Mini-Lesson

Functions

4 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 fountain jet traces a parabola. How far out does it land?
Mission 4Why does a drug's concentration never quite reach zero?

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.
On a no-calculator paper, set $f (x) = g (x)$ f(x) = g(x) and solve algebraically instead.

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.
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.
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 ( $y = f (x) + 2$ y = f(x) + 2 with $b = 2$ b = 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 fountain jet traces a parabola. How far out does it land?

Make a guess: A fountain jet traces a parabola. How far out does it land?

Functions · quadratics

Three forms of a quadratic

A quadratic graph is a parabola: $a > 0$ a > 0 opens upwards (minimum), $a < 0$ a < 0 downwards (maximum).
Standard form $f (x) = a x^{2} + b x + c$ f(x) = ax² + bx + c: $y$ y-intercept $(0, c)$ (0, c); axis of symmetry $x = - \frac{b}{2 a}$ x = −b/2a.
Intercept form $f (x) = a (x - p) (x - q)$ f(x) = a(x − p)(x − q): $x$ x-intercepts $(p, 0)$ (p, 0) and $(q, 0)$ (q, 0); axis halfway, $x = \frac{p + q}{2}$ x = (p + q)/2.
Vertex form $f (x) = a (x - h)^{2} + k$ f(x) = a(x − h)² + k: vertex $(h, k)$ (h, k); axis $x = h$ x = h.
$x^{2} - 6 x + 5 = (x - 1) (x - 5) = (x - 3)^{2} - 4$ x² − 6x + 5 = (x − 1)(x − 5) = (x − 3)² − 4: one graph, three views.

Worked example Functions · quadratics

Changing between forms

Question

f (x) = 2 x^{2} + 8 x - 10

. Write

f (x)

in the form

a (x - h)^{2} + k

and in the form

a (x - p) (x - q)

. Hence state the vertex and the

x

-intercepts.

1
Take out the factor 2 from the $x$ x terms: $2 (x^{2} + 4 x) - 10$ 2(x² + 4x) − 10
2
Complete the square inside: $2 [(x + 2)^{2} - 4] - 10$ 2[(x + 2)² − 4] − 10
3
Multiply out the bracket and tidy: $2 (x + 2)^{2} - 8 - 10 = 2 (x + 2)^{2} - 18$ 2(x + 2)² − 8 − 10 = 2(x + 2)² − 18
4
Factorise the standard form: $2 (x^{2} + 4 x - 5) = 2 (x - 1) (x + 5)$ 2(x² + 4x − 5) = 2(x − 1)(x + 5)

Answer

f (x) = 2 (x + 2)^{2} - 18 = 2 (x - 1) (x + 5)

; vertex

(- 2, - 18)

;

x

-intercepts

(1, 0)

and

(- 5, 0)

⚠️ Watch out

the

- 4

inside the square bracket is also multiplied by 2. Writing

2 (x + 2)^{2} - 4 - 10

gives the wrong vertex

(- 2, - 14)

.

Quick check

?

f (x) = - 3 (x + 2)^{2} + 5

. What are the coordinates of the vertex of the graph of

f

?

Functions · quadratics

Solving, and the discriminant

Solve $a x^{2} + b x + c = 0$ ax² + bx + c = 0 by factorising, completing the square or the quadratic formula:

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

The solutions are the roots, or zeros of $f$ f. Check them with your GDC where allowed.
The discriminant $Δ = b^{2} - 4 a c$ Δ = b² − 4ac: $Δ > 0$ Δ > 0 two distinct real roots; $Δ = 0$ Δ = 0 two equal real roots (the graph touches the $x$ x-axis); $Δ < 0$ Δ < 0 no real roots.

Sort it

?Use the discriminant to sort each equation by the nature of its roots.

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

Calculate

?The equation

3 k x^{2} + 2 x + k = 0

has two equal real roots. Find the positive value of

k

. Give your answer to 3 significant figures.

Worked example Functions · quadratics

A quadratic inequality

Question

Solve

2 x^{2} + x - 6 \leq 0

.

