Mini-Lesson

Functions & Modelling (AI HL)

5 missions. Guess, learn, then check. Your place is saved.

Mission 1Two taxis, two price lines: when does the cheaper one switch?
Mission 2Undo "double it, then add 3": why do you subtract 3 first?
Mission 3A phone loses 20% of its value every year. Is it ever worth nothing?
Mission 4The tide rises and falls like a wave. When is the harbour too shallow?
Mission 5Why do rumours and crazes spread fast, then level off?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Two taxis, two price lines: when does the cheaper one switch?

Make a guess: Two taxis, two price lines: when does the cheaper one switch?

Functions · lines and graphs

Gradient and the forms of a line

Gradient $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ m = (y₂ − y₁)/(x₂ − x₁): the rate of change of $y$ y per unit of $x$ x.
Gradient-intercept form: $y = m x + c$ y = mx + c, where $c$ c is the $y$ y-intercept.
Point-gradient form: $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁) through the point $(x_{1}, y_{1})$ (x₁, y₁).
General form: $a x + b y + d = 0$ ax + by + d = 0.
A road climbing 1.2 m over 15 m horizontally has gradient $\frac{1.2}{15} = 0.08$ 1.2/15 = 0.08, an 8% incline.

Worked example Functions · lines and graphs

The line through two points

Question

Find the equation of the line through

A (2, 5)

and

B (6, - 3)

. Give your answer in the form

a x + b y + d = 0

, where

a

,

b

,

d \in Z

.

1
Gradient: $m = \frac{- 3 - 5}{6 - 2} = \frac{- 8}{4} = - 2$ m = (−3 − 5)/(6 − 2) = (−8)/4 = −2
2
Point-gradient form through $A$ A: $y - 5 = - 2 (x - 2)$ y − 5 = −2(x − 2)
3
Expand: $y = - 2 x + 9$ y = −2x + 9
4
Move every term to one side: $2 x + y - 9 = 0$ 2x + y − 9 = 0

Answer

2 x + y - 9 = 0

⚠️ Watch out

subtract the coordinates in the same order.

\frac{5 - ( - 3 )}{6 - 2} = 2

has the wrong sign.

Functions · lines and graphs

Parallel and perpendicular lines

Parallel lines have equal gradients: $m_{1} = m_{2}$ m₁ = m₂.
Perpendicular lines: $m_{1} \times m_{2} = - 1$ m₁ × m₂ = −1, so $m_{2} = - \frac{1}{m _{1}}$ m₂ = −1/m₁ (the negative reciprocal).
To read a gradient from general form, rearrange to $y = m x + c$ y = mx + c. For $2 x + 5 y - 7 = 0$ 2x + 5y − 7 = 0: $y = - \frac{2}{5} x + \frac{7}{5}$ y = −2/5x + 7/5, so $m = - \frac{2}{5}$ m = −2/5.
Any line perpendicular to it has gradient $\frac{5}{2}$ 5/2.

Calculate

?The line

L

passes through

(3, 1)

and is perpendicular to

2 x + 5 y - 7 = 0

. Find the

y

-intercept of

L

. Give your answer as a decimal.

Functions · lines and graphs

Functions, domain, range and inverses

A function gives each input exactly one output. Notation: $f (x)$ f(x), $v (t)$ v(t), $C (n)$ C(n).
The domain is the set of allowed inputs; the range is the set of outputs. $f (x) = 2 - x$ f(x) = √(2 − x) has domain $x \leq 2$ x ≤ 2 and range $f (x) \geq 0$ f(x) ≥ 0.
The inverse function $f^{- 1}$ f⁻¹ undoes $f$ f: solving $f (x) = 10$ f(x) = 10 is the same as finding $f^{- 1} (10)$ f⁻¹(10).
The graph of $f^{- 1}$ f⁻¹ is the reflection of $f$ f in $y = x$ y = x. Only one-to-one functions have inverses; the domain of $f^{- 1}$ f⁻¹ is the range of $f$ f.

