Year 12 · Algebra
Quick tips — memory joggers
Parabola
\(y = ax^{2} + bx + c\)
The highest power of \(x\) is squared.
Hyperbola
\(y = \dfrac{k}{x}\)
\(x\) is written on the denominator.
Exponential
\(y = ka^{x}\)
\(x\) is written as a power.
Simultaneous & break-even
Exponential models
Quadratics & parabolas
Inverse variation
Two branches; the axes (dashed) are asymptotes.
Level 1 · Fluency
Two lines are drawn on the same axes. What does their point of intersection represent?
The solution (the \(x\) and \(y\) values that satisfy both equations at once).
Two lines are graphed and meet at the point \((5, 2)\). Write down the solution of the pair of simultaneous equations.
\(x = 5\) and \(y = 2\) (the coordinates of the intersection point).
If two lines are parallel and never cross, how many solutions does the pair of simultaneous equations have?
None — the lines never meet, so there is no point that satisfies both equations.
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.
\((2, 3)\)
\(x + 1 = -x + 5 \Rightarrow 2x = 4 \Rightarrow x = 2\), \(y = 3\). Intersection at \((2, 3)\).
The lines \(y = 2x - 1\) and \(y = x + 2\) are graphed. Find their point of intersection algebraically.
\((3, 5)\)
\(2x - 1 = x + 2 \Rightarrow x = 3\), then \(y = 3 + 2 = 5\). Intersection at \((3, 5)\).
The lines \(y = 3x + 2\) and \(y = -x + 6\) are graphed. Find their point of intersection algebraically.
\((1, 5)\)
\(3x + 2 = -x + 6 \Rightarrow 4x = 4 \Rightarrow x = 1\), then \(y = 3(1) + 2 = 5\). Intersection at \((1, 5)\).
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.
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.
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.
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 \(t\).
\(2c + t = 11\).
Three pens and two rulers cost $16. Write this as an equation using \(p\) and \(r\).
\(3p + 2r = 16\).
One adult ticket and four child tickets cost $50. Write this as an equation using \(a\) and \(c\).
\(a + 4c = 50\).
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).
\(3p + 2d = 23\) and \(p + 2d = 13\).
At a fruit stall, 5 apples and 2 bananas cost $9, and 3 apples and 2 bananas cost $7. Write a pair of simultaneous equations (\(a\) = apple, \(b\) = banana).
\(5a + 2b = 9\) and \(3a + 2b = 7\).
At a cinema, 2 movie tickets and 3 popcorns cost $61, and 2 movie tickets and 1 popcorn cost $43. Write a pair of simultaneous equations (\(m\) = movie ticket, \(p\) = popcorn).
\(2m + 3p = 61\) and \(2m + p = 43\).
Level 3 · Further Application
Using \(3p + 2d = 23\) and \(p + 2d = 13\):
$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 \(3a + 2b = 7\):
$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 \(2m + p = 43\):
$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?
They are equal — cost = revenue, so profit is zero.
At a business's break-even point, what is the value of the profit?
Zero — there is neither a profit nor a loss.
If a business's revenue is greater than its cost, is it making a profit or a loss?
A profit (revenue exceeds cost, so profit is positive).
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.
40 items
\(200 + 4n = 9n \Rightarrow 200 = 5n \Rightarrow n =\) 40 items.
A workshop’s costs are \(C = 300 + 5n\) and revenue is \(R = 8n\), where \(n\) is the number of items. Find the break-even number of items.
100 items
\(300 + 5n = 8n \Rightarrow 300 = 3n \Rightarrow n =\) 100 items.
A market seller’s costs are \(C = 150 + 3n\) and revenue is \(R = 6n\), where \(n\) is the number of items. Find the break-even number of items.
50 items
\(150 + 3n = 6n \Rightarrow 150 = 3n \Rightarrow n =\) 50 items.
Level 3 · Further Application
A business has costs \(C = 500 + 6n\) and revenue \(R = 11n\).
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 \(R = 15n\).
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 \(R = 25n\).
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 (\(t = 0\)).
3000
\(P = 3000(1.15)^{0} =\) 3000.
A quantity is modelled by \(A = 500(1.2)^{t}\). State the initial value (\(t = 0\)).
500
\(A = 500(1.2)^{0} =\) 500.
A quantity is modelled by \(N = 1200(0.9)^{t}\). State the initial value (\(t = 0\)).
1200
\(N = 1200(0.9)^{0} =\) 1200.
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.
6034
\(P = 3000(1.15)^{5} =\) 6034.
An investment follows \(V = 2000(1.08)^{t}\) dollars, where \(t\) is in years. Calculate its value after 4 years, to the nearest whole number.
