Year 12 · Algebra

A4Types of relationships 0 / 14

YR 11

A1 A2 M1 M2 F1 S1 S2

YR 12

A4 M6 M7 F4 F5 S4 S5 N2 N3

💡 Quick tips — memory joggers

For simultaneous equations, the solution is where the two lines cross; the break-even point is where cost $=$ revenue, so profit is zero.
Exponential $y = ka^{x}$: base $a > 1$ means growth, $0 < a < 1$ means decay. “Up $r\%$” gives base $1 + \tfrac{r}{100}$; “down $r\%$” gives $1 - \tfrac{r}{100}$.
A parabola’s axis of symmetry sits midway between the $x$-intercepts. It opens up (min) when $a > 0$ and down (max) when $a < 0$.
Inverse variation $y = \dfrac{k}{x}$: as $x$ grows, $y$ shrinks (more people → smaller share). Always find $k$ from a given pair first.

Level 1 · Fluency

Two lines are drawn on the same axes. What does their point of intersection represent?

Level 2 · Application

The lines $y = x + 1$ and $y = -x + 5$ are graphed. Find their point of intersection algebraically to check a graphical solution.
Level 3 · Further Application
A farm has only chickens ($x$) and sheep ($y$). Altogether there are 28 animals, so $x + y = 28$. Each chicken has 2 legs and each sheep 4 legs, giving 82 legs in total.
1. (a) Write a second equation for the legs. (1 mark)
2. (b) Solve the two equations to find the number of chickens and sheep. (2 marks)
15 chickens and 13 sheep

(a) $2x + 4y = 82$.
(b) From $x + y = 28$, $x = 28 - y$. Substitute: $2(28 - y) + 4y = 82 \Rightarrow 56 + 2y = 82 \Rightarrow y = 13$. So 15 chickens and 13 sheep.
A ground sells adult ($x$) and child ($y$) tickets. There are 40 tickets sold in all, so $x + y = 40$. Adult tickets cost $12 and child tickets $7, giving $405 in total.
1. (a) Write a second equation for the takings. (1 mark)
2. (b) Solve to find the number of each type of ticket. (2 marks)
25 adult and 15 child tickets

(a) $12x + 7y = 405$.
(b) From $x = 40 - y$: $12(40 - y) + 7y = 405 \Rightarrow 480 - 5y = 405 \Rightarrow 5y = 75 \Rightarrow y = 15$. So 25 adult and 15 child tickets.
A money box holds only $2 coins ($x$) and $1 coins ($y$). There are 24 coins in all, so $x + y = 24$, and their total value is $37.
1. (a) Write a second equation for the value. (1 mark)
2. (b) Solve to find how many of each coin there are. (2 marks)
13 two-dollar coins and 11 one-dollar coins

(a) $2x + y = 37$.
(b) Subtract $x + y = 24$: $(2x + y) - (x + y) = 37 - 24 \Rightarrow x = 13$, then $y = 11$. So 13 two-dollar coins and 11 one-dollar coins.
Level 1 · Fluency

Two coffees and one tea cost $11. Write this as an equation using $c$ and $t$.

Level 2 · Application

At a stall, 3 pies and 2 drinks cost $23, and 1 pie and 2 drinks cost $13. Write a pair of simultaneous equations ($p$ = pie, $d$ = drink).
Level 3 · Further Application
Using $3p + 2d = 23$ and $p + 2d = 13$:
1. (a) Solve for the price of a pie and a drink. (2 marks)
2. (b) State the cost of 2 pies and 3 drinks. (1 mark)
$22

(a) Subtract: $(3p + 2d) - (p + 2d) = 23 - 13 \Rightarrow 2p = 10 \Rightarrow p =$ $5. Then $5 + 2d = 13 \Rightarrow d =$ $4.
(b) $2 \times 5 + 3 \times 4 =$ $22.
Using $5a + 2b = 9$ and $3a + 2b = 7$:
1. (a) Solve for the price of an apple and a banana. (2 marks)
2. (b) State the cost of 4 apples and 5 bananas. (1 mark)
$14