Quick check

?The function

f

is defined by

f (x) = x - 3 + 1

, for

x \geq 3

. Which is the range of

f

?

Functions · lines and graphs

Key features of graphs, with technology

Sketch: the shape, with axes and key features labelled. Draw: accurate, to scale.
Key features: intercepts, zeros (roots of $f (x) = 0$ f(x) = 0), maximum and minimum points, vertex, symmetry, vertical and horizontal asymptotes.
To solve $f (x) = g (x)$ f(x) = g(x), graph both on your GDC and find the intersection points. A GDC also graphs sums and differences such as $y = f (x) - g (x)$ y = f(x) − g(x), whose zeros are the same $x$ x-values.

Worked example Functions · lines and graphs

Where two graphs meet

Question

f (x) = x^{2} - 3 x

and

g (x) = 2 x - 3

. Find the

x

-coordinates of the points where the graphs of

f

and

g

intersect. Give your answers to 3 significant figures.

1
Set the functions equal: $x^{2} - 3 x = 2 x - 3$ x² − 3x = 2x − 3
2
Rearrange to zero: $x^{2} - 5 x + 3 = 0$ x² − 5x + 3 = 0
3
Use the intersect tool on your GDC (or the quadratic formula): $x = \frac{5 \pm 13}{2}$ x = (5 ± √13)/2

Answer

x = 0.697

or

x = 4.30

⚠️ Watch out

a GDC finds one intersection at a time. Check the window shows every crossing so you do not miss one.

Calculate

?A ball is thrown upwards. Its height in metres after

t

seconds is modelled by

h (t) = - 4.9 t^{2} + 12 t + 2

, for

t \geq 0

. Use your GDC to find when the ball hits the ground. Give your answer to 3 significant figures.

seconds

Mission check

?Taxi A charges £3.50 plus £1.20 per km. Taxi B charges £6.00 plus £0.95 per km. When does the cheaper one switch?

Mission 2

Undo "double it, then add 3": why do you subtract 3 first?

Make a guess: Undo "double it, then add 3": why do you subtract 3 first?

Functions · composites, inverses and transformations

Composite functions

$(f \circ g) (x) = f (g (x))$ (f ° g)(x) = f(g(x)): apply $g$ g first, then $f$ f to the result.
With $f (x) = 2 x + 3$ f(x) = 2x + 3 and $g (x) = x^{2}$ g(x) = x²: $(f \circ g) (x) = 2 x^{2} + 3$ (f ° g)(x) = 2x² + 3 but $(g \circ f) (x) = (2 x + 3)^{2}$ (g ° f)(x) = (2x + 3)². Order matters.
In context: $n (t) = 50 t$ n(t) = 50t phones are made in $t$ t hours and $C (n) = 200 + 4 n$ C(n) = 200 + 4n is their cost in £. Then $(C \circ n) (t) = 200 + 200 t$ (C ° n)(t) = 200 + 200t is the cost after $t$ t hours.
$g (x)$ g(x) must lie in the domain of $f$ f for $f (g (x))$ f(g(x)) to exist.

Calculate

?

f (x) = 3 x - 1

and

g (x) = x^{2} + 2

. Find

(g \circ f) (2)

.

Functions · composites, inverses and transformations

Inverse functions

$f^{- 1}$ f⁻¹ undoes $f$ f: $(f \circ f^{- 1}) (x) = (f^{- 1} \circ f) (x) = x$ (f ° f⁻¹)(x) = (f⁻¹ ° f)(x) = x, the identity function.
To undo a chain of steps, undo the last step first, like taking off shoes before socks.
Method: write $y = f (x)$ y = f(x), rearrange to make $x$ x the subject, then swap $x$ x and $y$ y.
An inverse exists only if $f$ f is one-to-one. If it is not, restrict the domain. The domain of $f^{- 1}$ f⁻¹ is the range of $f$ f.