$2721
\(V = 2000(1.08)^{4} = 2000 \times 1.36049 =\) $2721.
A quantity follows \(Q = 4000(1.1)^{t}\), where \(t\) is in years. Calculate its value after 3 years.
5324
\(Q = 4000(1.1)^{3} = 4000 \times 1.331 =\) 5324.
Level 3 · Further Application
A medicine in the body follows \(A = 40(0.8)^{t}\) mg, where \(t\) is in hours.
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.
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.
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?
\((0, 1)\), since \(2^{0} = 1\).
At what point does the graph of \(y = 5^{x}\) cross the \(y\)-axis?
\((0, 1)\), since \(5^{0} = 1\).
What is the equation of the horizontal asymptote of the graph of \(y = 4^{x}\)?
\(y = 0\) (the \(x\)-axis) — the curve approaches it but never touches it.
Level 2 · Application
Describe the shape of the graph of \(y = 3^{x}\), including what happens as \(x\) becomes large negative.
It rises steeply for positive \(x\) (growth), passes through \((0, 1)\), and approaches (but never touches) the \(x\)-axis as \(x \to -\infty\) (a horizontal asymptote at \(y = 0\)).
Describe the shape of the graph of \(y = 2^{-x}\), including what happens as \(x\) becomes large positive.
It falls (decay), passing through \((0, 1)\), and approaches (but never touches) the \(x\)-axis as \(x \to +\infty\) (a horizontal asymptote at \(y = 0\)).
Does the graph of \(y = 10^{x}\) ever take a negative \(y\)-value? Explain.
No — \(10^{x} > 0\) for every value of \(x\), so the curve stays above the \(x\)-axis (its asymptote is \(y = 0\)).
Level 3 · Further Application
Consider \(y = 2^{x}\) and \(y = 2^{-x}\).
(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 \(y = 3^{-x}\).
(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 \(a > 0\)).
Replacing \(x\) with \(-x\) reflects any graph in the \(y\)-axis. Since \(y = a^{-x}\) is \(y = a^{x}\) with \(x\) replaced by \(-x\), the two curves are mirror images in the \(y\)-axis, meeting at \((0, 1)\).
Level 1 · Fluency
For a population model \(P = ka^{t}\), what does the value of \(P\) at \(t = 0\) represent?
The initial (starting) population.
For a car-value model \(V = ka^{t}\), what does the \(y\)-intercept (\(t = 0\)) represent?
The initial (purchase) value of the car.
In the model \(N = 250(1.3)^{t}\), what does the number 250 represent?
The initial amount (the value when \(t = 0\)).
Level 2 · Application
The value of a car is \(V = 30\,000(0.85)^{t}\) dollars, \(t\) years after purchase. Interpret the \(y\)-intercept.
At \(t = 0\), \(V = 30\,000\), so the \(y\)-intercept ($30 000) is the car’s purchase price.
The value of a phone is \(V = 900(0.8)^{t}\) dollars, \(t\) years after purchase. Interpret the \(y\)-intercept.
At \(t = 0\), \(V = 900\), so the \(y\)-intercept ($900) is the phone’s purchase price.
A tree’s height is \(H = 150(1.4)^{t}\) cm, \(t\) years after planting. Interpret the \(y\)-intercept.
At \(t = 0\), \(H = 150\), so the \(y\)-intercept (150 cm) is the tree’s height when it was planted.
Level 3 · Further Application
A bacterial culture is modelled by \(N = 500(2)^{t}\), where \(t\) is in hours.
(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.
(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.
(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.
\(V = 800(2)^{t}\).
A quantity starts at 250 and triples each year. Write an exponential model for its value \(V\) after \(t\) years.
\(V = 250(3)^{t}\).
A sample starts at 60 g and halves each day. Write an exponential model for its mass \(M\) after \(t\) days.
\(M = 60(0.5)^{t}\).
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.
$6945.75
\(V = 6000(1.05)^{t}\). After 3 years: \(6000(1.05)^{3} =\) $6945.75.
A $5000 investment grows by 4% each year. Write an exponential model for its value \(V\) after \(t\) years, and use it to find the value after 3 years.
$5624.32
\(V = 5000(1.04)^{t}\). After 3 years: \(5000(1.04)^{3} = 5000 \times 1.124864 =\) $5624.32.
A stamp worth $200 increases in value by 10% each year. Write an exponential model for its value \(V\) after \(t\) years, and use it to find the value after 2 years.
$242
\(V = 200(1.1)^{t}\). After 2 years: \(200(1.1)^{2} = 200 \times 1.21 =\) $242.
Level 3 · Further Application
A town’s population is 8000 and decreases by 4% each year.
