Mini-Lesson

Functions

6 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?
Mission 5How can you add a quartic's four roots without finding any of them?
Mission 6A function is its own inverse. What must its graph look like?

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?

Mission 5

How can you add a quartic's four roots without finding any of them?

Make a guess: How can you add a quartic's four roots without finding any of them?

Functions · polynomials

Polynomial graphs, zeros and factors

A degree $n$ n polynomial has at most $n$ n real zeros: the $x$ x-intercepts of $y = P (x)$ y = P(x), the real roots of $P (x) = 0$ P(x) = 0.
Each root $α$ α gives a factor $(x - α)$ (x − α). Counting complex and repeated roots, there are exactly $n$ n.
Single root: the graph crosses the axis. Double root: it touches and turns. Triple root: it crosses, flattening.
The sign of $a_{n}$ aₙ and the parity of $n$ n fix the ends: $y = x^{4} - 3 x^{3} - \dots$ y = x⁴ − 3x³ − … rises at both ends.

Sort it

?At each given root, does the graph cross or touch the

x

-axis?

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

Functions · polynomials

The remainder and factor theorems

Remainder theorem: when $P (x)$ P(x) is divided by $(x - a)$ (x − a), the remainder is $P (a)$ P(a).
Dividing by $(a x - b)$ (ax − b), the remainder is $P (\frac{b}{a})$ P(b/a): use the value of $x$ x that makes the divisor zero.
Factor theorem: $(x - a)$ (x − a) is a factor of $P (x)$ P(x) if and only if $P (a) = 0$ P(a) = 0.
To find unknown coefficients, turn each piece of information into an equation in them, then solve simultaneously.

Worked example Functions · polynomials

Finding unknown coefficients

Question

P (x) = x^{3} + a x^{2} + b x - 6

.

(x - 2)

is a factor of

P (x)

, and the remainder when

P (x)

is divided by

(x + 1)

is

- 12

. Find

a

and

b

.

1
Factor theorem, $P (2) = 0$ P(2) = 0: $8 + 4 a + 2 b - 6 = 0$ 8 + 4a + 2b − 6 = 0, so $2 a + b = - 1$ 2a + b = −1
2
Remainder theorem, $P (- 1) = - 12$ P(−1) = −12: $- 1 + a - b - 6 = - 12$ −1 + a − b − 6 = −12, so $a - b = - 5$ a − b = −5
3
Add the two equations: $3 a = - 6$ 3a = −6, so $a = - 2$ a = −2, then $b = 3$ b = 3
4
Check: $P (x) = x^{3} - 2 x^{2} + 3 x - 6 = (x - 2) (x^{2} + 3)$ P(x) = x³ − 2x² + 3x − 6 = (x − 2)(x² + 3)

Answer

a = - 2

,

b = 3

⚠️ Watch out

dividing by

(x + 1)

means substituting

x = - 1

, not

x = 1

.

Calculate

?Find the remainder when

2 x^{3} - 3 x^{2} + 4 x - 5

is divided by

(2 x - 1)

. Give your answer as a decimal.

Worked example Functions · polynomials

Factorising a cubic fully

Question

Factorise

P (x) = 2 x^{3} + 3 x^{2} - 11 x - 6

fully and solve

P (x) = 0

.

1
Try small factors of the constant term: $P (1) = - 12$ P(1) = −12, but $P (2) = 16 + 12 - 22 - 6 = 0$ P(2) = 16 + 12 − 22 − 6 = 0, so $(x - 2)$ (x − 2) is a factor
2
Divide (or compare coefficients): $P (x) = (x - 2) (2 x^{2} + 7 x + 3)$ P(x) = (x − 2)(2x² + 7x + 3)
3
Factorise the quadratic: $2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3)$ 2x² + 7x + 3 = (2x + 1)(x + 3)
4
Set each factor to zero: $x = 2$ x = 2, $x = - \frac{1}{2}$ x = −1/2, $x = - 3$ x = −3

Answer

P (x) = (x - 2) (2 x + 1) (x + 3)

; roots

- 3

,

- \frac{1}{2}

, 2

⚠️ Watch out

with a leading coefficient of 2, a root can be a fraction such as

- \frac{1}{2}

: keep the factor as

(2 x + 1)

.