(a) Subtract: $(5a + 2b) - (3a + 2b) = 9 - 7 \Rightarrow 2a = 2 \Rightarrow a =$ $1. Then $3(1) + 2b = 7 \Rightarrow b =$ $2.
(b) $4 \times 1 + 5 \times 2 =$ $14.
Using $2m + 3p = 61$ and $2m + p = 43$:
1. (a) Solve for the price of a movie ticket and a popcorn. (2 marks)
2. (b) State the cost of 3 movie tickets and 2 popcorns. (1 mark)
$69

(a) Subtract: $(2m + 3p) - (2m + p) = 61 - 43 \Rightarrow 2p = 18 \Rightarrow p =$ $9. Then $2m + 9 = 43 \Rightarrow 2m = 34 \Rightarrow m =$ $17.
(b) $3 \times 17 + 2 \times 9 =$ $69.
Level 1 · Fluency

At the break-even point of a business, how do total cost and total revenue compare?

Level 2 · Application

A stall’s costs are $C = 200 + 4n$ and revenue is $R = 9n$, where $n$ is the number of items. Find the break-even number of items.
Level 3 · Further Application
A business has costs $C = 500 + 6n$ and revenue $R = 11n$.
1. (a) Find the break-even number of items. (2 marks)
2. (b) State whether making 80 items gives a profit or loss, and by how much. (1 mark)
loss of $100

(a) $500 + 6n = 11n \Rightarrow 500 = 5n \Rightarrow n = 100$ items.
(b) At $n = 80$ (below break-even): $R = 880$, $C = 980$, so a loss of $100.
A business has costs $C = 800 + 7n$ and revenue $R = 15n$.
1. (a) Find the break-even number of items. (2 marks)
2. (b) State whether making 150 items gives a profit or loss, and by how much. (1 mark)
profit of $400

(a) $800 + 7n = 15n \Rightarrow 800 = 8n \Rightarrow n = 100$ items.
(b) At $n = 150$ (above break-even): $R = 2250$, $C = 1850$, so a profit of $400.
A business has costs $C = 1200 + 10n$ and revenue $R = 25n$.
1. (a) Find the break-even number of items. (2 marks)
2. (b) State whether making 60 items gives a profit or loss, and by how much. (1 mark)
loss of $300

(a) $1200 + 10n = 25n \Rightarrow 1200 = 15n \Rightarrow n = 80$ items.
(b) At $n = 60$ (below break-even): $R = 1500$, $C = 1800$, so a loss of $300.
Level 1 · Fluency

A population is modelled by $P = 3000(1.15)^{t}$. State the initial population ($t = 0$).

Level 2 · Application

A population follows $P = 3000(1.15)^{t}$, where $t$ is in years. Calculate the population after 5 years, to the nearest whole number.
Level 3 · Further Application
A medicine in the body follows $A = 40(0.8)^{t}$ mg, where $t$ is in hours.
1. (a) Find the amount remaining after 4 hours, to one decimal place. (2 marks)
2. (b) State whether this is growth or decay, and how you can tell from the equation. (1 mark)
16.4 mg

(a) $A = 40(0.8)^{4} =$ 16.4 mg.
(b) Decay — the base 0.8 is less than 1, so the amount decreases each hour.
A medicine in the body follows $A = 60(0.75)^{t}$ mg, where $t$ is in hours.
1. (a) Find the amount remaining after 3 hours, to one decimal place. (2 marks)
2. (b) State whether this is growth or decay, and how you can tell from the equation. (1 mark)
25.3 mg

(a) $A = 60(0.75)^{3} = 60 \times 0.421875 =$ 25.3 mg.
(b) Decay — the base 0.75 is less than 1, so the amount decreases each hour.
A medicine in the body follows $A = 100(0.85)^{t}$ mg, where $t$ is in hours.
1. (a) Find the amount remaining after 5 hours, to one decimal place. (2 marks)
2. (b) State whether this is growth or decay, and how you can tell from the equation. (1 mark)
44.4 mg

(a) $A = 100(0.85)^{5} = 100 \times 0.443705 =$ 44.4 mg.
(b) Decay — the base 0.85 is less than 1, so the amount decreases each hour.
Level 1 · Fluency

The graph of $y = 2^{x}$ cuts the $y$-axis at what point?

