3 missions. Guess, learn, then check. Your place is saved.
Mission 1Two taxis, two price lines: when does the cheaper one switch?
Mission 2A phone loses 20% of its value every year. Is it ever worth nothing?
Mission 3The tide rises and falls like a wave. When is the harbour too shallow?
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
Gradientm=x2−x1y2−y1m = (y₂ − y₁)/(x₂ − x₁): the rate of change of yy per unit of xx.
Gradient-intercept form: y=mx+cy = mx + c, where cc is the yy-intercept.
Point-gradient form: y−y1=m(x−x1)y − y₁ = m(x − x₁) through the point (x1,y1)(x₁, y₁).
General form: ax+by+d=0ax + by + d = 0.
A road climbing 1.2 m over 15 m horizontally has gradient 151.2=0.081.2/15 = 0.08, an 8% incline.
Worked exampleFunctions · lines and graphs
The line through two points
Question
Find the equation of the line through A(2,5)A(2, 5) and B(6,−3)B(6, −3). Give your answer in the form ax+by+d=0ax + by + d = 0, where aa, bb, d∈Zd ∈ ℤ.
Point-gradient form through AA: y−5=−2(x−2)y − 5 = −2(x − 2)
3
Expand: y=−2x+9y = −2x + 9
4
Move every term to one side: 2x+y−9=02x + y − 9 = 0
Answer
2x+y−9=02x + y − 9 = 0
⚠️ Watch out
subtract the coordinates in the same order. 6−25−(−3)=2(5 − (−3))/(6 − 2) = 2 has the wrong sign.
Functions · lines and graphs
Parallel and perpendicular lines
Parallel lines have equal gradients: m1=m2m₁ = m₂.
Perpendicular lines: m1×m2=−1m₁ × m₂ = −1, so m2=−m11m₂ = −1/m₁ (the negative reciprocal).
To read a gradient from general form, rearrange to y=mx+cy = mx + c. For 2x+5y−7=02x + 5y − 7 = 0: y=−52x+57y = −2/5x + 7/5, so m=−52m = −2/5.
Any line perpendicular to it has gradient 255/2.
Calculate
?The line LL passes through (3,1)(3, 1) and is perpendicular to 2x+5y−7=02x + 5y − 7 = 0. Find the yy-intercept of LL. 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−xf(x) = √(2 − x) has domain x≤2x ≤ 2 and range f(x)≥0f(x) ≥ 0.
The inverse functionf−1f⁻¹ undoes ff: solving f(x)=10f(x) = 10 is the same as finding f−1(10)f⁻¹(10).
The graph of f−1f⁻¹ is the reflection of ff in y=xy = x. Only one-to-one functions have inverses; the domain of f−1f⁻¹ is the range of ff.
Quick check
?The function ff is defined by f(x)=x−3+1f(x) = √(x − 3) + 1, for x≥3x ≥ 3. Which is the range of ff?
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)=0f(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 xx-values.
Worked exampleFunctions · lines and graphs
Where two graphs meet
Question
f(x)=x2−3xf(x) = x² − 3x and g(x)=2x−3g(x) = 2x − 3. Find the xx-coordinates of the points where the graphs of ff and gg intersect. Give your answers to 3 significant figures.
1
Set the functions equal: x2−3x=2x−3x² − 3x = 2x − 3
2
Rearrange to zero: x2−5x+3=0x² − 5x + 3 = 0
3
Use the intersect tool on your GDC (or the quadratic formula): x=25±13x = (5 ± √13)/2
Answer
x=0.697x = 0.697 or x=4.30x = 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 tt seconds is modelled by h(t)=−4.9t2+12t+2h(t) = −4.9t² + 12t + 2, for t≥0t ≥ 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?
Each cost is a linear function of distance: the fixed fee is the yy-intercept and the price per km is the gradient. The cheaper taxi switches where the lines intersect: 3.50+1.20d=6.00+0.95d3.50 + 1.20d = 6.00 + 0.95d gives d=10d = 10 km, where both cost £15.50. Below 10 km A's lower fee wins; beyond it B's smaller gradient wins.
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
A linear modelf(x)=mx+cf(x) = mx + c fits a quantity that changes by the same amount per unit of xx: mm is the rate, cc 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
An exponential model fits a quantity that changes by the same percentage each step: f(x)=kax+cf(x) = ka^x + c or f(x)=ka−x+cf(x) = ka^(−x) + c (with a>0a > 0), or f(x)=kerx+cf(x) = ke^rx + c.
In kaxka^x: a>1a > 1 is growth, 0<a<10 < a < 1 is decay. A 20% fall each year means a=0.8a = 0.8; a 5% rise means a=1.05a = 1.05.
The starting value is f(0)=k+cf(0) = k + c.
The graph has a horizontal asymptotey=cy = c: decay models get ever closer to cc without reaching it.
Worked exampleFunctions · modelling
A cooling drink
Question
The temperature of a cup of tea, in °C, tt minutes after it is made, is modelled by T(t)=20+70(0.92)tT(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.
(c) As tt grows, 70(0.92)t→070(0.92)^t → 0, so T=20T = 20: the room temperature
4
(d) Use your GDC to solve 20+70(0.92)t=4020 + 70(0.92)^t = 40 (intersect with y=40y = 40): t=15.0t = 15.0 minutes
Answer
(a) 90 °C (b) 50.4 °C (c) T=20T = 20, room temperature (d) after 15.0 minutes
⚠️ Watch out
the initial value is k+c=90k + c = 90, not k=70k = 70.
Quick check
?A car bought for €24,000 loses 15% of its value each year. Which model gives its value, in euros, after tt years?
