Study Tip: Sing Songs

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1 Study Tip: Sing Songs It s corny but it really works! Take facts or ideas you need to learn. Transform them into a goofy poem, song or rap. It s: Easier to learn Easier to recall Not so boring Best for learning facts PAGE 38

2 GRAPHING GRAPHING INVESTIGATE TABLES, EQUATIONS & GRAPHS 4 CREDITS (908) THE SKILLS YOU NEED TO KNOW: LINEAR FEATURES p 4 There are two features for any linear graph that are important:. The y-intercept This is the value when the graph starts and is the initial condition.. The gradient This is the slope of the graph and gives the speed/rate etc. of what is being graphed. GRAPHING LINEAR EQUATIONS p 47 Method ( y = mx+ c). Plot the y-intercept. Then plot the points according to the gradient Method - Cover-up method. To find the x-intercept make y = 0 and solve.. To find the y-intercept make x = 0 and solve. 3. Plot these two points and draw a line through them. LINEAR SEQUENCES p 45. Draw a 5 columned table. The x column is always,, 3, 4,... or the identifier of your sequence. 3. The y column is the terms of your sequence 4. The m column is the difference between the y values (they should all be the same) 5. The mx column is the difference multiplied by the x value. 6. The final column c is the difference between the mx and y values. It should always be the same number. 7. The final step is to put the m and c values into the equation; y = mx+ c WRITING LINEAR EQUATIONS p 5. The general equation is: y = mx+ c. c is the y-intercept 3. m is the gradient = rise run INTERPRETING LINEAR GRAPHS p 55 Drawing a Graph Interpreting a Graph. Plotting the points. y-intercept. Discrete or continuous. Gradient 3. Other interesting features State the reasoning behind any answer giving actual values from the graph such as the gradient(s), starting point(s), or value(s) at a specific point. Note: Problems may rely on knowledge from earlier graphing sections PAGE 39

3 QUADRATIC SEQUENCES p 63. Draw a 9 columned table. The x column is always,, 3, 4,... or the identifier of your sequence. 3. The y column is the terms of your sequence 4. The d column is the difference between the y values 5. The d column is the difference between the d values (they should all be the same). a is half d. 6. The ax column is half the second difference multiplied by the x value squared. 7. The L column is the difference between the y and ax values. This gives us a linear sequence. 8. The final three columns are solving a linear sequence. 9. The final step is to put the a, m and c values into the equation y = ax + mx+ c TRANSLATING PARABOLAS p 77. To translate an equation by (c,d) simply replace x with (x - c) and y with (y - d) and simplify. SCALING PARABOLAS p 80 Parabolas can be made taller or shorter by putting a number in front of the equation. Steps:. Write equation using either vertex or x-intercept method. Find a point not used to write the equation and substitute it into the equation with k 3. Rearrange to find the value of k 4. Write out the final equation GRAPHING PARABOLAS: X-INTERCEPTS p 65. The general equation is: y = ( x+ a)( x+ b). a and b are the two x-intercepts. They are the negative of the numbers in the equation. The vertex of the parabola is half way between the x-intercepts. 3. Substitute the x value of the vertex into the original equation to find the v value of the vertex. Note: A negative sign at the front of the equation means the parabola is upside down. GRAPHING PARABOLAS: VERTEX p 7. The general equation is y = ( x c) + d. c is how far the vertex has been horizontally shifted from the origin. 3. d is how far the vertex has been vertically shifted from the origin. 4. Use this form if there are no x-intercepts or if you are given a vertex and the x-intercepts are unclear. Note: A negative sign at the front of the equation means the parabola is upside down. QUADRATIC PROBLEMS p 84 These are almost always Excellence questions. These common themes often exist in the questions:. Write an equation. Substitute in known values 3. Solve 4. Put in context Note: There are often different ways of coming to the same answer. PAGE 40

4 GRAPHING Study Tip: Difficult Areas If you are struggling: Don t spend hours trying to understand Do: Write the problem down as precisely as possible so someone can help Ask a teacher they will be pleased to help, it s their job! Or ask a parent, sibling or friend Don t be afraid to ask! PAGE 4

5 LINEAR FEATURES SUMMARY There are two features for any linear graph that are important: The y-intercept This is the value when x = 0 and is the initial condition. e.g. y =. The gradient This is the slope of the graph and gives the speed/rate etc. of what is being graphed. Gradient= rise rise run e.g. Gradient = = = run 4 For a complete tutorial on this topic visit PRACTICE QUESTIONS. Sam needs to hire a car and he gets quotes from two different companies. He uses the graph below to work out the cost of a trip for either company. 3. Jamie and Amanda both leave school at the same time, Jamie rides a bike and Amanda skates. The following graph shows their distances from home and the time it takes them to get home. a. How much does Rob s Rides charge per kilometre? b. Charges by Cam s Cars include a flat charge for every trip. How much is it?. James is doing some home renovations and needs a concrete mixer. Two companies hire them out: Hire Co. and Garden Equipment. The cost for up to 30 days is shown. a. How far is Amanda s home from the school? b. How fast (in meters/min) did Jamie ride on her way home? c. How long did it take Jamie to ride home? 4. A dairy company is testing two different types of pump for pumping milk from its tankers into the factory tanks. By attaching the pumps to the milk tankers and measuring how long it takes to empty them it can be determined how well they perform. The graph below shows what happened with the first pump. a. How much does Garden Equipment charge per day? b. What is the initial fixed fee at Hire Co.? a. How many litres of milk did the tanker hold initially? b. How many litres were emptied in one minute? PAGE 4

6 GRAPHING 5. Will and Leah were travelling overseas. They wanted to go online to confirm bookings and the next stage of their travel. There were three internet cafes near their hostel: Cyber Time, Gateway, and Cafe Cyber. The graph below shows the connection costs for Cyber Time and Gateway Cafes for up to two hours (0 minutes). 6. Brian and Lisa are at a mountain bike park where there are various tracks, jumps and drop-offs. At one point two different jumps are side by side: Hang Time and Baby Steps. They can be seen below in the graph where d = distance from the start (in meters) and h = height above the ground (in meters). Cafe Cyber costs 7 cents per minute plus an initial connection fee of 50 cents. The cost to use Cafe Cyber can be described by the equation C = 7t+ 50 where t = time the internet was used (in minutes), and C = cost of the time on the internet (in cents). a. Which one of the three cafes has the highest connection fee? b. How is this shown by the graphs? a. What is the height of the Baby steps jump at the start? b. What is the increase in height of the Hang Time over the 5 m? Study Tip: Believe in Yourself Self belief is the biggest indicator of achievement: Set yourself a goal that you really want Then understand and genuinely believe it is possible to pass (or get Excellence) Stop worrying, just patiently follow the steps outlined for you When you meet barriers, don t be discouraged, simply find a way through If you can t make progress yourself, just ask for help. PAGE 43

