer: 5.8 metres 6. ( ) ( ) 18 5 = Answers: 7 cm, 55 cm, and 8 cm. 7. ( ) ( ) = Answers: 0 cm, cm, and 58 cm NSSAL 185 Draft 009 C. D. Pilmer