Math Functions and Graphs Exploration

1. A function is defined as f (x) = 3x − 1. Find: (i) f (−3) f (x) = 3x – 1 f (–3) = 3(–3) – 1 = – 9 – 1 f (–3) = – 10

1. A function is defined as f (x) = 3x − 1. Find: (ii) f (0) f (x) = 3x – 1 f (0) = 3(0) – 1 = 0 – 1 f (0) = – 1

1. A function is defined as f (x) = 3x − 1. Find: (iii) f (2) f (x) = 3x – 1 f (2) = 3(2) – 1 = 6 – 1 f (2) = 5

1. A function is defined as f (x) = 3x − 1. Find: (iv) f (4) f (x) = 3x – 1 f (4) = 3(4) – 1 = 12 – 1 f (4) = 11

1. A function is defined as f (x) = 3x − 1. Find: (v) Hence, or otherwise, graph the function f (x).

2. A function is defined as g(x) = 5 − 2x. Find: (i) g(− 2) g(x) = 5 – 2x g(– 2) = 5 – 2(– 2) = 5 + 4 g(– 2) = 9

2. A function is defined as g(x) = 5 − 2x. Find: (ii) g(−1) g(x) = 5 – 2x g(–1) = 5 – 2(–1) = 5 + 2 g(–1) = 7

2. A function is defined as g(x) = 5 − 2x. Find: (iii) g(2) g(2) = 5 – 2(2) = 5 – 4 g(2) = 1

2. A function is defined as g(x) = 5 − 2x. Find: (iv) g(3) g(3) = 5 – 2(3) = 5 – 6 g(3) = –1

2. A function is defined as g(x) = 5 − 2x. Find: (v) Hence, or otherwise, graph the function g(x).

3. By finding points on the function, or otherwise, graph the following functions: (i) f (x) = x + 3 Let x = – 2: f (– 2) = – 2 + 3 f (– 2) = 1 (– 2, 1) Let x = 2: f (2) = 2 + 3 f (2) = 5 (2, 5)

3. By finding points on the function, or otherwise, graph the following functions: (ii) g(x) = 2x + 4 Let x = – 2: g(– 2) = 2(– 2) + 4 = – 4 + 4 g(– 2) = 0 (– 2, 0) Let x = 2: g(x) = 2x + 4 g(2) = 2(2) + 4 = 4 + 4 g(2) = 8 (2, 8)

3. By finding points on the function, or otherwise, graph the following functions: (iii) h(x) = 6 − x Let x = – 2: h(– 2) = 6 – (– 2) = 6 + 2 h(– 2) = 8 (– 2, 8) Let x = 2: h(x) = 6 – x h(2) = 6 – 2 h(2) = 4 (2, 4)

3. By finding points on the function, or otherwise, graph the following functions: (iv) k(x) = − 2x − 3 Let x = – 2: k(– 2) = – 2(– 2) – 3 = 4 – 3 k(– 2) = 1 (– 2, 1) Let x = 2: k(2) = – 2(2) – 3 = – 4 – 3 k(2) = – 7 (2, – 7)

4. By finding the point where the line crosses the y-axis (y-intercept) and the rate of change of each line, hence or otherwise, write down the function represented by each of the following graphs: (i) Intersect on y-axis = 1 = c y = mx + c y = x + 1

4. By finding the point where the line crosses the y-axis (y-intercept) and the rate of change of each line, hence or otherwise, write down the function represented by each of the following graphs: (ii) Intersect on y-axis = 2 y = mx + c

4. By finding the point where the line crosses the y-axis (y-intercept) and the rate of change of each line, hence or otherwise, write down the function represented by each of the following graphs: (iii) Intersect on y-axis = 4 y = mx + c

4. By finding the point where the line crosses the y-axis (y-intercept) and the rate of change of each line, hence or otherwise, write down the function represented by each of the following graphs: (iv) Intersect on y-axis = – 1 y = mx + c