Worked example Functions · composites, inverses and transformations

An inverse with a restricted domain

Question

f (x) = (x - 3)^{2} - 2

, for

x \geq 3

. Find

f^{- 1} (x)

and state its domain.

1
Write $y = (x - 3)^{2} - 2$ y = (x − 3)² − 2 and add 2: $y + 2 = (x - 3)^{2}$ y + 2 = (x − 3)²
2
Square-root, taking the positive root because $x \geq 3$ x ≥ 3: $x - 3 = y + 2$ x − 3 = √(y + 2)
3
Make $x$ x the subject: $x = 3 + y + 2$ x = 3 + √(y + 2)
4
Swap $x$ x and $y$ y: $f^{- 1} (x) = 3 + x + 2$ f⁻¹(x) = 3 + √(x + 2)
5
The range of $f$ f is $f (x) \geq - 2$ f(x) ≥ −2, so the domain of $f^{- 1}$ f⁻¹ is $x \geq - 2$ x ≥ −2

Answer

f^{- 1} (x) = 3 + x + 2

,

x \geq - 2

⚠️ Watch out

without the restriction

x \geq 3

,

f

is many-to-one (

f (1) = f (5) = 2

) and has no inverse.

Quick check

?

f (x) = \frac{x + 4}{3}

. Which is

f^{- 1} (x)

?

Functions · composites, inverses and transformations

Transformations of graphs

Translations: $y = f (x) + b$ y = f(x) + b moves the graph up $b$ b; $y = f (x - a)$ y = f(x − a) moves it right $a$ a.
Translation by the vector $(3 - 2)$ ( 3 ; −2 ) (3 right, 2 down): $y = f (x - 3) - 2$ y = f(x − 3) − 2.
Reflections: $y = - f (x)$ y = −f(x) in the $x$ x-axis; $y = f (- x)$ y = f(−x) in the $y$ y-axis.
Stretches: $y = p f (x)$ y = pf(x) is vertical, scale factor $p$ p; $y = f (q x)$ y = f(qx) is horizontal, scale factor $\frac{1}{q}$ 1/q.
Invariant axes: $y = p f (x)$ y = pf(x) fixes the $x$ x-axis; $y = f (q x)$ y = f(qx) fixes the $y$ y-axis.
Changes outside $f$ f act on $y$ y-values; changes inside act on $x$ x-values, the opposite way to how they look.

Worked example Functions · composites, inverses and transformations

Order in a composite transformation

Question

Describe a sequence of transformations that maps

y = x^{2}

onto

y = 3 x^{2} + 2

.

1
Multiply the outputs by 3: vertical stretch, scale factor 3, gives $y = 3 x^{2}$ y = 3x²
2
Add 2 to the outputs: translation 2 units up (by the vector $(02)$ ( 0 ; 2 )), gives $y = 3 x^{2} + 2$ y = 3x² + 2
3
Track a point: $(1, 1) \to (1, 3) \to (1, 5)$ (1, 1) → (1, 3) → (1, 5)

Answer

a vertical stretch with scale factor 3, followed by a translation of 2 units up

⚠️ Watch out

in the other order,

y = x^{2} + 2

stretched by 3 gives

y = 3 x^{2} + 6

. The order matters.

Calculate

?The point

P (6, - 4)

lies on the graph of

y = f (x)

. The graph is transformed to

y = - 2 f (x) + 5

. Find the

y

-coordinate of the image of

P

.

Sort it

?Each equation is a transformation of

y = f (x)

. Does it change the

x

-coordinates or the

y

-coordinates of points on the graph?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?

f (x) = 2 x + 3

doubles, then adds 3. Why does

f^{- 1}

subtract 3 first?

Mission 3

A phone loses 20% of its value every year. Is it ever worth nothing?

Make a guess: A phone loses 20% of its value every year. Is it ever worth nothing?