7 ANSWERS PRACTICE.. 3. a. It goes through the points (0,0) and (40,60) rise 60 0 gradient = = run = = 5. Therefore there is a charge of $.50 per kilometre. b. Flat charge is $30 (y-intercept). a = $ 0 per day (from gradient) b. $50 (from y-intercept) a. 00 m or. km (y-intercept) b = 300 m per minute. c. 6 minutes (x-intercept) a L (y-intercept) b = 500 litres per minute 0 a. Gateway b. It has the highest C-intercept. i.e. the cost is highest when time is zero. a. m (h-intercept) b =. 5 m Study Tip: Study Location Choose an ideal study place: Can be the library, your room, anywhere quiet Should have none of the distractions that could slow you down Should have all the resources you need Should enable more effective study and give better results PAGE 44

8 GRAPHING LINEAR SEQUENCES SUMMARY y = mx+ c When given a linear sequence, e.g. 0, 3, 6, 9, a 5 columned table is drawn The x column is always,, 3, 4,... or the identifier of your sequence. The y column is the terms of your sequence The m column is the difference between the y values (they should all be the same) The mx column is the difference multiplied by the x value. In this case 3, 3,... The final column c is the difference between the mx and y values. It should always be the same number. The final step is to put the m and c values into the equation; y = mx+ c In this case giving: y = 3x + 7 OLD NCEA QUESTIONS. Sarah starts making a pattern of houses using toothpicks as shown in the diagram below. She begins a table for the number of toothpicks she uses for the number of houses in the pattern. Give the rule for calculating the number of toothpicks T that Sarah will need to make the nth design.. Kiri decides to make a different pattern involving separate houses. She begins with the same design as Sarah, as shown in the diagram to the right. Each new shape adds one more toothpick to each side of the previous design, as shown in the diagram to the far right. a. Give the rule for the number of toothpicks required to make the nth house in Kiri s pattern. b. Use this rule to find the number of toothpicks needed for the 6th house in the pattern. PRACTICE QUESTIONS For each of the following patterns find the linear equation that represent them. 3. 6, 5, 4, 33, , 6, 8, 0, 5., -, -6, -0, , 8, 53, 78, 03 x y m mx c For a complete tutorial on this topic visit Design (n) Number of toothpicks used in the design (T) Jo grouped marbles together into groups. Each group had a different number, the first had 4, the second had, the third had 8, the fourth had 5, and the fifth had 3. Write an equation which models this pattern. 8. Sam had a bag of lollies. If in the first minute he had 37 and each minute thereafter it decreased in the following pattern: 37, 9,, 3, 5. Write an equation which models the number of lollies, N, and the time passed, t PAGE 45

9 ANSWERS NCEA.. n T m mx c Equation is: T = 4n+ a. n T m mx c Equation is: T = 5 n b. T = 5 6= 30 PRACTICE 3. x y m mx c Equation is: y = 9x 3 x y m mx c Equation is: y = x x y m mx c Equation is: y = 4x+ 6 x y m mx c Equation is: y = 5x x y m mx c Equation is: y = 7x 3 t N m mx c Equation is: N = 8t+ 45 PAGE 46

10 GRAPHING GRAPHING LINEAR EQUATIONS SUMMARY Method ( y = mx+ c) 3 e.g. y = x Plot the y-intercept (in this case it is +3). Then plot the points according to the gradient (in this case move up 3 and across 4) Method - Coverup method e.g. 4y 3x=. To find the x-intercept make y = 0 and solve x = 3x = x = 4. To find the y-intercept make x = 0 and solve. 4y 3 0= 4y == y 3 3. Plot these two points and draw a line through them. For a complete tutorial on this topic visit OLD NCEA QUESTIONS. Sketch graphs of the equations below on the grids given. a. y = 6 x b. 3y x+ 6= 0. Sketch graphs of the equations below on the grids given. a. y = x+ b. 3x+ y = 8 There are more blank sets of axes on the next page. PAGE 47

11 PRACTICE QUESTIONS Draw the following graphs: 3. x = 5 4. y= x 5 5. y= x 5 6. y = 7 7. y= 4x 8. y= x 6 9. y+ 3x= y 8x= 8. y= x+ 5 3 PAGE 48

12 GRAPHING PAGE 49

13 ANSWERS NCEA. a. Rearranged: y = x+ 6 y-intercept = c = 6 gradient = m = - b. Not in the form y = mx + c therefore cover-up method: y = 0: 3x + 0= 8 3x = 8 x = 6 x = 0: 3 0+ y = 8 y = 8 y = 9 b. Not in the form y = mx + c therefore cover-up method: y = 0: 3 0 x + 6= 0 x + 6= 0 x = 6 x = 3 x = 0: 3y 0+ 6= 0 3y + 6= 0 3y = 6 y = PRACTICE y-intercept = c = -5 gradient = m =. a. y-intercept = c = gradient = m = / PAGE 50

14 GRAPHING 5. y-intercept = c = - gradient = m = / 9. Not in the form y = mx + c therefore cover-up method: y = 0: 0+ 3x = 9 3x = 9 x = 3 x = 0: y + 3 0= 9 y = 9 y = y-intercept = c = - gradient = m = 4 0. Not in the form y = mx + c therefore cover-up method: y = 0: 4 0 8x = 8 8x = 8 x = x = 0: 4y 8 0= 8 4y = 8 y = 8. y-intercept = c = -6 gradient = m = -5/. y-intercept = c = 5 gradient = m = /3 PAGE 5