Level 2 · Application

Describe the shape of the graph of $y = 3^{x}$, including what happens as $x$ becomes large negative.
Level 3 · Further Application
Consider $y = 2^{x}$ and $y = 2^{-x}$.
1. (a) State the point both graphs share. (1 mark)
2. (b) Describe how the two graphs differ. (2 marks)
(a) Both pass through $(0, 1)$.
(b) $y = 2^{x}$ increases as $x$ increases (growth), while $y = 2^{-x}$ decreases as $x$ increases (decay); each is the mirror image of the other in the $y$-axis.
Consider $y = 3^{x}$ and $y = 3^{-x}$.
1. (a) State the point both graphs share. (1 mark)
2. (b) Describe how the two graphs differ. (2 marks)
(a) Both pass through $(0, 1)$.
(b) $y = 3^{x}$ increases as $x$ increases (growth), while $y = 3^{-x}$ decreases as $x$ increases (decay); each is the mirror image of the other in the $y$-axis.
Explain why the graph of $y = a^{-x}$ is the reflection of $y = a^{x}$ in the $y$-axis (for $a > 0$).
Level 1 · Fluency

For a population model $P = ka^{t}$, what does the value of $P$ at $t = 0$ represent?

Level 2 · Application

The value of a car is $V = 30\,000(0.85)^{t}$ dollars, $t$ years after purchase. Interpret the $y$-intercept.
Level 3 · Further Application
A bacterial culture is modelled by $N = 500(2)^{t}$, where $t$ is in hours.
1. (a) Interpret the value 500 in the model. (1 mark)
2. (b) Interpret the base 2 in the context. (2 marks)
(a) 500 is the initial number of bacteria (at $t = 0$).
(b) The base 2 means the number of bacteria doubles every hour.
An investment is modelled by $A = 4000(1.06)^{t}$ dollars, where $t$ is in years.
1. (a) Interpret the value 4000 in the model. (1 mark)
2. (b) Interpret the base 1.06 in the context. (2 marks)
(a) 4000 is the initial investment ($4000 at $t = 0$).
(b) The base 1.06 means the investment grows by 6% each year.
A radioactive sample is modelled by $M = 80(0.5)^{t}$ grams, where $t$ is in days.
1. (a) Interpret the value 80 in the model. (1 mark)
2. (b) Interpret the base 0.5 in the context. (2 marks)
(a) 80 is the initial mass of the sample (80 g at $t = 0$).
(b) The base 0.5 means the mass halves each day.
Level 1 · Fluency

A quantity starts at 800 and doubles each year. Write an exponential model for its value $V$ after $t$ years.

Level 2 · Application

A $6000 investment grows by 5% each year. Write an exponential model for its value $V$ after $t$ years, and use it to find the value after 3 years.
Level 3 · Further Application
A town’s population is 8000 and decreases by 4% each year.
1. (a) Write an exponential model for the population $P$ after $t$ years. (1 mark)
2. (b) Find the population after 6 years, to the nearest whole number. (2 marks)
6264

(a) $P = 8000(0.96)^{t}$.
(b) $P = 8000(0.96)^{6} =$ 6264.
A car worth $24 000 loses 15% of its value each year.
1. (a) Write an exponential model for the value $V$ after $t$ years. (1 mark)
2. (b) Find the value after 4 years, to the nearest dollar. (2 marks)
$12 528

(a) $V = 24\,000(0.85)^{t}$.
(b) $V = 24\,000(0.85)^{4} = 24\,000 \times 0.522006 =$ $12 528.
A colony of 5000 insects decreases by 20% each week.
1. (a) Write an exponential model for the number $N$ after $t$ weeks. (1 mark)
2. (b) Find the number after 3 weeks. (2 marks)
2560 insects

(a) $N = 5000(0.8)^{t}$.
(b) $N = 5000(0.8)^{3} = 5000 \times 0.512 =$ 2560 insects.
Level 1 · Fluency

For the quadratic $y = x^{2} - 4x + 3$, state the value of $y$ when $x = 0$.