Functions · modelling

Linear and piecewise linear models

A linear model $f (x) = m x + c$ f(x) = mx + c fits a quantity that changes by the same amount per unit of $x$ x: $m$ m is the rate, $c$ c the start value.
A piecewise model uses different rules on different parts of the domain, like a phone plan with a data allowance, then a charge per GB.
Always state a reasonable domain: data used cannot be negative.
To make the pieces join up (a continuous model), choose the parameter so both rules agree at the boundary.

Calculate

?A phone plan costs €12 a month, which includes 5 GB of data. Each extra GB costs €2.50. Find the cost, in euros, of a month in which 8.4 GB is used.

Calculate

?The function

f (x) = 2 + 0.5 x

for

0 \leq x < 4

, and

f (x) = a x^{2} - x

for

x \geq 4

, is continuous. Find the value of

a

.

Functions · modelling

Exponential growth and decay models

An exponential model fits a quantity that changes by the same percentage each step: $f (x) = k a^{x} + c$ f(x) = ka^x + c or $f (x) = k a^{- x} + c$ f(x) = ka^(−x) + c (with $a > 0$ a > 0), or $f (x) = k e^{r x} + c$ f(x) = ke^rx + c.
In $k a^{x}$ ka^x: $a > 1$ a > 1 is growth, $0 < a < 1$ 0 < a < 1 is decay. A 20% fall each year means $a = 0.8$ a = 0.8; a 5% rise means $a = 1.05$ a = 1.05.
The starting value is $f (0) = k + c$ f(0) = k + c.
The graph has a horizontal asymptote $y = c$ y = c: decay models get ever closer to $c$ c without reaching it.

Worked example Functions · modelling

A cooling drink

Question

The temperature of a cup of tea, in °C,

t

minutes after it is made, is modelled by

T (t) = 20 + 70 (0.92)^{t}

. Find (a) the initial temperature, (b) the temperature after 10 minutes, (c) the equation of the horizontal asymptote and what it represents, (d) when the tea reaches 40 °C.

1
(a) Substitute $t = 0$ t = 0: $T (0) = 20 + 70 = 90$ T(0) = 20 + 70 = 90 °C
2
(b) Substitute $t = 10$ t = 10: $T (10) = 20 + 70 (0.92)^{10} = 50.4$ T(10) = 20 + 70(0.92)¹⁰ = 50.4 °C (3 sf)
3
(c) As $t$ t grows, $70 (0.92)^{t} \to 0$ 70(0.92)^t → 0, so $T = 20$ T = 20: the room temperature
4
(d) Use your GDC to solve $20 + 70 (0.92)^{t} = 40$ 20 + 70(0.92)^t = 40 (intersect with $y = 40$ y = 40): $t = 15.0$ t = 15.0 minutes

Answer

(a) 90 °C (b) 50.4 °C (c)

T = 20

, room temperature (d) after 15.0 minutes

⚠️ Watch out

the initial value is

k + c = 90

, not

k = 70

.

Quick check

?A car bought for €24,000 loses 15% of its value each year. Which model gives its value, in euros, after

t

years?

Functions · modelling

Direct and inverse variation

Variation models have the form $f (x) = a x^{n}$ f(x) = axⁿ, $n \in Z$ n ∈ ℤ.
$n > 0$ n > 0: direct variation, such as $y = a x^{2}$ y = ax² ( $y$ y varies as the square of $x$ x).
$n < 0$ n < 0: inverse variation, such as $y = \frac{a}{x ^{2}} = a x^{- 2}$ y = a/x² = ax⁻². The $y$ y-axis is a vertical asymptote.
Find $a$ a from one data point: if $y = a x^{- 2}$ y = ax⁻² and $y = 45$ y = 45 when $x = 2$ x = 2, then $a = 45 \times 2^{2} = 180$ a = 45 × 2² = 180, and $y = 20$ y = 20 when $x = 3$ x = 3.
Choose a model from the context and the shape of the data, and beware of extrapolating far beyond the data.