15 WRITING LINEAR EQUATIONS SUMMARY The general equation is: y = mx+ c c is the y-intercept e.g. from graph y-intercept = m is the gradient = rise run e.g. from any two points on the graph m = = 4 Combined together the equation for this line is y= x+ For a complete tutorial on this topic visit OLD NCEA QUESTIONS. Jake is hiring a builder to help with some odd jobs around his house. Jake agrees to pay the builder for his travel costs and an hourly rate for work on the job. A graph is drawn that shows how much Jake pays for the work. Find the equation for the line on the graph that shows the total cost for doing the work. Write the equation of the graph shown below. 3. Write the equations of the lines drawn on the grid to the right. PAGE 5

16 GRAPHING PRACTICE QUESTIONS Write the equations of the lines drawn below Study Tip: Follow the Steps Think logically to increase your achievement level: Learn the steps to achieving every type of question Follow the steps EVERY time! PAGE 53

17 ANSWERS NCEA. y-intercept is at y = 40. Another point in the line is (4, 40) rise gradient = = = = 5 run therefore equation is C = 5h+ 40 h. It is a vertical line that crosses at x = -3 so the equation is x = a. y-intercept is at y = 0. another point in the line is (4, -3) rise gradient = = 3 0 = 3 run therefore equation is y = 3 x 4 b. y-intercept is at y = 4. h another point in the line is (-5, 0) rise 0 4 gradient = = run = = therefore equation is y = x+ 4 5 h PRACTICE 4. a. y-intercept is at y = 8. Gradient is 4 y= 4x+ 8 h b. y-intercept is at y =. Gradient is / y= x+ c. y-intercept is at y = -5. Gradient is 0 y = 5 h a. y-intercept is at y = -6. Gradient is -/3 y= x 3 6 b. No y-intercept as line is vertical. Gradient is infinite. x = 6 c. y-intercept is at y = 6. Gradient is - y= x+ 6 a. y-intercept is at y =. Gradient is - y= x+ b. y-intercept is at y = 0. Gradient is 3 y= 3 x c. No y-intercept as line is vertical. Gradient is infinite x = 6 a. y-intercept is at y =. Gradient is -4/3 4 y= x+ 3 b. y-intercept is at y = -8. Gradient is 3 y= 3x 8 c. y-intercept is at y = 0. Gradient is /5 y= x 5 PAGE 54

18 GRAPHING INTERPRET LINEAR GRAPHS SUMMARY Drawing a Graph Interpreting a Graph. Plotting the points. y-intercept. Discrete or continuous. Gradient 3. Other interesting features State the reasoning behind any answer giving actual values from the graph such as the gradient(s), starting point(s), or value(s) at a specific point. For a complete tutorial on this topic visit OLD NCEA QUESTIONS. Sarah starts making a pattern of houses using toothpicks as shown in the diagram below. She writes a table for the number of toothpicks she uses for the number of houses in the pattern and works out an equation as:t = 4n+ Number of toothpicks Design (n) used in the design (T) a. On the grid below, sketch a graph showing the number of toothpicks required for up to the 0th design. b. Kiri decides to make a different pattern involving separate houses. She begins with the same design as Sarah, as shown in the diagram to the near right. Each new shape adds one more toothpick to each side of the previous design, as shown in the diagram to the far right. The equation for the new pattern is T = 5 n. Describe how the graph for the number of toothpicks Kiri used for n houses relates to Sarah s graph.. The table below gives the adult single train fares for travel from the centre of a city. Number of stations, n Stage number Adult single fare 3 $ $ $ $5.75 a. On the grid below, sketch the graph of the adult train fares against the number of stations from the centre of the city. PAGE 55

19 b. A child s fare is $.50 for the first stage. Each additional stage, for a child, increases the fare by 75 cents. If a graph was drawn for the child s fares, describe the similarities and differences between the graphs of the child s fare and the adult s fare. 3. Blake receives a copy of his bank statement and finds he is overdrawn (he has a negative amount in the bank). He starts a saving plan. The graph below shows the amount of money Blake hopes to have in his bank account, S, if he follows his savings plan for n weeks. To help Emma know how much she can expect to pay, Ian provides the following table: Number of hours Payment (P) worked (h) 4 $60 5 $85 6 $0 7 $35 8 $60 9 $345 0 $370 a. On the grid below, plot a graph showing the payment required for the number of hours worked. a. How much does Blake plan to bank each week? b. Give the equation for the graph of Blake s saving plan in terms of S, the amount in Blake s account, and n, the number of weeks after the start of his saving plan. c. Blake s grandmother thinks he should be saving more. At the end of 4 weeks she tells him that if the amount in his bank account at the end of 9 weeks is $300, she will give him $50. He increases the fixed amount he saves each week from the end of week 4. He reaches his grandmother s target of $300 in his account and banks the $50 from his grandmother. He continues saving at the increased rate after banking the $50 from his grandmother. Describe how the graph changes from week 4 onwards. Hints: You can do this by giving equations for some parts of the graph. You may find it helpful to sketch the graph using the grid above. 4. Emma is employing Ian to build a deck at her house. She provides all the building material. She pays Ian $P for the number of hours, h, that he works. She also pays for Ian s travel to her home each day. Ian works for 8 hours each day. He knows the deck will take more than 4 hours to build. b. How much does Ian charge for his travel each day? c. Explain why the graph rises more steeply after 8 hours. d. What would Emma expect to pay if the work took 30 hours? Explain your calculation. e. Zarko lives next door to Emma and says he could build the deck for her. He does not need to be paid for his travel, but he charges $35 an hour. How long would the work take if the payments to Zarko and Ian were the same? Explain how you calculated your answer. Hint: there may be more than one solution. f. Another builder gives Emma a graph, showing the amount she would charge for 8 hours work. Give the rule for the payment that this builder would receive for the first 8 hours that she worked. PAGE 56

20 GRAPHING 5. Brad and Zara were arranging hexagonal shaped tables for their wedding. They try different arrangements of tables to see how many people they can fit when the tables are put next to each other, as shown in the diagram. They mark where each person could sit with an X. 6. The graph below shows Mark and Katie s journeys from their homes to school. They leave home at the same time. Katie rides her bike and Mark walks. Brad says the equation for the total number of people, p, if there are n tables, is given by the equation p = 4n+. Zara plots a graph of the number of people, p, against the number of tables. a. How much further does Katie live from school than Mark? b. Write a full detailed comparison of Mark and Katie s journeys from their homes to school. Explain how the equation and graph relate to the number of people at the tables. PRACTICE QUESTIONS 7. Sam needs to hire a car and he gets quotes from two different companies. He uses the graph below to work out the cost of a trip for either company. 8. James is doing some home renovations and needs a concrete mixer. Two companies hire them out: Hire Co. and Garden Equipment. The cost for up to 30 days is shown. a. For what distance would the cost of hiring be the same? b. Write out the equation for the Cam s Cars graph. James figures out that he will need the concrete mixer for 7 days. Which hire place would be cheaper? Explain your answer. PAGE 57