Level 2 · Application

The braking distance $d$ (metres) of a car varies with the square of its speed $v$ (km/h): $d = 0.01v^{2}$. Find the braking distance at 50 km/h and at 100 km/h.
Level 3 · Further Application
A ball’s height is $h = -5t^{2} + 30t$ metres, where $t$ is in seconds.
1. (a) Find the height after 1 second. (1 mark)
2. (b) Find the two times when the ball is at height 40 m. (2 marks)
25 m; 4 s

(a) $h = -5(1)^{2} + 30(1) =$ 25 m.
(b) $-5t^{2} + 30t = 40 \Rightarrow 5t^{2} - 30t + 40 = 0 \Rightarrow t^{2} - 6t + 8 = 0 \Rightarrow (t - 2)(t - 4) = 0$, so $t = 2$ s and $t =$ 4 s.
A ball’s height is $h = -5t^{2} + 40t$ metres, where $t$ is in seconds.
1. (a) Find the height after 2 seconds. (1 mark)
2. (b) Find the two times when the ball is at height 60 m. (2 marks)
60 m; 6 s

(a) $h = -5(2)^{2} + 40(2) = -20 + 80 =$ 60 m.
(b) $-5t^{2} + 40t = 60 \Rightarrow 5t^{2} - 40t + 60 = 0 \Rightarrow t^{2} - 8t + 12 = 0 \Rightarrow (t - 2)(t - 6) = 0$, so $t = 2$ s and $t =$ 6 s.
A ball’s height is $h = -5t^{2} + 25t$ metres, where $t$ is in seconds.
1. (a) Find the height after 1 second. (1 mark)
2. (b) Find the two times when the ball is at height 30 m. (2 marks)
20 m; 3 s

(a) $h = -5(1)^{2} + 25(1) = -5 + 25 =$ 20 m.
(b) $-5t^{2} + 25t = 30 \Rightarrow 5t^{2} - 25t + 30 = 0 \Rightarrow t^{2} - 5t + 6 = 0 \Rightarrow (t - 2)(t - 3) = 0$, so $t = 2$ s and $t =$ 3 s.
Level 1 · Fluency

A parabola has its lowest point at $(3, -2)$. What is this point called?

Level 2 · Application

A parabola has $x$-intercepts at $x = 2$ and $x = 8$. State the equation of its axis of symmetry.
Level 3 · Further Application
A parabola passes through $x$-intercepts at $x = -1$ and $x = 5$, opening upwards.
1. (a) State the equation of the axis of symmetry. (1 mark)
2. (b) State the $x$-coordinate of the turning point and whether it is a maximum or minimum. (2 marks)
minimum

(a) $x = \dfrac{-1 + 5}{2} = x = 2$.
(b) The turning point is at $x = 2$; since the parabola opens upwards it is a minimum.
A parabola passes through $x$-intercepts at $x = 0$ and $x = 8$, opening downwards.
1. (a) State the equation of the axis of symmetry. (1 mark)
2. (b) State the $x$-coordinate of the turning point and whether it is a maximum or minimum. (2 marks)
maximum

(a) $x = \dfrac{0 + 8}{2} = x = 4$.
(b) The turning point is at $x = 4$; since the parabola opens downwards it is a maximum.
A parabola passes through $x$-intercepts at $x = -3$ and $x = 1$, opening upwards.
1. (a) State the equation of the axis of symmetry. (1 mark)
2. (b) State the $x$-coordinate of the turning point and whether it is a maximum or minimum. (2 marks)
minimum

(a) $x = \dfrac{-3 + 1}{2} = x = -1$.
(b) The turning point is at $x = -1$; since the parabola opens upwards it is a minimum.
Level 1 · Fluency

Does the parabola $y = -x^{2} + 4$ open upwards or downwards?