Sort it

?Which type of model fits each situation best?

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?The number of members of a sports club

t

years after it opened is modelled by

P (t) = k a^{t}

. When

t = 1

there are 360 members, and when

t = 4

there are 622.08 (as given by the model). Use the model to predict the number of members when

t = 6

. Give your answer to the nearest whole number.

Mission check

?A phone bought for €900 loses 20% of its value every year. Is it ever worth nothing?

Mission 4

The tide rises and falls like a wave. When is the harbour too shallow?

Make a guess: The tide rises and falls like a wave. When is the harbour too shallow?

Functions · modelling

Quadratic models

$f (x) = a x^{2} + b x + c$ f(x) = ax² + bx + c, $a \neq = 0$ a ≠ 0, fits data that rise then fall (or fall then rise): a thrown ball, an arch, the area of a fenced plot.
The vertex gives the maximum ( $a < 0$ a < 0) or minimum ( $a > 0$ a > 0); the axis of symmetry is $x = - \frac{b}{2 a}$ x = −b/2a.
The zeros (use your GDC) often mark where the model starts or ends, such as when a ball lands.
Choose a reasonable domain: a height cannot be negative, so the model stops at the zero.

Worked example Functions · modelling

A thrown ball

Question

A ball is thrown so that its height, in metres,

t

seconds later is

h (t) = - 4.9 t^{2} + 14 t + 1.5

. Find (a) the maximum height, (b) when the ball lands, (c) a reasonable domain for the model.

1
(a) Axis of symmetry: $t = - \frac{14}{2 ( - 4.9 )} = 1.43$ t = −14/(2(−4.9)) = 1.43 s (3 sf)
2
Substitute: $h (1.43) = 11.5$ h(1.43) = 11.5 m (using the exact value $t = \frac{10}{7}$ t = 10/7)
3
(b) Use your GDC to solve $h (t) = 0$ h(t) = 0: $t = - 0.103$ t = −0.103 or $t = 2.96$ t = 2.96; reject the negative root
4
(c) From the throw to the landing: $0 \leq t \leq 2.96$ 0 ≤ t ≤ 2.96

Answer

(a) 11.5 m (b) after 2.96 s (c)

0 \leq t \leq 2.96

⚠️ Watch out

the

t

-value of the vertex, 1.43, is a time. The question asks for the height.

Functions · modelling

Cubic models and finding parameters

$f (x) = a x^{3} + b x^{2} + c x + d$ f(x) = ax³ + bx² + cx + d fits data with up to two turning points, such as the volume of an open box cut from a sheet.
A box from a 30 cm by 20 cm sheet with squares of side $x$ x cut out has $V = x (30 - 2 x) (20 - 2 x)$ V = x(30 − 2x)(20 − 2x), domain $0 < x < 10$ 0 < x < 10. A GDC gives the maximum: $V \approx 1, 056$ V ≈ 1,056 cm³ at $x \approx 3.92$ x ≈ 3.92.
Finding parameters: substitute known points into the model. Three points give three linear equations in $a$ a, $b$ b, $c$ c: solve them with your GDC.

Calculate

?A ball's path is modelled by

y = a x^{2} + b x + c

, where

x

m is the horizontal distance and

y

m the height. The path passes through

(0, 2)

,

(2, 8)

and

(4, 6)

. Find the horizontal distance at which the ball hits the ground. Give your answer to 3 significant figures.

Calculate

?An open box is made from a sheet of card 24 cm by 15 cm by cutting a square of side

x

cm from each corner and folding up the sides. Use your GDC to find the maximum possible volume of the box, in cm³. Give your answer to 3 significant figures.