1
Solve the equation first: $2 x^{2} + x - 6 = (2 x - 3) (x + 2) = 0$ 2x² + x − 6 = (2x − 3)(x + 2) = 0, so $x = 1.5$ x = 1.5 or $x = - 2$ x = −2
2
Sketch: $a = 2 > 0$ a = 2 > 0, so the parabola opens upwards and is below the $x$ x-axis between its zeros
3
Include the endpoints, since the inequality is $\leq$ ≤

Answer

- 2 \leq x \leq 1.5

⚠️ Watch out

for

> 0

or

\geq 0

with

a > 0

, the solution is the two outside parts,

x < p

or

x > q

, not one interval.

Quick check

?The equation

3 k x^{2} + 2 x + k = 0

has two distinct real roots. Find the possible values of

k

.

Calculate

?A farmer builds a rectangular pen against a long straight wall, using 40 m of fencing for the other three sides. The two sides at right angles to the wall are each

x

m long. Show that the area is

A = x (40 - 2 x)

, then find the maximum possible area.

m²

Mission check

?The jet follows

h = - 0.5 x^{2} + 2 x + 0.5

, starting 0.5 m above the water. How far from the nozzle does it land in the water?

Mission 4

Why does a drug's concentration never quite reach zero?

Make a guess: Why does a drug's concentration never quite reach zero?

Functions · rational, exponential and log functions

Reciprocal and rational functions

The reciprocal function $f (x) = \frac{1}{x}$ f(x) = 1/x, $x \neq = 0$ x ≠ 0, has asymptotes $x = 0$ x = 0 and $y = 0$ y = 0.
It is self-inverse: $f^{- 1} (x) = \frac{1}{x}$ f⁻¹(x) = 1/x, so its graph is symmetric in $y = x$ y = x.
For $f (x) = \frac{a x + b}{c x + d}$ f(x) = (ax + b)/(cx + d): vertical asymptote $x = - \frac{d}{c}$ x = −d/c (denominator zero); horizontal asymptote $y = \frac{a}{c}$ y = a/c (value as $x$ x grows large).
A sketch shows every asymptote, as a dashed line, and both axis intercepts.

Worked example Functions · rational, exponential and log functions

Asymptotes and intercepts

Question

Sketch

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

, stating its asymptotes and axis intercepts.

1
Vertical asymptote, where the denominator is zero: $x + 3 = 0$ x + 3 = 0, so $x = - 3$ x = −3
2
Horizontal asymptote, $y = \frac{a}{c}$ y = a/c: $y = \frac{2}{1} = 2$ y = 2/1 = 2
3
$x$ x-intercept, where the numerator is zero: $2 x - 1 = 0$ 2x − 1 = 0, so $(0.5, 0)$ (0.5, 0)
4
$y$ y-intercept, $f (0) = \frac{- 1}{3}$ f(0) = (−1)/3: $(0, - \frac{1}{3})$ (0, −1/3)

Answer

asymptotes

x = - 3

and

y = 2

; intercepts

(0.5, 0)

and

(0, - \frac{1}{3})

⚠️ Watch out

the vertical asymptote is

x = - \frac{d}{c}

; reading

x = 3

from "

x + 3

" is a sign slip.

Quick check

?Which gives the asymptotes of the graph of

f (x) = \frac{3 x + 2}{2 x - 4}

?

Functions · rational, exponential and log functions

Exponential and logarithmic functions

Exponential: $f (x) = a^{x}$ f(x) = a^x ( $a > 0$ a > 0) and $f (x) = e^{x}$ f(x) = e^x pass through $(0, 1)$ (0, 1), have range $f (x) > 0$ f(x) > 0 and horizontal asymptote $y = 0$ y = 0.
Logarithmic: $f (x) = lo g_{a} x$ f(x) = log_a x and $f (x) = ln x$ f(x) = ln x, domain $x > 0$ x > 0, pass through $(1, 0)$ (1, 0) and have vertical asymptote $x = 0$ x = 0.
They are inverses: $lo g_{a} a^{x} = x$ log_a a^x = x and $a^{l o g_{a} x} = x$ a^(log_a x) = x, so their graphs are reflections in $y = x$ y = x.
Any base can be changed to e: $a^{x} = e^{x l n a}$ a^x = e^(x ln a).
Exponential decay $e^{- k x}$ e^(−kx) ( $k > 0$ k > 0): each increase of 1 in $x$ x multiplies the value by the same factor $e^{- k}$ e^(−k), between 0 and 1.