Functions · modelling
Direct and inverse variation
Variation models have the form f(x)=axnf(x) = axⁿ, n∈Zn ∈ ℤ.
n>0n > 0: direct variation, such as y=ax2y = ax² (yy varies as the square of xx).
n<0n < 0: inverse variation, such as y=x2a=ax−2y = a/x² = ax⁻². The yy-axis is a vertical asymptote.
Find aa from one data point: if y=ax−2y = ax⁻² and y=45y = 45 when x=2x = 2, then a=45×22=180a = 45 × 2² = 180, and y=20y = 20 when x=3x = 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 tt years after it opened is modelled by P(t)=katP(t) = ka^t. When t=1t = 1 there are 360 members, and when t=4t = 4 there are 622.08 (as given by the model). Use the model to predict the number of members when t=6t = 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?
Losing 20% each year multiplies the value by 0.8, so V=900(0.8)tV = 900(0.8)^t: an exponential decay model with horizontal asymptote V=0V = 0. In the model the phone is never worth nothing. In reality a buyer may pay only scrap value, so a model V=k(0.8)t+cV = k(0.8)^t + c with c>0c > 0 could fit better: that is reflecting on the model.
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+cf(x) = ax² + bx + c, a=0a ≠ 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<0a < 0) or minimum (a>0a > 0); the axis of symmetry is x=−2abx = −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 exampleFunctions · modelling
A thrown ball
Question
A ball is thrown so that its height, in metres, tt seconds later is h(t)=−4.9t2+14t+1.5h(t) = −4.9t² + 14t + 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=−2(−4.9)14=1.43t = −14/(2(−4.9)) = 1.43 s (3 sf)
2
Substitute: h(1.43)=11.5h(1.43) = 11.5 m (using the exact value t=710t = 10/7)
3
(b) Use your GDC to solve h(t)=0h(t) = 0: t=−0.103t = −0.103 or t=2.96t = 2.96; reject the negative root
4
(c) From the throw to the landing: 0≤t≤2.960 ≤ t ≤ 2.96
Answer
(a) 11.5 m (b) after 2.96 s (c) 0≤t≤2.960 ≤ t ≤ 2.96
⚠️ Watch out
the tt-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+df(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 xx cut out has V=x(30−2x)(20−2x)V = x(30 − 2x)(20 − 2x), domain 0<x<100 < x < 10. A GDC gives the maximum: V≈1,056V ≈ 1,056 cm³ at x≈3.92x ≈ 3.92.
Finding parameters: substitute known points into the model. Three points give three linear equations in aa, bb, cc: solve them with your GDC.
Calculate
?A ball's path is modelled by y=ax2+bx+cy = ax² + bx + c, where xx m is the horizontal distance and yy m the height. The path passes through (0,2)(0, 2), (2,8)(2, 8) and (4,6)(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 xx 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)=asin(bx)+dorf(x)=acos(bx)+df(x) = asin (bx) + d or f(x) = acos (bx) + d
Amplitude∣a∣=2max−min|a| = (max − min)/2; principal axisy=d=2max+miny = d = (max + min)/2.
Period=b360∘= (360°)/b, with xx in degrees: the time for one full cycle.
acos(bx)+dacos (bx) + d with a>0a > 0 starts at a maximum; with a<0a < 0 it starts at a minimum.
Worked exampleFunctions · 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)+dh(t) = acos (bt) + d, with tt in minutes. (b) For how long in each revolution is the seat above 34 m?
1
Amplitude 244−4=20(44 − 4)/2 = 20; it starts at a minimum, so a=−20a = −20
2
Principal axis d=244+4=24d = (44 + 4)/2 = 24; period 30=b36030 = 360/b, so b=12b = 12
3
So h(t)=−20cos(12t∘)+24h(t) = −20cos (12t°) + 24
4
Use your GDC to solve h(t)=34h(t) = 34: t=10t = 10 and t=20t = 20, and h>34h > 34 between them
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, tt hours after high tide is modelled by d(t)=3cos(30t∘)+5d(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?
Model the depth with a sinusoidal function: amplitude 3, principal axis d=5d = 5, period 12 hours. The harbour is too shallow while the graph is below d=4d = 4: solve 3cos(30t∘)+5=43cos (30t°) + 5 = 4 with a GDC to get t≈3.65t ≈ 3.65 and t≈8.35t ≈ 8.35, about 4.70 hours around each low tide.
Recap
Big ideas
A line has gradient m=x2−x1y2−y1m = (y₂ − y₁)/(x₂ − x₁) and can be written y=mx+cy = mx + c, y−y1=m(x−x1)y − y₁ = m(x − x₁) or ax+by+d=0ax + by + d = 0; parallel means m1=m2m₁ = m₂, perpendicular means m1m2=−1m₁ m₂ = −1.
A function has a domain (inputs) and a range (outputs); f−1f⁻¹ undoes ff, exists for one-to-one functions and is the reflection of ff in y=xy = 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).
Same amount each step: linear f(x)=mx+cf(x) = mx + c (piecewise if the rule changes). Same percentage each step: exponential kax+cka^x + c.
An exponential model kax+cka^x + c has horizontal asymptote y=cy = c; decay approaches it but never reaches it.
Variation axnaxⁿ: find aa 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 2max−min(max − min)/2, principal axis y=2max+miny = (max + min)/2, period b360∘(360°)/b.
🏆
Mini-lesson complete!
⭐⭐⭐
You've worked through Functions & Modelling (AI SL) for IB Diploma Mathematics: Applications & Interpretation SL. 🎉
Mission 1Two taxis, two price lines: when does the cheaper one switch?
Mission 2A phone loses 20% of its value every year. Is it ever worth nothing?
Mission 3The tide rises and falls like a wave. When is the harbour too shallow?