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IB Diploma Mathematics: Applications & Interpretation SL · Functions & Modelling
Mini-Lesson

Functions & Modelling (AI SL)

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

  1. Mission 1Two taxis, two price lines: when does the cheaper one switch?
  2. Mission 2A phone loses 20% of its value every year. Is it ever worth nothing?
  3. Mission 3The tide rises and falls like a wave. When is the harbour too shallow?
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 and intercept−1123456−112345Oxyy=0.5x+142c=1
  • Gradient m = (y₂ − y₁)/(x₂ − x₁): the rate of change of y per unit of x.
  • Gradient-intercept form: y = mx + c, where c is the y-intercept.
  • Point-gradient form: y − y₁ = m(x − x₁) through the point (x₁, y₁).
  • General form: ax + by + d = 0.
  • A road climbing 1.2 m over 15 m horizontally has gradient 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 ax + by + d = 0, where a, b, d ∈ ℤ.
  1. 1
    Gradient: m = (−3 − 5)/(6 − 2) = (−8)/4 = −2
  2. 2
    Point-gradient form through A: y − 5 = −2(x − 2)
  3. 3
    Expand: y = −2x + 9
  4. 4
    Move every term to one side: 2x + y − 9 = 0
Answer
2x + y − 9 = 0
⚠️ Watch out
subtract the coordinates in the same order. (5 − (−3))/(6 − 2) = 2 has the wrong sign.
Functions · lines and graphs

Parallel and perpendicular lines

  • Parallel lines have equal gradients: m₁ = m₂.
  • Perpendicular lines: m₁ × m₂ = −1, so m₂ = −1/m₁ (the negative reciprocal).
  • To read a gradient from general form, rearrange to y = mx + c. For 2x + 5y − 7 = 0: y = −2/5x + 7/5, so m = −2/5.
  • Any line perpendicular to it has gradient 5/2.
Calculate
?The line L passes through (3, 1) and is perpendicular to 2x + 5y − 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 and its inverse−6−4−2246−6−4−22468Oxyff−1y=x(2,3)(3,2)
  • A function gives each input exactly one output. Notation: f(x), v(t), C(n).
  • The domain is the set of allowed inputs; the range is the set of outputs. f(x) = √(2 − x) has domain x ≤ 2 and range f(x) ≥ 0.
  • The inverse function f⁻¹ undoes f: solving f(x) = 10 is the same as finding f⁻¹(10).
  • The graph of f⁻¹ is the reflection of f in y = x. Only one-to-one functions have inverses; the domain of f⁻¹ is the range of f.
Quick check
?The function f is defined by f(x) = √(x − 3) + 1, for x ≥ 3. Which is the range of f?
Functions · lines and graphs

Key features of graphs, with technology

Parabola and linex=1.5−11234−4−3−2−11234567xyy=x2−3xy=2x−3(0,0)(3,0)minimum (1.5,−2.25)
  • Sketch: the shape, with axes and key features labelled. Draw: accurate, to scale.
  • Key features: intercepts, zeros (roots of f(x) = 0), maximum and minimum points, vertex, symmetry, vertical and horizontal asymptotes.
  • To solve 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), whose zeros are the same x-values.
Worked example Functions · lines and graphs

Where two graphs meet

Question
f(x) = x² − 3x and g(x) = 2x − 3. Find the x-coordinates of the points where the graphs of f and g intersect. Give your answers to 3 significant figures.
  1. 1
    Set the functions equal: x² − 3x = 2x − 3
  2. 2
    Rearrange to zero: x² − 5x + 3 = 0
  3. 3
    Use the intersect tool on your GDC (or the quadratic formula): 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) = −4.9t² + 12t + 2, for t ≥ 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

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

Phone plan cost01234567891004812162024Data used g (GB)Cost (€)€12 fixed€2.50 perextra GB
  • A linear model f(x) = mx + c fits a quantity that changes by the same amount per unit of x: m is the rate, 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.
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.
Functions · modelling