5. Using the same axes and scale, graph the functions: f (x) = x + 2 and g(x) = 2x − 1 in the domain − 2 ≤ x ≤ 5. g(x) = 2x – 1 f (x) = x + 2 f (– 2) = – 2 + 2 g(–2) = 2(– 2) – 1 f (– 2) = 0 (– 2, 0) = – 4 – 1 g(–2) = – 5 (– 2, – 5) f (5) = 5 + 2 f (5) = 7 (5, 7) g(5) = 2(5) – 1 = 10 – 1 g(5) = 9 (5, 9)

5. Using the same axes and scale, graph the functions: f (x) = x + 2 and g(x) = 2x − 1 in the domain − 2 ≤ x ≤ 5.

5. Using the same axes and scale, graph the functions: f (x) = x + 2 and g(x) = 2x − 1 in the domain − 2 ≤ x ≤ 5. Hence, or otherwise, find the coordinates of the point where f (x) = g(x). f (x) = g(x) at the point where the two lines intersect. This occurs at (3, 5).

6. Using the same axes and scale, graph the functions: h(x) = 2x − 3 and k(x) = 2 − x in the domain − 1 ≤ x ≤ 5. h(x) = 2x – 3 h(x) = 2x – 3 h(– 1) = 2(– 1) – 3 h(5) = 2(5) – 3 = –2 – 3 = 10 – 3 h(– 1) = –5 (– 1, – 5) h(5) = 7 (5, 7)

6. Using the same axes and scale, graph the functions: h(x) = 2x − 3 and k(x) = 2 − x in the domain − 1 ≤ x ≤ 5.

6. Using the same axes and scale, graph the functions: h(x) = 2x − 3 and k(x) = 2 − x in the domain − 1 ≤ x ≤ 5. Hence, or otherwise, find the coordinates of the point where h(x) = k(x). h(x) = k(x) at the point where the two lines intersect. This occurs at (2, 1).

7. Sakthi is saving for a holiday. He opens a savings account and deposits €100. He then deposits €30 per week for the next 25 weeks (week number 1 is the first week). (i) Copy and complete the following table: The account starts with €100 at week 0 and then €30 is added each week. Week 1: €100 + €30 = €130 Week 2: €130 + €30 = €160, etc

7. Sakthi is saving for a holiday. He opens a savings account and deposits €100. He then deposits €30 per week for the next 25 weeks (week number 1 is the first week). (ii) Write a function to represent the account balance, in terms of the week number.Explain any letters used. Initial value = 100 Rate of change = 30 W = Week number, B = Balance B = 100 + 30W

7. Sakthi is saving for a holiday. He opens a savings account and deposits €100. He then deposits €30 per week for the next 25 weeks (week number 1 is the first week). (iii) Hence, or otherwise, graph the function for the 25 weeks. W = 25 B = 100 + 30(25) = 100 + 750 B = 850

7. Sakthi is saving for a holiday. He opens a savings account and deposits €100. He then deposits €30 per week for the next 25 weeks (week number 1 is the first week). (iv) The point (9,370) is on the graph. Write a sentence that describes the meaning of this ordered pair. After 9 weeks, the account balance will be €370.

7. Sakthi is saving for a holiday. He opens a savings account and deposits €100. He then deposits €30 per week for the next 25 weeks (week number 1 is the first week). (v) Use your graph to find the account balance after seven weeks. Go to 7 on the x-axis. Go up to the graph (7, 310)  After 7 weeks, account balance = €310.

7. Sakthi is saving for a holiday. He opens a savings account and deposits €100. He then deposits €30 per week for the next 25 weeks (week number 1 is the first week). (vi) If the holiday costs €610, after how many weeks will Sakthi have enough money saved? (17, 610)  Balance = €610 after 17 weeks.