21 9. Jamie and Amanda both leave school at the same time, Jamie rides a bike and Amanda skates. The following graph shows their distances from home and the time it takes them to get home.. Sarah has been measuring her height since 003. Below is a graph showing her height, a trend line has also been added to the graph. a. Write an equation for Amanda s distance from home, d, with respect to the time taken, t. b. They both left school at 3.7 pm. At what time would the both be the same distance away from home? 0. A dairy company is testing two different types of pump for pumping milk from its tankers into the factory tanks. By attaching the pumps to the milk tankers and measuring how long it takes to empty them it can be determined how well they perform. The graph below shows what happened with the first pump. a. Find the equation of the growth trend line, where y = Sarah s height, in cm and x = the number of years since 003. b. What does the gradient of the trend line show about how Sarah is growing? c. From this trend line, estimate Sarah s height in the year Sam was a painter and noticed that the more tins of pain he put in the back of his Ute the more it sagged. He measured this for a few different quantities of paint tins and measured the gap between the tyre and the body of the Ute. a. What is the equation of the graph shown? b. The pump can be operated at a slower speed if necessary. Add a line to the graph above to show this. c. A similar test, with the other pump on a different tanker was also performed. What happened can be modelled by the equation: y = x Sketch the graph of this equation a. What is the equation of the line? Use y = gap size, in cm and x = number of paint tins. b. With how many paint tins would it take for the Ute to sag and hit the tyre? c. Sam changed to using paint tins that are twice the weight of the old ones. Draw the graph of what you would expect to happen on the above graph. PAGE 58

22 GRAPHING 3. Will and Leah were travelling overseas. They wanted to go online to confirm bookings and the next stage of their travel. There were three internet cafes near their hostel: Cyber Time, Gateway, and Cafe Cyber. The graph below shows the connection costs for Cyber Time and Gateway Cafes for up to two hours (0 minutes). 4. Brian and Lisa are at a mountain bike park where there are various tracks, jumps and drop-offs. At one point two different jumps are side by side: Hang Time and Baby Steps. They can be seen below in the graph where d = distance from the start (in meters) and h = height above the ground (in meters). Cafe Cyber costs 7 cents per minute plus an initial connection fee of 50 cents. The cost to use Cafe Cyber can be described by the equation C = 7t+ 50 where t = time the internet was used (in minutes), and C = cost of the time on the internet (in cents). a. Draw the graph for Cafe Cyber from t = 0 to t = 0 minutes. b. i. How long would Will and Leah need to use the internet for, for the cost to be the same in Cyber Time and Gateway. ii. How is this shown by the graphs? c. i. Use the graph to write the equation for the a. What is the equation of the graph of the Hang Time jump? b. What is the equation of the graph of the Baby Steps jump? Cyber Time cafe. ii. Use the graph to write the equation for the Gateway cafe. Study Tip: Motivation Bored? If you do feel bored, spend some time focusing on what you want to achieve. If a grade isn t enough, think of a reward that you will treat yourself to if you reach your target grade. PAGE 59

23 ANSWERS NCEA. a.. (Achieved: Correct graph as a line) (Merit: Correct graph as a line, starting at and finishing at 0) (Excellence: Correct Graph as discrete variables, starting at and finishing at 0) b. The graph for the number of toothpicks Kiri uses for the nth house will differ from Sarah s in the gradient. The gradient will be steeper, having increased from 4 in Sarah s to 5 in Kiri s graph. Both graphs will still begin at the same place (,5). (Excellence) a. b. (Achieved: Sloping line connecting steps) (Merit: Show horizontal sets of points) (Excellence: Correct Graph as discrete variables) both increase as the stage number increases (every three stations). However, the adult single fare increases by $.5 per stage while the child single fare increases by 75 c per stage. This results in increasing vertical distances between the points plotted as the stage number increases (every three stations). (Excellence) a. When n increases by one, S increases by $40, i.e. Blake plans to bank $40 per week. b. S-intercept= -00, gradient = 40 so equation is S = 40n 00 c. At the end of four weeks, Blake s bank balance is = $ 40. At the end of 9 weeks, Blake wants his bank balance to be $300. We calculate a new gradient using the points (4,-40) and (9,300): rise 300 ( 40) New Gradient = = run = = 68 5 So Blake increases his savings to $68 each week and the graph will have a steeper gradient from n = 5. Let S = 68 n+ c. When n = 9, we want s = c= 300. i.e. c = = 3. Therefore S = 68n 3 for n = 5,..., 9 When n = 0, Blake banks the $50 from his grandmother and continues to save at the increased rate. This will result in a steeper rise in the graph at n = 0, but the same gradient of 68 thereafter. Therefore: S = 68n = 68n 6 for n 0 (Achieved - a correct) (Merit - b correct) (Excellence - c correct) a. PAGE 60 The graphs of the adult and child single fares are similar in that they remain constant for the stations within each of the four stages and they (Achieved: 4-8 hrs plotted somehow) (Merit: 4-8 and 8 - plotted somehow) (Excellence: Continuous line for 4-8 and 8 -, make sure 8 is $60 not $30)