Level 2 · Application

For $y = x^{2} - 2x - 3$, find the $y$-intercept and the $x$-intercepts.
Level 3 · Further Application
Consider $y = x^{2} - 6x + 5$.
1. (a) Find the $x$-intercepts. (1 mark)
2. (b) Find the coordinates of the turning point. (2 marks)
$(3, -4)$

(a) $(x - 1)(x - 5) = 0 \Rightarrow x = 1$ and $x = 5$.
(b) Axis of symmetry $x = \dfrac{1 + 5}{2} = 3$; $y = 3^{2} - 6(3) + 5 = -4$. Turning point $(3, -4)$.
Consider $y = x^{2} - 4x - 5$.
1. (a) Find the $x$-intercepts. (1 mark)
2. (b) Find the coordinates of the turning point. (2 marks)
$(2, -9)$

(a) $(x - 5)(x + 1) = 0 \Rightarrow x = 5$ and $x = -1$.
(b) Axis of symmetry $x = \dfrac{-1 + 5}{2} = 2$; $y = 2^{2} - 4(2) - 5 = -9$. Turning point $(2, -9)$.
Consider $y = x^{2} + 2x - 3$.
1. (a) Find the $x$-intercepts. (1 mark)
2. (b) Find the coordinates of the turning point. (2 marks)
$(-1, -4)$

(a) $(x + 3)(x - 1) = 0 \Rightarrow x = -3$ and $x = 1$.
(b) Axis of symmetry $x = \dfrac{-3 + 1}{2} = -1$; $y = (-1)^{2} + 2(-1) - 3 = -4$. Turning point $(-1, -4)$.
Level 1 · Fluency

A revenue parabola has its maximum at $(50, 8000)$. What does the $y$-value 8000 represent?

Level 2 · Application

A ball’s height is $h = -5t^{2} + 20t$ metres. The turning point is at $(2, 20)$. Interpret this point.
Level 3 · Further Application
A concert promoter’s takings are $P = -40x^{2} + 3200x + 5000$, where $x$ is the ticket price rise in dollars.
1. (a) The turning point is at $x = 40$. Interpret this value. (1 mark)
2. (b) Calculate the maximum takings. (2 marks)
$69 000

(a) A ticket price rise of $40 gives the maximum takings.
(b) $P = -40(40)^{2} + 3200(40) + 5000 = -64\,000 + 128\,000 + 5000 =$ $69 000.
A promoter’s takings are $P = -20x^{2} + 800x + 3000$, where $x$ is the ticket price rise in dollars.
1. (a) The turning point is at $x = 20$. Interpret this value. (1 mark)
2. (b) Calculate the maximum takings. (2 marks)
$11 000

(a) A ticket price rise of $20 gives the maximum takings.
(b) $P = -20(20)^{2} + 800(20) + 3000 = -8000 + 16\,000 + 3000 =$ $11 000.
A promoter’s takings are $P = -10x^{2} + 600x + 2000$, where $x$ is the ticket price rise in dollars.
1. (a) The turning point is at $x = 30$. Interpret this value. (1 mark)
2. (b) Calculate the maximum takings. (2 marks)
$11 000

(a) A ticket price rise of $30 gives the maximum takings.
(b) $P = -10(30)^{2} + 600(30) + 2000 = -9000 + 18\,000 + 2000 =$ $11 000.
Level 1 · Fluency

A ball’s height model gives $h = -5$ m at $t = 5$ s. Why is this value not sensible?

Level 2 · Application

A ball’s height is $h = -5t^{2} + 20t$. It lands when $h = 0$. Find the sensible domain of $t$.
Level 3 · Further Application
For the takings model $P = -40x^{2} + 3200x + 5000$ ($x$ = ticket price rise in dollars), takings must stay above zero and the price rise cannot be negative.
1. (a) State a sensible lower bound for $x$. (1 mark)
2. (b) Explain why very large values of $x$ are not sensible. (2 marks)
(a) $x \ge 0$ (you cannot have a negative price rise).
(b) For large $x$ the parabola turns downward and takings fall (eventually below zero), because raising the price too far means too few tickets sold — so only $x$-values near the turning point are realistic.
A rectangular garden has area $A = x(20 - x)$ m², where $x$ is one side length in metres.
1. (a) State a sensible lower bound for $x$. (1 mark)
2. (b) Explain why $x$ cannot be 20 or more. (2 marks)
(a) $x > 0$ (a side length must be positive).
(b) At $x = 20$ the area is $20(20 - 20) = 0$, and for $x > 20$ the factor $(20 - x)$ is negative, giving a negative (impossible) area — so a sensible domain is $0 < x < 20$.
A projectile’s height is $h = -5t^{2} + 40t$ metres, where $t$ is in seconds.
1. (a) State a sensible lower bound for $t$. (1 mark)
2. (b) Explain why very large values of $t$ are not sensible. (2 marks)
(a) $t \ge 0$ (time cannot be negative).
(b) The projectile lands when $h = 0$: $-5t(t - 8) = 0 \Rightarrow t = 8$ s. Beyond this the model gives negative height, which is impossible, so only $0 \le t \le 8$ is realistic.
Level 1 · Fluency

In the relationship $y = \dfrac{k}{x}$, what happens to $y$ as $x$ increases?