Exponential growth and decay models

A cup of tea coolingT=20010203040020406080100Time t (minutes)T (°C)(0,90)
  • An exponential model fits a quantity that changes by the same percentage each step: f(x) = ka^x + c or f(x) = ka^(−x) + c (with a > 0), or f(x) = ke^rx + c.
  • In ka^x: a > 1 is growth, 0 < a < 1 is decay. A 20% fall each year means a = 0.8; a 5% rise means a = 1.05.
  • The starting value is f(0) = k + c.
  • The graph has a horizontal asymptote y = c: decay models get ever closer to 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. 1
    (a) Substitute t = 0: T(0) = 20 + 70 = 90 °C
  2. 2
    (b) Substitute t = 10: T(10) = 20 + 70(0.92)¹⁰ = 50.4 °C (3 sf)
  3. 3
    (c) As t grows, 70(0.92)^t → 0, so T = 20: the room temperature
  4. 4
    (d) Use your GDC to solve 20 + 70(0.92)^t = 40 (intersect with y = 40): 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) = axⁿ, n ∈ ℤ.
  • n > 0: direct variation, such as y = ax² (y varies as the square of x).
  • n < 0: inverse variation, such as y = a/x² = ax⁻². The y-axis is a vertical asymptote.
  • Find a from one data point: if y = ax⁻² and y = 45 when x = 2, then a = 45 × 2² = 180, and y = 20 when 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) = ka^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 3

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) = ax² + bx + c, 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) or minimum (a > 0); the axis of symmetry is 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) = −4.9t² + 14t + 1.5. Find (a) the maximum height, (b) when the ball lands, (c) a reasonable domain for the model.
  1. 1
    (a) Axis of symmetry: t = −14/(2(−4.9)) = 1.43 s (3 sf)
  2. 2
    Substitute: h(1.43) = 11.5 m (using the exact value t = 10/7)
  3. 3
    (b) Use your GDC to solve h(t) = 0: t = −0.103 or t = 2.96; reject the negative root
  4. 4
    (c) From the throw to the landing: 0 ≤ t ≤ 2.96
Answer
(a) 11.5 m (b) after 2.96 s (c) 0 ≤ t ≤ 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) = 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 cut out has V = x(30 − 2x)(20 − 2x), domain 0 < x < 10. A GDC gives the maximum: V ≈ 1,056 cm³ at x ≈ 3.92.
  • Finding parameters: substitute known points into the model. Three points give three linear equations in a, b, c: solve them with your GDC.
Calculate
?A ball's path is modelled by y = ax² + bx + 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

Tide depth061218240123456789Time t (hours)d (m)d=5amplitude 3period 12
f(x) = asin (bx) + d or f(x) = acos (bx) + d
  • Amplitude |a| = (max − min)/2; principal axis y = d = (max + min)/2.
  • Period = (360°)/b, with x in degrees: the time for one full cycle.
  • acos (bx) + d with a > 0 starts at a maximum; with a < 0 it starts at a minimum.
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) = acos (bt) + d, with t in minutes. (b) For how long in each revolution is the seat above 34 m?
  1. 1
    Amplitude (44 − 4)/2 = 20; it starts at a minimum, so a = −20
  2. 2
    Principal axis d = (44 + 4)/2 = 24; period 30 = 360/b, so b = 12
  3. 3
    So h(t) = −20cos (12t°) + 24
  4. 4
    Use your GDC to solve h(t) = 34: t = 10 and t = 20, and h > 34 between them
Answer
(a) h(t) = −20cos (12t°) + 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) = 3cos (30t°) + 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.
Mission check

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

Big ideas

  • A line has gradient m = (y₂ − y₁)/(x₂ − x₁) and can be written y = mx + c, y − y₁ = m(x − x₁) or ax + by + d = 0; parallel means m₁ = m₂, perpendicular means m₁ m₂ = −1.
  • A function has a domain (inputs) and a range (outputs); f⁻¹ undoes f, exists for one-to-one functions and is the reflection of f in y = x.
  • Key features (intercepts, zeros, maximum and minimum, asymptotes) and intersections come from a GDC or from solving f(x) = g(x).
  • Same amount each step: linear f(x) = mx + c (piecewise if the rule changes). Same percentage each step: exponential ka^x + c.
  • An exponential model ka^x + c has horizontal asymptote y = c; decay approaches it but never reaches it.
  • Variation axⁿ: find a from one point. Choose models from context and data shape; beware extrapolation.
  • 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 (max − min)/2, principal axis y = (max + min)/2, period (360°)/b.
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Mini-lesson complete!

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

  1. Mission 1Two taxis, two price lines: when does the cheaper one switch?
  2. Mission 2A phone loses 20% of its value every year. Is it ever worth nothing?
  3. Mission 3The tide rises and falls like a wave. When is the harbour too shallow?

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Specification points and sources for this lesson are listed at the bottom of the page.

Sources (2)

The specification points and sources behind this lesson.

  1. 1 IB Mathematics: applications and interpretation guide (first assessment 2021), SL 2.1, SL 2.2, SL 2.3, SL 2.4
  2. 2 IB Mathematics: applications and interpretation guide (first assessment 2021), SL 2.5, SL 2.6