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.
$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.
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 \(x = 0\).
\(y = 3\) (the constant term).
For the quadratic \(y = x^{2} + 5x - 2\), state the value of \(y\) when \(x = 0\).
\(y = -2\) (the constant term).
A braking-distance model is \(d = 0.008v^{2}\) m, where \(v\) is speed in km/h. Find \(d\) when \(v = 0\).
\(d = 0.008 \times 0^{2} = 0\) m (a stationary car needs no braking distance).
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.
At 50: \(0.01 \times 50^{2} = 25\) m. At 100: \(0.01 \times 100^{2} = 100\) m (four times as far for double the speed).
The braking distance is \(d = 0.006v^{2}\) m, where \(v\) is speed in km/h. Find the braking distance at 40 km/h and at 80 km/h.
At 40: \(0.006 \times 40^{2} = 9.6\) m. At 80: \(0.006 \times 80^{2} = 38.4\) m (four times as far for double the speed).
The stopping distance is \(d = 0.007v^{2}\) m, where \(v\) is speed in km/h. Find the stopping distance at 60 km/h and at 120 km/h.
At 60: \(0.007 \times 60^{2} = 25.2\) m. At 120: \(0.007 \times 120^{2} = 100.8\) m (four times as far for double the speed).
Level 3 · Further Application
A ball’s height is \(h = -5t^{2} + 30t\) metres, where \(t\) is in seconds.
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.
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.
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?
The turning point (or vertex) — here a minimum.
A parabola has its highest point at \((-1, 6)\). What is this point called?
The turning point (or vertex) — here a maximum.
A parabola’s axis of symmetry is a vertical line that passes through which special point of the curve?
The turning point (vertex).
Level 2 · Application
A parabola has \(x\)-intercepts at \(x = 2\) and \(x = 8\). State the equation of its axis of symmetry.
\(x = 5\)
The axis of symmetry is midway between the intercepts: \(x = \dfrac{2 + 8}{2} =\) \(x = 5\).
A parabola has \(x\)-intercepts at \(x = 1\) and \(x = 7\). State the equation of its axis of symmetry.
\(x = 4\)
The axis of symmetry is midway between the intercepts: \(x = \dfrac{1 + 7}{2} =\) \(x = 4\).
A parabola has \(x\)-intercepts at \(x = -2\) and \(x = 6\). State the equation of its axis of symmetry.
\(x = 2\)
The axis of symmetry is midway between the intercepts: \(x = \dfrac{-2 + 6}{2} =\) \(x = 2\).
Level 3 · Further Application
A parabola passes through \(x\)-intercepts at \(x = -1\) and \(x = 5\), opening upwards.
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.
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.
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?
Downwards (the coefficient of \(x^{2}\) is negative).
Does the parabola \(y = 2x^{2} - 5\) open upwards or downwards?
Upwards (the coefficient of \(x^{2}\) is positive).
State the \(y\)-intercept of the parabola \(y = x^{2} + 3x - 6\).
\((0, -6)\) — the constant term gives the \(y\)-intercept.
Level 2 · Application
For \(y = x^{2} - 2x - 3\), find the \(y\)-intercept and the \(x\)-intercepts.
\(y\)-intercept: \((0, -3)\). \(x\)-intercepts: \(x^{2} - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0 \Rightarrow x = 3\) and \(x = -1\).
For \(y = x^{2} - x - 6\), find the \(y\)-intercept and the \(x\)-intercepts.
\(y\)-intercept: \((0, -6)\). \(x\)-intercepts: \(x^{2} - x - 6 = 0 \Rightarrow (x - 3)(x + 2) = 0 \Rightarrow x = 3\) and \(x = -2\).
For \(y = x^{2} + 2x - 8\), find the \(y\)-intercept and the \(x\)-intercepts.
\(y\)-intercept: \((0, -8)\). \(x\)-intercepts: \(x^{2} + 2x - 8 = 0 \Rightarrow (x + 4)(x - 2) = 0 \Rightarrow x = -4\) and \(x = 2\).
Level 3 · Further Application
Consider \(y = x^{2} - 6x + 5\).
\((3, -4)\)
(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 \(y = x^{2} - 4x - 5\).
\((2, -9)\)
(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 \(y = x^{2} + 2x - 3\).
\((-1, -4)\)
(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?
The maximum revenue ($8000), achieved at the value \(x = 50\).
A ball’s height parabola has its maximum at \((3, 45)\). What does the \(y\)-value 45 represent?
The greatest height (45 m) reached by the ball, at \(t = 3\) seconds.
A profit parabola crosses the \(x\)-axis at \(x = 20\). What does this \(x\)-intercept represent?