24 GRAPHING b. Between h = 4 and h = 8, the payments increase by $5. Therefore Ian charges $5 per hour. Let P= 5 h+ c be the equation of the line. When h = 4 we want P= c= 60 Therefore c = = 60 so P= 5h+ 60 When h = 0, P = 60 i.e. Ian s travel cost must be $60. Alternative Solution: When h = 9, the payment increases by = $85. Since this includes one hours work ($5) and a second days travel, the cost of the travel must be: 85-5 = $60. c. The steeper rise in the graph after 8 hours reflects the cost of travel included for a second day s work. d. 30 hrs = 3 8 hrs + 6 hrs, so Ian will make 4 trips to Emma s house. Cost of 30 hours work at $5 per hour = 30 5 = $ 750 Cost of 4 trips at $60 per trip = 4 60 = $ 40 Therefore total cost = = $ 990 e. To model Zarko s payments at $35 per hour and no charge for travel, let P= 35 h. To equate these payments solve: 35h= 5h h 5h= 60 0h = 60 h = 6 So Zarko s and Ian s payments would be the same if they worked 6 hours. NB: There is more than one solution. As Ian works for a lower hourly wage but charges a travel allowance, the graph of his costs zigzags above and below Zarko s giving multiple times where their costs are the same. See the graph of Ian s and Zarko s payments in the next column. This graph suggests that other solutions are at and 8 hours. This is because the graphs coincide at these h values (and also at h = 6). f. The graph passes through (0,60) and (6,300) y-intercept = 60. rise gradient = = = = 40 run therefore P= 40h+ 60 (Achieved - b or c correct) (Merit - d correct) (Excellence - e and f correct) 5. We can see in the diagram that for each arrangement of tables, four people are seated at the top and bottom sides of each table, and two people are seated at the ends. This gives the equation p = 4n+, where 4 represents the increase of 4 people each time a table is added, while the represents the two people on the ends of the row of tables. This is also mirrored in the graph where n =, p = 4 + = 6, and each time n increases by the p value is increased by 4. (Excellence) 6. a. Katie s y-intercept is y = 6 Mark s y-intercept is y =.5. Difference = = 4.5 km. b. Mark and Katie both leave home at the same time for school. Mark has only.5 km to travel (y-intercept) while Katie has 6 km to travel (y-intercept). As Katie is riding her bike she travels faster than Mark. This can be seen in the gradients of their respective journeys: Katie has a gradient of -/3 or a speed of 3 minutes per kilometre while Mark has a gradient of -/ or a speed of minutes per kilometre. Mark s speed is a quarter of Katie s. They both arrive at school at the same time 8 minutes after leaving home (x-intercept). (Excellence) PRACTICE 7. a. The charges would be the same for a journey of 60 km. b. It goes through the points (0,30) and (60,90). rise gradient = = = = run therefore equation is C = d Hire Co. would be cheaper because their cost for 7 days is lower than Garden Equipment s. The y value 9. for 7 days is lower for Hire Co. than for Garden Equipment. a. y-intercept = 00. rise 0 gradient = = 00 run 8 0 = 00 = 50 8 therefore the equation is d = 50t+ 00 b. The graphs cross at 4 minutes so 4 minutes after they leave they are the same distance from home. 3.7 pm + 4 minutes = 3. pm. (Achieved - a or b correct) (Merit - a and b correct) PAGE 6

25 0. a. y-intercept = b. Gradient = -500 Therefore M = 500t c. c. 3. a. (Achieved - a or c correct) (Merit - two of a, b or c correct) (Excellence - all correct) (Achieved - b or c correct) (Merit - two of a, b or c correct) (Excellence - all correct). a. y-intercept = 00 and another point on the line is (4,40) rise gradient = = = = 0 run Equation: y = 0x+ 00 b. Sarah has average growth of approximately 0 cm per year. c. The year 000 would have a x value of -3 so substituting that into the equation should give Sarah s height in 000. y = = 70 cm (Achieved - a or b correct) (Merit - a and b correct) (Excellence - all correct). a. y-intercept occurs at (0,0) and another point occurs at (30,5) rise gradient = run = 5 0 = = Equation of the line is y = x+ 0 b. It would hit when y = 0 x + 0 = 0 x = 0 x = paint tins would be required to make it hit the wheel. b. i. 50 minutes ii. It is shown by the intersection of the Cyber Time and Gateway graphs. c. i. C-intercept is (0,0) rise 00 0 gradient = = run = = 0 0 Therefore C =0 t ii. C-intercept is (0,00) rise gradient = = run = = 8 00 Therefore C = 8t a. h-intercept is at (0,0.5) rise gradient = run = 5. 5 = Therefore h= d b. h-intercept =. rise gradient = = run 5 Therefore h= d + 5 PAGE 6

26 QUADRATIC SEQUENCES GRAPHING SUMMARY y = ax + mx+ c When given a quadratic sequence, e.g. 3,, 5, 4, 63. We then draw a 9 columned table: The x column is always,, 3, 4,... or the identifier of your sequence. The y column is the terms of your sequence x y d d ax L m mx c The d column is the difference between the y values The d column is the difference between the d values (they should all be the same) The a value is half d in this case a =. 3 Therefore the ax column is half the second difference multiplied by the x value squared. In this case,,... The L column is the difference between the y and ax values. This gives us a linear sequence. The final three columns are solving a linear sequence. The final step is to put the a, m and c values into the equation y = ax + mx+ c In this case giving: y = x + 3x For a complete tutorial on this topic visit OLD NCEA QUESTIONS. Sarah starts making a pattern of houses using toothpicks as shown in the diagram below. She made a table for the number of toothpicks she uses for the number of houses in the pattern. Give the rule for the total number of toothpicks that Sarah would need if she was to continue following the pattern and complete n designs. PRACTICE QUESTIONS The following patterns are quadratic in nature. Find the quadratic equation that represent them.., 3, 9, 9, 33 3., 8, 40, 68, 0 4., 6,, 0, , -5, -45, -73, Graeme was doing a running drill at hockey practice where he had to run a 0 m distance a set number of times in each set. Set :, Set : 0, Set 3: 6, Set 4: 0, Set 5:. The number of times he has to run can be modelled by a quadratic equation. Find the equation from the information given, let s be the set number and T be the times run. Use this rule to find the total number of toothpicks needed to complete the first designs using Sarah s pattern. Design (n) Number of toothpicks used in the design (T) Lisa was plaing a game on the computer where she had to level up. Each time she went up a level it got harder, and she needed more experience than the previous level to do it. Using the information given work out the quadratic formula for the experience needed for each level. Let l be her level and E be the experience needed. Level (l) Experience (E) PAGE 63

27 ANSWERS NCEA. x y d d ax L m mx c Therefore the equation for the total number of toothpicks can be written: T = n + 3n (Excellence) PRACTICE. x y d d ax L m mx c Equation is therefore: y = x 4x + 3 x y d d ax L m mx c Equation is therefore: y = 3x + 7x 8 x y d d ax L m mx c Equation is therefore: y = x + x x y d d ax L m mx c Equation is therefore: y = 4x 9 s T d d ax L m mx c Equation is therefore:t = s + s + (Excellence) l E d d ax L m mx c Equation is therefore: E = 3l + l 3 (Excellence) PAGE 64