8. A gym can take a maximum of 2,500 members. When it first opens, 670 people join up immediately. The manager finds that a further 30 people join up each week. (i) What is the rate of change of this function? The membership will increase by 30 members each week. Rate of change = 30

8. A gym can take a maximum of 2,500 members. When it first opens, 670 people join up immediately. The manager finds that a further 30 people join up each week. (ii) Write a function to represent the membership of the gym, in terms of the number of weeks.Clearly explain any letters you use. Initial value = 670 Rate of change = 30 W = week number, N = number of members. N = 670 + 30W

8. A gym can take a maximum of 2,500 members. When it first opens, 670 people join up immediately. The manager finds that a further 30 people join up each week. (iii) What will the total membership be after 12 weeks? N = 670 + 30(12) = 670 + 360 N = 1030

8. A gym can take a maximum of 2,500 members. When it first opens, 670 people join up immediately. The manager finds that a further 30 people join up each week. (iv) How many weeks will it take before the gym is at maximum membership? Maximum membership = 2500: N = 670 + 30W 2500 = 670 + 30W 1830 = 30W 61 = W 61 Weeks, 1 Year 9 Weeks, before full membership is achieved.

8. A gym can take a maximum of 2,500 members. When it first opens, 670 people join up immediately. The manager finds that a further 30 people join up each week. (v) Do you think the function you found in part (i) is an accurate description for the membership numbers of a gym? Give a reason for your answer. No. Some members will leave the gym during the period which is not accounted for. Also, certain months, for example January, may be busier than others.

9. A taxi company charges a flat fee of €4·50 and then €1·50 per kilometre travelled. (i) What is the rate of change (slope) of this function? The cost is increasing by €1·50 per kilometre, so the rate of change is 1·5.

9. A taxi company charges a flat fee of €4·50 and then €1·50 per kilometre travelled. (ii) Write a function to represent the cost of a journey with this taxi company. Clearly explain any letters used. Initial value = €4·50 Rate of change = €1·50 C = Cost of Journey, D = distance travelled. C = 4·5 + 1·5D

9. A taxi company charges a flat fee of €4·50 and then €1·50 per kilometre travelled. (iii) Draw a graph of this function, to represent the cost of a journey up to 30 km. Cost for 30 km: C = 4·5 + 1·5(30) = 4·5 + 45 = 49·5

9. A taxi company charges a flat fee of €4·50 and then €1·50 per kilometre travelled. (iv) The point (5, 12) is on the graph. Write a sentence that describes the meaning of this ordered pair. A 5 km taxi journey has a cost of €12.

9. A taxi company charges a flat fee of €4·50 and then €1·50 per kilometre travelled. (v) Use this graph to find the cost of a 16 km journey. (16, 28·5)  16 km journey = €28·50 cost.

9. A taxi company charges a flat fee of €4·50 and then €1·50 per kilometre travelled. (vi) Bertrand takes a trip with this taxi company and he is charged €37·50. Use your graph to find the length of Bertrand’s trip. (22, 37·5)  €37·50 cost = 22 km journey.

10. Cooking instructions for roasting a chicken are 20 minutes plus 25 minutes per 500 g weight of the chicken. (i) Write a function to represent the cooking time for a chicken, in terms of its weight in kilograms. Explain any letters used. Initial value = 20 minutes Rate of change = 25 minutes per 500 g  50 minutes per 1 kg T = cooking time in minutes; W = weight in kilograms T = 20 + 50W

10. Cooking instructions for roasting a chicken are 20 minutes plus 25 minutes per 500 g weight of the chicken. (ii) Draw a graph of this function to represent the cooking time for a chicken up to the weight of 3 kg. Find the time for a 3kg chicken: T = 20 + 50W T = 20 + 50(3) = 20 + 150 T = 170 minutes

10. Cooking instructions for roasting a chicken are 20 minutes plus 25 minutes per 500 g weight of the chicken. (iii) Use your graph to find how long it would take to cook a 1·5 kg chicken. Go to 1·5 on the x-axis and find the corresponding value on the y-axis: 1·5 kg → 95 minutes

10. Cooking instructions for roasting a chicken are 20 minutes plus 25 minutes per 500 g weight of the chicken. (iv) Use your graph to estimate the weight of a chicken, which takes 130 minutes to cook. Go to 130 on the y-axis and find the corresponding value on thex-axis. 130 minutes → 2·2 kg

Math Functions and Graphs Exploration