Functions · modelling

Sinusoidal models

f (x) = a sin (b x) + d or f (x) = a cos (b x) + d

Amplitude $∣ a ∣ = \frac{max - min}{2}$ |a| = (max − min)/2; principal axis $y = d = \frac{max + min}{2}$ y = d = (max + min)/2.
Period $= \frac{36 0 ^{\circ}}{b}$ = (360°)/b, with $x$ x in degrees: the time for one full cycle.
$a cos (b x) + d$ acos (bx) + d with $a > 0$ a > 0 starts at a maximum; with $a < 0$ a < 0 it starts at a minimum.
In general, $f (x) = a sin (b (x - c)) + d$ f(x) = asin (b(x − c)) + d. Use radians unless degrees are shown: period $\frac{2 π}{b}$ 2π/b. The horizontal translation $c$ c is the phase shift.

Worked example Functions · modelling

A Ferris wheel

Question

A seat on a Ferris wheel is 4 m above the ground at its lowest and 44 m at its highest. One revolution takes 30 minutes, and the seat starts at its lowest point. (a) Find a model

h (t) = a cos (b t) + d

, with

t

in minutes. (b) For how long in each revolution is the seat above 34 m?

1
Amplitude $\frac{44 - 4}{2} = 20$ (44 − 4)/2 = 20; it starts at a minimum, so $a = - 20$ a = −20
2
Principal axis $d = \frac{44 + 4}{2} = 24$ d = (44 + 4)/2 = 24; period $30 = \frac{360}{b}$ 30 = 360/b, so $b = 12$ b = 12
3
So $h (t) = - 20 cos (12 t^{\circ}) + 24$ h(t) = −20cos (12t°) + 24
4
Use your GDC to solve $h (t) = 34$ h(t) = 34: $t = 10$ t = 10 and $t = 20$ t = 20, and $h > 34$ h > 34 between them

Answer

(a)

h (t) = - 20 cos (12 t^{\circ}) + 24

(b) 10 minutes

⚠️ Watch out

the amplitude is half of the max minus the min, 20, not the full 40.

Calculate

?The depth of water in a harbour, in metres,

t

hours after high tide is modelled by

d (t) = 3 cos (30 t^{\circ}) + 5

. A boat needs a depth of at least 4 m. Use your GDC to find for how many hours in each 12-hour cycle the harbour is too shallow for the boat. Give your answer to 3 significant figures.

Calculate

?The height of a wave, in metres, is modelled by

f (t) = 2.5 sin (\frac{π}{6} (t - 3)) + 12

,

t \geq 0

, where

t

is in seconds. Find the first time

t

at which the height is a maximum.

Mission check

?The tide rises and falls like a wave. When is the harbour too shallow?

Mission 5

Why do rumours and crazes spread fast, then level off?

Make a guess: Why do rumours and crazes spread fast, then level off?

Functions · further modelling

Logistic models

f (x) = \frac{L}{1 + C e ^{- k x}}, L, C, k > 0

Use it when growth is restricted: a population on an island, bacteria in a dish, a rumour in a closed group.
Early on, growth is close to exponential. As $f$ f nears $L$ L there is less room left, so growth slows.
The horizontal asymptote $y = L$ y = L is the carrying capacity. The starting value is $f (0) = \frac{L}{1 + C}$ f(0) = L/(1 + C).
Growth is fastest when $f = \frac{L}{2}$ f = L/2, halfway to the limit.

Worked example Functions · further modelling

How a rumour spreads

Question

The number of students who have heard a rumour

t

days after it starts is modelled by

N (t) = \frac{1200}{1 + 59 e ^{- 0.8 t}}

. Find (a)

N (0)

, (b)

N (5)

, (c) the long-term number, (d) when 1,000 students have heard it.