28 GRAPHING PARABOLAS: X-INTERCEPTS SUMMARY OLD NCEA QUESTIONS GRAPHING A parabola is a graph of a quadratic equation. The general equation is: y = ( x+ a)( x+ b) e.g. y = ( x+ 3)( x + ) a and b are the two x-intercepts. They are the negative of the numbers in the equation. e.g. x-intercepts are -3 and - The vertex of the parabola is the turning point and is half way between the x-intercepts. e.g. The vertex will lay on the line x = - Substitute the x value of the vertex into the original equation to find the y value of the vertex. e.g. y = ( + 3)( + ) = ()( ) = The vertex is at (-,-) An equation written y = x( x + ) is the same. The x-intercepts are just 0 and - this time. Note: A negative sign at the front of the equation means the parabola is upside down. For a complete tutorial on this topic visit For the graph below give:. For the graph below, give: a. the intercepts b. the function 3. Sketch the graph of the equation: y = ( 4 x)( x + ) a. the x and y intercepts b. the equation of the graph. 4. On the grid below, sketch the graph of y = ( x )( x + 4) PAGE 65

29 PRACTICE QUESTIONS For the following equations factorise (if necessary), write out the x-intercepts and vertex location, and plot the graphs. There are blank graphs on the next page to draw on. 5. y= ( x )( x + 3) 6. y= x( x 3) 7. y= x 0x + 8. y= 9 x 9. y= ( x+ 5)( x + 7) 0. y= ( 3 x)( x 4). y= x + 6x. y= ( x+ 3)( x) For the following graphs give the x-intercepts and write out the equation PAGE 66

30 GRAPHING PAGE 67

31 ANSWERS NCEA.. a. x-intercepts are at x = and x = - y-intercept is at y =. b. We know the intercepts so sub into equation giving: y = ( x )( x + ). As the graph is upside down it is necessary to add a negative to the front of the equation. The final equation is therefore: y = ( x )( x + ) = x x+ a. x-intercepts are at x = -3 and x = 3 y-intercept is at y = -9 b. From the intercepts: y = ( x 3)( x+ 3) or y = x 9 3. It is a negative parabola with x-intercepts at x = - and x = 4 and a y-intercept at y = 8. Vertex is midway between x-intercepts at x =. The y value is: y = ( 4 )( + ) = ()() 3 3 = 9 PRACTICE 5. x-intercepts at x = and x = -3 Vertex is midway between x-intercepts at x = -. The y value is: y = ( )( + 3) = 4 6. x-intercepts at x = 0 and x = 3 Vertex is midway between x-intercepts at x =.5. The y value is: y = 55.(. 3) = It is a negative parabola with x-intercepts x = and x = -4 and y-intercept y = 8. Vertex is midway between x-intercepts at x = -. The y value is: y = ( )( + 4) = ( 3)() 3 = 9 7. y = x 0x + = ( x 7)( x 3) x-intercepts at x = 7 and x = 3 Vertex is midway between x-intercepts at x = 5. The y value is: y = ( 5 7)( 5 3) = 4 PAGE 68

32 GRAPHING 8. y = 9 x = ( x 9) = ( x 3)( x+ 3) x-intercepts at x = 3 and x = -3 Vertex is midway between x-intercepts at x = 0. The y value is: y = ( 0 3)( 0+ 3) = 9. y = x + 6x = xx ( + 6) x-intercepts at x = 0 and x = -6 Vertex is midway between x-intercepts at x = -3. The y value is: y = 3( 3+ 6) = 9 9. x-intercepts at x = -7 and x = -5 Vertex is midway between x-intercepts at x = -6. The y value is: y = ( 6+ 5)( 6+ 7) =. x-intercepts at x = and x = -3 Vertex is midway between x-intercepts at x = The y value is: y = ( )( ( 05. )) = x-intercepts at x = 3 and x = 4 Vertex is midway between x-intercepts at x = 3.5. The y value is: y = ( )( ) = x-intercepts at x = -8 and x = -5 Equation is therefore: y = ( x+ 5)( x + 8) 4. x-intercepts at x = and x = -4 Equation is therefore: y = ( x )( x + 4) 5. x-intercepts at x = 3 and x = 9 Upside down so negative in front. Equation is therefore: y = ( x 3)( x 9) 6. x-intercepts at x = 0 and x = 3 Equation is therefore: y = x( x 3) 7. x-intercepts at x = 0 and x = -4 Upside down so negative in front. Equation is therefore: y = x( x + 4) 8. x-intercepts at x = - and x = 4 Equation is therefore: y = ( x+ )( x 4) PAGE 69

33 Study Tip: Make Sure You Understand Understanding is important. Wait until you have a clear understanding of the topic before moving on to the next section. When you get stuck, ask for clarification (see your teacher the next day). PAGE 70

34 GRAPHING PARABOLAS: VERTEX SUMMARY OLD NCEA QUESTIONS GRAPHING The general equation is y = ( x c) + d e.g. y = ( x ) + c is how far the vertex has been horizontally shifted from the origin. e.g. it has been shifted to the right d is how far the vertex has been vertically shifted from the origin. e.g. it has been shifted up If given the equation, the vertex is located at the point (c, d). e.g. (, ) Use this form if there are no x-intercepts or if you are given a vertex and the x-intercepts are unclear. Note: A negative sign at the front of the equation means the parabola is upside down. For a complete tutorial on this topic visit Sketch the graph of the equation: y = x 9. Sketch the graph of the equation: y = x PAGE 7

35 PRACTICE QUESTIONS For the following equations write out the vertex locations, and plot the graphs. There are blank graphs to draw on, on the next page. 3. y = ( x+ 7) y = x 5 5. y = ( x 3) y = ( x+ 5) 6 7. y = ( x 7) 8. y = ( x ) For the following graphs give the vertex locations and write out the equation PAGE 7

36 GRAPHING PAGE 73

37 ANSWERS NCEA. Looking at the equation it has no horizontal movement and vertical movement of Vertex is at (3, 5). Looking at the equation it has no horizontal movement and vertical movement of Vertex is at (-5, -6) PRACTICE 3. Vertex is at (-7, 3) 7. Vertex is at (7, 0) 8. Vertex is at (, -) 4. Vertex is at (0, -5) PAGE 74