A break-even point — the number of items where the profit is zero.
Level 2 · Application
A ball’s height is \(h = -5t^{2} + 20t\) metres. The turning point is at \((2, 20)\). Interpret this point.
The ball reaches its greatest height of 20 m at \(t = 2\) seconds.
A ball’s height is \(h = -5t^{2} + 30t\) metres. The turning point is at \((3, 45)\). Interpret this point.
The ball reaches its greatest height of 45 m at \(t = 3\) seconds.
A rocket’s height model has a turning point at \((4, 80)\). Interpret this point.
The rocket reaches its maximum height of 80 m at \(t = 4\) seconds.
Level 3 · Further Application
A concert promoter’s takings are \(P = -40x^{2} + 3200x + 5000\), where \(x\) is the ticket price rise in dollars.
$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.
$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.
$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?
Height cannot be negative — the ball has already landed, so the model no longer applies.
A ball’s height model gives \(h = -3\) m at some time \(t\). Why is this value not sensible?
A height cannot be negative — the ball has already landed, so the model no longer applies.
A quadratic model gives the number of items sold as \(n = -12\). Why is this value not sensible?
You cannot sell a negative number of items, so the value has no real meaning.
Level 2 · Application
A ball’s height is \(h = -5t^{2} + 20t\). It lands when \(h = 0\). Find the sensible domain of \(t\).
\(h = 0 \Rightarrow -5t(t - 4) = 0 \Rightarrow t = 0\) or \(t = 4\). Sensible domain: \(0 \le t \le 4\) seconds.
A ball’s height is \(h = -5t^{2} + 30t\). It lands when \(h = 0\). Find the sensible domain of \(t\).
\(h = 0 \Rightarrow -5t(t - 6) = 0 \Rightarrow t = 0\) or \(t = 6\). Sensible domain: \(0 \le t \le 6\) seconds.
A ball’s height is \(h = -5t^{2} + 15t\). It lands when \(h = 0\). Find the sensible domain of \(t\).
\(h = 0 \Rightarrow -5t(t - 3) = 0 \Rightarrow t = 0\) or \(t = 3\). Sensible domain: \(0 \le t \le 3\) seconds.
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.
(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.
(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.
(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?
\(y\) decreases (they are inversely related).
In the relationship \(y = \dfrac{k}{x}\), if \(x\) is halved, what happens to \(y\)?
\(y\) doubles (dividing by half the value doubles the result).
For the inverse variation \(y = \dfrac{k}{x}\), what is the value of the product \(xy\)?
\(xy = k\), a constant.
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 \(x = 6\).
4
\(k = xy = 3 \times 8 = 24\), so \(y = \dfrac{24}{x}\). When \(x = 6\), \(y = \dfrac{24}{6} =\) 4.
\(y\) varies inversely with \(x\), and \(y = 5\) when \(x = 4\). Find \(k\) in \(y = \dfrac{k}{x}\), and hence \(y\) when \(x = 10\).
2
\(k = xy = 4 \times 5 = 20\), so \(y = \dfrac{20}{x}\). When \(x = 10\), \(y = \dfrac{20}{10} =\) 2.
\(y\) varies inversely with \(x\), and \(y = 12\) when \(x = 2\). Find \(k\) in \(y = \dfrac{k}{x}\), and hence \(y\) when \(x = 8\).
3
\(k = xy = 2 \times 12 = 24\), so \(y = \dfrac{24}{x}\). When \(x = 8\), \(y = \dfrac{24}{8} =\) 3.
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.
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 \(R = 5\), \(I = 12\).
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.
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?
Less — more people sharing the same amount means a smaller share each (inverse variation).
A fixed prize is shared equally among the winners. If the number of winners doubles, does each share get larger or smaller?
Smaller — the same prize split among more people gives a smaller share each (inverse variation).
6 workers finish a task in a certain time. Assuming the same total work, will 3 workers take more or less time?
More time — fewer workers means the job takes longer (inverse variation).
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?
15 days
\(k = 6 \times 20 = 120\) worker-days. Time \(= \dfrac{120}{8} =\) 15 days.
5 taps fill a tank in 12 minutes. Assuming the time varies inversely with the number of taps, how long would 4 taps take?
15 minutes
\(k = 5 \times 12 = 60\) tap-minutes. Time \(= \dfrac{60}{4} =\) 15 minutes.
3 painters finish a room in 8 hours. Assuming the time varies inversely with the number of painters, how long would 4 painters take?
6 hours
\(k = 3 \times 8 = 24\) painter-hours. Time \(= \dfrac{24}{4} =\) 6 hours.
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.
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.
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.
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.