1
(a) $N (0) = \frac{1200}{1 + 59} = 20$ N(0) = 1200/(1 + 59) = 20
2
(b) $N (5) = \frac{1200}{1 + 59 e ^{- 4}} = 577$ N(5) = 1200/(1 + 59e⁻⁴) = 577 (3 sf)
3
(c) As $t \to \infty$ t → ∞, $e^{- 0.8 t} \to 0$ e^(−0.8t) → 0, so $N \to 1200$ N → 1200
4
(d) Use your GDC to solve $N (t) = 1000$ N(t) = 1000 (or $59 e^{- 0.8 t} = 0.2$ 59e^(−0.8t) = 0.2, so $t = \frac{l n 295}{0.8}$ t = (ln 295)/0.8): $t = 7.11$ t = 7.11 days

Answer

(a) 20 (b) 577 (c) 1,200 (d) 7.11 days

⚠️ Watch out

the starting value is

\frac{L}{1 + C}

, not

L

and not

C

.

Calculate

?The population of rabbits on an island

t

years after they arrive is modelled by

P (t) = \frac{5000}{1 + 24 e ^{- 0.3 t}}

. Use your GDC to find how many years it takes for the population to reach 4,000. Give your answer to 3 significant figures.

Functions · further modelling

Half-life and natural log models

Exponential decay $f (t) = k e^{- r t}$ f(t) = ke^(−rt): the half-life is the time for $f$ f to halve. Solve $k e^{- r t} = \frac{k}{2}$ ke^(−rt) = k/2, so $t = \frac{l n 2}{r}$ t = (ln 2)/r, whatever the starting amount.
If $r$ r is unknown, find it from a data point first.
Natural log model $f (x) = a + b ln x$ f(x) = a + bln x, $x > 0$ x > 0: it keeps growing (for $b > 0$ b > 0) but ever more slowly, with no horizontal asymptote. In $2 + 3 ln x$ 2 + 3ln x, each time $x$ x is multiplied by $e$ e, $f$ f rises by 3.

Calculate

?A patient is given 80 mg of a medicine. The amount left after

t

hours is modelled by

m (t) = 80 e^{- r t}

. After 6 hours, 60 mg remain. Find the half-life of the medicine, in hours, to 3 significant figures.

Calculate

?The height of a young tree, in cm,

t

months after planting is modelled by

h (t) = a + b ln t

,

t \geq 1

. After 1 month it is 40 cm tall, and after 10 months it is 86 cm tall. Use the model to find the height after 30 months. Give your answer to 3 significant figures.

Functions · further modelling

Scaling and linearising with logs

Data from 10 to 1,000,000 are hard to plot. Using $lo g_{10}$ log₁₀ gives 1 to 6: each step is a factor of 10.
If $y = a b^{x}$ y = ab^x then $ln y = ln a + x ln b$ ln y = ln a + xln b: a semi-log graph ( $ln y$ ln y against $x$ x) is a straight line.
If $y = a x^{b}$ y = ax^b then $ln y = ln a + b ln x$ ln y = ln a + bln x: a log-log graph ( $ln y$ ln y against $ln x$ ln x) is a straight line.
Fit a straight line; read off gradient and intercept, then undo the logs. The plot with $r$ r closer to $\pm 1$ ± 1 points to the better model.

Worked example Functions · further modelling

From a semi-log line to a model

Question

For some data, a graph of

ln y

against

x

is a straight line,

ln y = 0.4 x + 1.2

. Find a model of the form

y = a b^{x}

.

1
Compare with $ln y = ln a + x ln b$ ln y = ln a + xln b: $ln a = 1.2$ ln a = 1.2 and $ln b = 0.4$ ln b = 0.4
2
Undo the logs: $a = e^{1.2} = 3.32$ a = e^1.2 = 3.32, $b = e^{0.4} = 1.49$ b = e^0.4 = 1.49 (3 sf)
3
Check: $ln (3.32 \times 1.4 9^{x}) = ln 3.32 + x ln 1.49 \approx 1.2 + 0.4 x$ ln (3.32 × 1.49^x) = ln 3.32 + xln 1.49 ≈ 1.2 + 0.4x

Answer

y = 3.32 \times 1.4 9^{x}

⚠️ Watch out

the intercept is

ln a

, not

a

: writing

a = 1.2

is the most common slip.