Calculate

?Write

5^{x}

in the form

e^{k x}

. Find

k

, giving your answer to 3 significant figures.

Functions · rational, exponential and log functions

Solving equations analytically and graphically

To bring down a power, take logs: $a^{x} = b$ a^x = b gives $x = lo g_{a} b = \frac{ln b}{ln a}$ x = log_a b = (ln b)/(ln a).
Spot hidden quadratics: $e^{2 x} = (e^{x})^{2}$ e^2x = (e^x)², so substitute $u = e^{x}$ u = e^x.
When there is no analytic method, as in $e^{x} = sin x$ e^x = sin x or $x^{4} + 5 x - 7 = 0$ x⁴ + 5x − 7 = 0, use your GDC: graph both sides and find the intersections, or graph one side minus the other and find the zeros.

Worked example Functions · rational, exponential and log functions

A hidden quadratic

Question

Solve

e^{2 x} - 5 e^{x} + 4 = 0

, giving exact answers.

1
Substitute $u = e^{x}$ u = e^x: $u^{2} - 5 u + 4 = 0$ u² − 5u + 4 = 0
2
Factorise: $(u - 1) (u - 4) = 0$ (u − 1)(u − 4) = 0, so $u = 1$ u = 1 or $u = 4$ u = 4
3
Return to $x$ x: $e^{x} = 1$ e^x = 1 gives $x = 0$ x = 0; $e^{x} = 4$ e^x = 4 gives $x = ln 4$ x = ln 4

Answer

x = 0

or

x = ln 4

⚠️ Watch out

e^{x} > 0

for every

x

, so a negative value of

u

would give no solution and must be rejected.

Calculate

?A cup of coffee cools so that its temperature,

t

minutes after it is poured, is

T (t) = 20 + 70 e^{- 0.05 t}

°C. Find the time taken for it to cool to 40 °C. Give your answer to 3 significant figures.

minutes

Calculate

?Use your GDC to solve

2^{x} = x + 3

. Give the positive solution to 3 significant figures.

Mission check

?A model gives

C (t) = 20 e^{- 0.3 t}

mg per litre. Why does the model's concentration never reach zero?

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.
Standard form gives the $y$ y-intercept, intercept form the $x$ x-intercepts, vertex form the vertex $(h, k)$ (h, k); the axis of symmetry is $x = - \frac{b}{2 a}$ x = −b/2a.
Solve by factorising, completing the square or $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ x = (−b ± √(b² − 4ac))/2a; $Δ = b^{2} - 4 a c$ Δ = b² − 4ac decides two, one repeated or no real roots.
For a quadratic inequality, find the roots, sketch the parabola and read off where it is above or below the axis.
$f (x) = \frac{a x + b}{c x + d}$ f(x) = (ax + b)/(cx + d) has vertical asymptote $x = - \frac{d}{c}$ x = −d/c and horizontal asymptote $y = \frac{a}{c}$ y = a/c; $\frac{1}{x}$ 1/x is its own inverse.
$a^{x}$ a^x and $lo g_{a} x$ log_a x are inverse functions: $a^{x} > 0$ a^x > 0 with asymptote $y = 0$ y = 0; $lo g_{a} x$ log_a x needs $x > 0$ x > 0 and has asymptote $x = 0$ x = 0; $a^{x} = e^{x l n a}$ a^x = e^(x ln a).
Solve with logs or a substitution such as $u = e^{x}$ u = e^x when you can; otherwise find intersections or zeros on your GDC.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Functions for IB Diploma Mathematics: Analysis & Approaches SL. 🎉

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 fountain jet traces a parabola. How far out does it land?
Mission 4Why does a drug's concentration never quite reach zero?

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