Level 2 · Application

$y$ varies inversely with $x$, and $y = 8$ when $x = 3$. Find $k$ in $y = \dfrac{k}{x}$, and hence $y$ when $x = 6$.
Level 3 · Further Application
The time $t$ to fill a pool varies inversely with the number of pumps $p$. With 4 pumps it takes 9 hours.
1. (a) Find the relationship between $t$ and $p$. (2 marks)
2. (b) Find the time taken with 6 pumps. (1 mark)
6 hours

(a) $t = \dfrac{k}{p}$; $9 = \dfrac{k}{4} \Rightarrow k = 36$, so $t = \dfrac{36}{p}$.
(b) $t = \dfrac{36}{6} =$ 6 hours.
The current $I$ varies inversely with the resistance $R$. When $R = 5$, $I = 12$.
1. (a) Find the relationship between $I$ and $R$. (2 marks)
2. (b) Find the current when $R = 15$. (1 mark)
4

(a) $I = \dfrac{k}{R}$; $12 = \dfrac{k}{5} \Rightarrow k = 60$, so $I = \dfrac{60}{R}$.
(b) $I = \dfrac{60}{15} =$ 4.
The loudness $L$ heard varies inversely with the distance $d$ from a speaker. At 3 m the loudness reads 24 units.
1. (a) Find the relationship between $L$ and $d$. (2 marks)
2. (b) Find the loudness at 8 m. (1 mark)
9 units

(a) $L = \dfrac{k}{d}$; $24 = \dfrac{k}{3} \Rightarrow k = 72$, so $L = \dfrac{72}{d}$.
(b) $L = \dfrac{72}{8} =$ 9 units.
Level 1 · Fluency

8 people share a pizza equally, each getting 2 slices’ worth. If 16 people share the same pizza, does each get more or less?

Level 2 · Application

A job takes 6 workers 20 days. Assuming the time varies inversely with the number of workers, how long would 8 workers take?
Level 3 · Further Application
The thickness of a fixed volume of dough varies inversely with the area it is spread over. Spread over 600 cm² it is 4 mm thick.
1. (a) Find the relationship between thickness $T$ and area $A$. (2 marks)
2. (b) Find the thickness if the same dough is spread over 800 cm². (1 mark)
3 mm

(a) $T = \dfrac{k}{A}$; $4 = \dfrac{k}{600} \Rightarrow k = 2400$, so $T = \dfrac{2400}{A}$.
(b) $T = \dfrac{2400}{800} =$ 3 mm.
For a fixed amount of gas, the pressure $P$ varies inversely with the volume $V$. At 5 L the pressure is 60 kPa.
1. (a) Find the relationship between $P$ and $V$. (2 marks)
2. (b) Find the pressure when the volume is 4 L. (1 mark)
75 kPa

(a) $P = \dfrac{k}{V}$; $60 = \dfrac{k}{5} \Rightarrow k = 300$, so $P = \dfrac{300}{V}$.
(b) $P = \dfrac{300}{4} =$ 75 kPa.
A fixed amount of paint is spread on a wall; its thickness $T$ varies inversely with the area $A$. Over 500 cm² it is 6 mm thick.
1. (a) Find the relationship between $T$ and $A$. (2 marks)
2. (b) Find the thickness if the same paint is spread over 750 cm². (1 mark)
4 mm

(a) $T = \dfrac{k}{A}$; $6 = \dfrac{k}{500} \Rightarrow k = 3000$, so $T = \dfrac{3000}{A}$.
(b) $T = \dfrac{3000}{750} =$ 4 mm.