38 GRAPHING 9. Vertex is at (-5, 5) Upside down so negative in front. y = ( x+ 5) Vertex is at (0, 7) Upside down so negative in front. y = ( x ) + 7. Vertex is at (6, 3) y = ( x 6) + 3. Vertex is at (-3, -8) y = ( x+ 3) 8 3. Vertex is at (-4, 0) y = ( x+ 4) 4. Vertex is at (5, -) Upside down so negative in front. y = ( x 5) Study Tip: Make it Colourful Colours help you remember! Use highlighters to mark the important points. Use different colours for different categories. PAGE 75

39 Exam Tip: Read the Test Scanning the test before you start can help by: Giving an overview Is essential for time planning Helps prevent basic errors (e.g. not staying on topic) PAGE 76

40 GRAPHING TRANSLATING PARABOLAS SUMMARY Translating a quadratic is moving it along and/or up a certain amount. To translate an equation by (c,d) simply replace x with (x - c) and y with (y - d) and simplify. e.g. Translate y = x 4x + 3 by (4,3) y 3= ( x 4) 4( x 4) + 3 y 3= x 8x+ 6 4x y 3= x x + 35 y = x x + 38 For a complete tutorial on this topic visit OLD NCEA QUESTIONS. The parabola shown is moved 3 units to the right and 5 units up. It has an equation of y = x x + Give the equation of the parabola in simplified form its new position AND give the y-intercept.. The parabola shown to the right is moved unit to the right and units up. It has an equation of y = x 9 Give the equation of the parabola in simplified form in its new position AND give the y-intercept. 3. The equation for the parabola drawn on the graph to the right can be written as y = ( x+ 3)( x 5) If this parabola is moved 3 units to the left and 0 units up, give the equation for the parabola in its new position AND give the coordinates of the y-intercept. PRACTICE QUESTIONS Translate the following equations by the given amounts, simplify, and also give the y-intercept. 4. y = x + 3x 4 by 6 to the right and 4 down 5. y = x 5x + 5 by to the right and 0 up 6. y = x + 5 by to the left and 3 down 7. y = ( x 4)( x + 3 ) by to the right and down 8. y = ( x+ 6)( x ) by 4 to the left 9. y = 4x x + 0 by 6 to the right and 8 up 0. y = x + 6 x by 5 to the left and up. y = 6x + x + 8 by down PAGE 77

41 ANSWERS NCEA. The parabola is translated by a vector of (3, 5). Replacing the values and simplifying the equation: y = x x + ( y 5) = ( x 3) ( x 3) + y 5= ( x 6x+ 9) x+ 3+ y 5 = x + 6x 9 x+ 3+ y 5= x + 5x 4 y = x + 5x + y-intercept when x = 0: y = + + = (Excellence). The parabola is translated by a vector of (, ). Replacing the values and simplifying the equation: y = x 9 ( y ) = ( x ) 9 y = x x + 9 y = x x 6 y-intercept when x = 0: y = = (Excellence) 3. The parabola is translated by a vector of (-3, 0). Replacing the values and simplifying the equation: y = ( x+ 3)( x 5) ( y 0) = (( x+ 3) + 3)(( x + 3) 5) y 0 = ( x+ 6)( x ) y 0 = x + 4x y = x + 4x + 8 y-intercept when x = 0: y = = 8 (Excellence) PRACTICE 4. Translated by (6, -4) y = x + 3x 4 ( y+ 4) = ( x 6) + 3( x 6) 4 y+ 4= x x x 8 4 y+ 4 = x 9x + 4 y = x 9x + 0 y-intercept when x = 0: y = + = (Excellence) 5. Translated by (, 0) y = x 5x + 5 ( y 0) = ( x ) 5( x ) + 5 y 0 = ( x 4x+ 4) 5x y 0 = x 8x+ 8 5x y 0 = x 3x + 3 y = x 3x + 33 y-intercept when x = 0: y = = 33 (Excellence) 6. Translated by (-, -3) y = x + 5 ( y+ 3) = ( x+ ) + 5 y+ 3= x + x y+ 3= x + x + 6 y = x + x + 3 y-intercept when x = 0: y = + + = (Excellence) 7. Translated by (, -) y = ( x 4)( x + 3) ( y+ ) = (( x ) 4)(( x ) + 3) y+ = ( x 6)( x 9) y + = x 45x+ 494 y = x 45x y-intercept when x = 0: y = = 493 (Excellence) 8. Translated by (-4, 0) y = ( x+ 6)( x ) y = (( x+ 4) + 6)(( x + 4) ) y = ( x+ 0)( x + ) y = x + x + 0 y-intercept when x = 0: y = = 0 (Excellence) 9. Translated by (6, 8) y = 4x x + 0 ( y 8) = 4( x 6) ( x 6) + 0 y 8= 4( x x+ 36) x y 8= 4x 48x+ 44 x y 8= 4x 60x + 36 y = 4x 60x + 44 y-intercept when x = 0: y = = 44 (Excellence) 0. Translated by (-5, ) y = x + 6x ( y ) = ( x+ 5) + 6( x + 5) y = x + 30x x+ 90 y = x + 36x + 35 y = x + 36x + 36 y-intercept when x = 0: y = = 36 (Excellence). Translated by (0, -) y = 6x + x + 8 ( y+ ) = 6x + x + 8 y = 6x + x 4 y-intercept when x = 0: y = + = (Excellence) PAGE 78

42 GRAPHING Exam Tip: Use the Clues Bullet points and key words are clues in a question. Use them! If you are unsure about the answer, don t give up. Find the clues and write about closely related word or ideas. PAGE 79