Quick check

?A graph of

ln y

against

ln x

is a straight line with gradient 1.5 and vertical intercept 0.8. Which model fits the data?

Mission check

?Why do rumours and crazes spread fast, then level off?

Recap

Big ideas

A line has gradient $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ m = (y₂ − y₁)/(x₂ − x₁) and can be written $y = m x + c$ y = mx + c, $y - y_{1} = m (x - x_{1})$ y − y₁ = m(x − x₁) or $a x + b y + d = 0$ ax + by + d = 0; parallel means $m_{1} = m_{2}$ m₁ = m₂, perpendicular means $m_{1} m_{2} = - 1$ m₁ m₂ = −1.
A function has a domain (inputs) and a range (outputs); $f^{- 1}$ f⁻¹ undoes $f$ f, exists for one-to-one functions and is the reflection of $f$ f in $y = x$ y = x.
Key features (intercepts, zeros, maximum and minimum, asymptotes) and intersections come from a GDC or from solving $f (x) = g (x)$ f(x) = g(x).
$(f \circ g) (x) = f (g (x))$ (f ° g)(x) = f(g(x)): apply $g$ g first; in general $f \circ g \neq = g \circ f$ f ° g ≠ g ° f.
$f^{- 1}$ f⁻¹ undoes $f$ f (last step first); it needs $f$ f to be one-to-one, possibly after restricting the domain, and the domain of $f^{- 1}$ f⁻¹ is the range of $f$ f.
$f (x) + b$ f(x) + b, $- f (x)$ −f(x), $p f (x)$ pf(x) act on $y$ y-values; $f (x - a)$ f(x − a), $f (- x)$ f(−x), $f (q x)$ f(qx) act on $x$ x-values; the order of transformations matters.
Same amount each step: linear $f (x) = m x + c$ f(x) = mx + c (piecewise if the rule changes). Same percentage each step: exponential $k a^{x} + c$ ka^x + c.
An exponential model $k a^{x} + c$ ka^x + c has horizontal asymptote $y = c$ y = c; decay approaches it but never reaches it.
Variation $a x^{n}$ axⁿ: find $a$ a from one point. Choose models from context and data shape; beware extrapolation.
For a continuous piecewise model, make the rules agree at each boundary.
Quadratic models: vertex for the maximum or minimum, zeros for start and end, and a domain that makes sense.
Find a model's parameters by substituting points and solving the linear equations with your GDC.
Sinusoidal models: amplitude $\frac{max - min}{2}$ (max − min)/2, principal axis $y = \frac{max + min}{2}$ y = (max + min)/2, period $\frac{36 0 ^{\circ}}{b}$ (360°)/b.
$a sin (b (x - c)) + d$ asin (b(x − c)) + d in radians: period $\frac{2 π}{b}$ 2π/b, phase shift $c$ c.
Logistic $\frac{L}{1 + C e ^{- k x}}$ L/(1 + Ce^(−kx)): near-exponential at first, fastest at $\frac{L}{2}$ L/2, levels off at the carrying capacity $L$ L.
Half-life of $k e^{- r t}$ ke^(−rt) is $\frac{l n 2}{r}$ (ln 2)/r; $a + b ln x$ a + bln x keeps growing ever more slowly.
Straight semi-log graph: $y = a b^{x}$ y = ab^x. Straight log-log graph: $y = a x^{b}$ y = ax^b. Gradient and intercept give the parameters once you undo the logs.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Functions & Modelling (AI HL) for IB Diploma Mathematics: Applications & Interpretation HL. 🎉

Mission 1Two taxis, two price lines: when does the cheaper one switch?
Mission 2Undo "double it, then add 3": why do you subtract 3 first?
Mission 3A phone loses 20% of its value every year. Is it ever worth nothing?
Mission 4The tide rises and falls like a wave. When is the harbour too shallow?
Mission 5Why do rumours and crazes spread fast, then level off?

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