43 SCALING PARABOLAS SUMMARY Parabolas can be made wider or narrower by putting a number in front of the equation. e.g. y = 3( x ) + has a 3 in front, making the parabola 3 times narrower Or y = ( x+ 3)( x )has a 4 4 in front making it as narrow (4 times wider). 4 Steps:. Write equation using either vertex y = k( x ) + or x-intercept method. Find a point not used to write the equation and substitute it into the equation with k 5= k( 0 ) + 3. Rearrange to find the value of k 5= k( ) + 5= k + k = 3 4. Write out the final equation y = 3( x ) + Note: A negative sign at the front of the equation means the parabola is upside down. For a complete tutorial on this topic visit OLD NCEA QUESTIONS. In a children s play park, a ball is kicked so that its flight path can be modelled by the equation h= kxx ( 6 ) where h metres is the height of the ball when it is x metres from the point from where it is kicked. If the maximum height of the ball is m, what is the value of k? PRACTICE QUESTIONS For the following equations find the scaling factor using the information given.. Equation: y k x x = ( + 3)( 4) Passes through the point (3, -) 4. Equation: y k x = ( + 4) Passes through the point (-0, ) 6. Equation: y k x x = ( + )( + 4) Passes through the point (-6, ) 3. Equation: y k x = ( ) + 4 Passes through the point (0, 4) 5. Equation: y k x x = ( )( 5) Passes through the point (6, 6) 7. Equation: y k x = ( + ) + Passes through the point (-4,3) PAGE 80

44 GRAPHING For the following graphs find the equations including the scaling factor PAGE 8

45 ANSWERS NCEA. h is a negative parabola with x-intercepts at x = 0 and x = 6. The maximum height of the ball is reached when 0 x = + 6 = 3 When x= 3, h= k()( 3 3 6) = 9k We want this height to be m i.e. 9k = k = 9 PRACTICE. Substituting (3, -) into the equation: kx ( + 3)( x 4) = y k( 3+ 3)( 3 4) = k( 6)( ) = 6k = k = 3. Substituting (0, 4) into the equation: kx ( ) + 4= y k( 0 ) + 4= 4 k( ) + 4= 4 4k == 0 k 5 4. Substituting (-0, ) into the equation: kx ( + 4) = y k( 0 + 4) = k( 6) = 36k = k = 3 5. Substituting (6, 6) into the equation: kx ( )( x 5) = y k( 6 )( 6 5) = 6 k( 4)() == 6 4k 6 k = 4 6. Substituting (-6, ) into the equation: kx ( + )( x+ 4) = y k( 6+ )( 6+ 4) = k( 5)( ) = 0k = k = 5 7. Substituting (-4, 3) into the equation: kx ( + ) + = y k( 4+ ) + = 3 k( ) + = 3 4k = k = 8. Information from the graph: x-intercepts at x = -7 and x = Vertex at (-3, -8) Equation can be of the form y = k( x a)( x b): y = k( x+ 7)( x ) Solving for k (substitute in vertex): 8= k( 3+ 7)( 3 ) 8= k( 4)( 4) 8= 6k = k Equation is therefore: y = ( x+ 7)( x ) OR of the form y = k( x c) + d : y = k( x + 3) 8 solving for k (substitute in x intercept): 0= k( + 3) 8 8= 6k = k Equation is therefore: y = ( x+ 3) 8 7 Both simplify to: y = x + 3x 9. Information from the graph: x-intercepts at x = -8 and x = -4 Vertex at (-6, 8) Equation can be of the form y = k( x a)( x b): y = k( x+ 8)( x+ 4) solving for k (substitute in vertex): 8= k( 6+ 8)( 6+ 4) 8= k( )( ) 8= 4 k = k Equation is therefore: y = ( x+ 8)( x + 4) OR of the form y = k( x c) + d : y = k( x + 6) + 8 solving for k (substitute in x intercept): 0= k( 4+ 6) + 8 8= k( ) 8= 4k = k Equation is therefore: y = ( x+ 6) + 8 Both simplify to: y = x 4x Information from the graph: y-intercept at y = 6 Vertex at (, -0) Equation is of the form y = k( x c) + d : y = k( x ) 0 Continued on next page... PAGE 8

46 GRAPHING solving for k (substitute in y-intercept): 6= k( 0 ) 0 6 = k( ) 6 = 4k 4 = k Equation is therefore: y = 4( x ) 0. Information from the graph: x-intercept at x = 5 and x = - Vertex at (-3, 8) Equation can be of the form y = k( x c) + d : y = k( x + 3) + 8 solving for k (substitute in x-intercept): 0= k( 5+ 3) + 8 8= k() 8 8= 64 k = k 8 Equation is therefore: y = ( x+ 3) OR of the form y = k( x a)( x b): y = k( x 5)( x+ ) solving for k (substitute in vertex): 8= k(( -3) 5)(( -3) + ) 8= k( 8)() 8 8= 64k = k 8 Equation is therefore: y = ( x 5)( x + ) Both simplify to: y = x x Information from the graph: x-intercepts at x = and x = 5 Vertex at (3, 6) Equation can be of the form y = k( x a)( x b): y = k( x )( x 5) solving for k (substitute in vertex): 6= k( 3 )( 3 5) 6= k( )( ) 6= 4k 3 = k 3 Equation is therefore: y = ( x )( x 5) OR of the form y = k( x c) + d : y = k( x 3) + 6 solving for k (substitute in x-intercept): 0= k( 3) + 6 6= k( ) 6= 4 k 3 = k 3 Equation is therefore: y = ( x 3) Both simplify to: y = x + 9x 3. Information from the graph: x-intercept at x = 3 and x = -3 Vertex at (-5, -8) Equation can be the form y = k( x c) + d : y = k( x + 5) 8 solving for k (substitute in x-intercept): 0= k( 3+ 5) 8 8= k() 8 8= 64k = k 8 Equation is therefore: y = ( x+ 5) 8 8 OR of the form y = k( x a)( x b): y = k( x 3)( x+ 3) solving for k (substitute in vertex): 8= k(( -5) 3)(( -5) + 3) 8= k ( 8)() 8 8= 64k = k 8 Equation is therefore: y = ( x 3)( x + 3) Both simplify to: y = x + x Information from the graph: x-intercepts at x = -5 and x = - y-intercept at y = 8 Equation is of the form y = k( x a)( x b): y = k( x+ )( x+ 5) solving for k (substitute in y-intercept): 8= k( 0+ )( 0+ 5) 8= k()() 5 8= 5k 8 = k 5 8 Equation is therefore: y = ( x+ )( x + 5) 5 5. Information from the graph: vertex at (4, 7) y-intercept at y = Equation is of the form y = k( x c) + d : y = k( x 4) + 7 solving for k (substitute in y-intercept): = k( 0 4) + 7 5= k( 4) 5= 6 k 5 = k 6 5 Equation is therefore: y = ( x 4) PAGE 83