Functions · polynomials

Sum and product of the roots

For $a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{1} x + a_{0} = 0$ aₙ xⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁ x + a₀ = 0 with roots $α_{1}, \dots, α_{n}$ α₁, …, αₙ: $sum = \frac{- a _{n - 1}}{a _{n}} product = \frac{( - 1 ) ^{n} a _{0}}{a _{n}}$ sum = (−aₙ₋₁)/aₙ product = ((−1)ⁿ a₀)/aₙ

Why: $a_{n} (x - α_{1}) \dots (x - α_{n})$ aₙ(x − α₁)⋯(x − αₙ) expands to $a_{n} x^{n} - a_{n} (α_{1} + \dots + α_{n}) x^{n - 1} + \dots + a_{n} (- 1)^{n} α_{1} \dots α_{n}$ aₙ xⁿ − aₙ(α₁ + … + αₙ)xⁿ⁻¹ + … + aₙ(−1)ⁿα₁⋯αₙ.
Count every root: repeated roots as often as they occur, and complex roots too. $x^{3} - 2 x^{2} + 3 x - 6 = 0$ x³ − 2x² + 3x − 6 = 0 has roots 2 and $\pm 3 i$ ±√3 i: sum 2, product 6.
The formula is in the formula booklet; the sign $(- 1)^{n}$ (−1)ⁿ is where marks are lost.

Quick check

?What is the product of the roots of

3 x^{4} - 5 x^{3} + 2 x - 12 = 0

?

Calculate

?The equation

x^{3} - 6 x^{2} + k x - 6 = 0

has three roots, one of which is equal to the sum of the other two. Find the value of

k

.

Mission check

?How can you add a quartic's four roots without finding any of them? Take

2 x^{4} - 14 x^{3} + 18 x^{2} + 14 x - 20 = 0

.

Mission 6

A function is its own inverse. What must its graph look like?

Make a guess: A function is its own inverse. What must its graph look like?

Functions · rational functions and transformations

Rational functions and their asymptotes

Vertical asymptotes: where the denominator is zero (and the numerator is not).
$\frac{a x + b}{c x ^{2} + d x + e}$ (ax + b)/(cx² + dx + e): the denominator's degree is higher, so $y \to 0$ y → 0: horizontal asymptote $y = 0$ y = 0. Up to two vertical asymptotes.
$\frac{a x ^{2} + b x + c}{d x + e}$ (ax² + bx + c)/(dx + e): divide to get $m x + k + \frac{r}{d x + e}$ mx + k + r/(dx + e). The last term $\to 0$ → 0, so $y = m x + k$ y = mx + k is an oblique asymptote.
A full sketch shows every asymptote (with its equation) and every axis intercept.

Worked example Functions · rational functions and transformations

Finding an oblique asymptote

Question

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

. Find the asymptotes and the axis intercepts of the graph of

f

.

1
Vertical asymptote where $x - 2 = 0$ x − 2 = 0: $x = 2$ x = 2
2
Divide: $2 x^{2} - x - 3 = (x - 2) (2 x + 3) + 3$ 2x² − x − 3 = (x − 2)(2x + 3) + 3, so $f (x) = 2 x + 3 + \frac{3}{x - 2}$ f(x) = 2x + 3 + 3/(x − 2)
3
As $x \to \pm \infty$ x → ±∞, $\frac{3}{x - 2} \to 0$ 3/(x − 2) → 0: oblique asymptote $y = 2 x + 3$ y = 2x + 3
4
Intercepts: $(2 x - 3) (x + 1) = 0$ (2x − 3)(x + 1) = 0 gives $x = \frac{3}{2}$ x = 3/2 and $x = - 1$ x = −1; $f (0) = \frac{- 3}{- 2} = \frac{3}{2}$ f(0) = (−3)/(−2) = 3/2

Answer

x = 2

and

y = 2 x + 3

; intercepts

(- 1, 0)

,

(\frac{3}{2}, 0)

,

(0, \frac{3}{2})

⚠️ Watch out

the remainder is not part of the asymptote.

y = 2 x + 3

is the quotient only.

Quick check

?Which are all the asymptotes of

y = \frac{3 x - 6}{x ^{2} - x - 6}

?

Functions · rational functions and transformations

Odd, even and self-inverse functions

Even: $f (- x) = f (x)$ f(−x) = f(x). The graph is symmetric in the $y$ y-axis: $x^{2}$ x², $cos x$ cos x.
Odd: $f (- x) = - f (x)$ f(−x) = −f(x). Rotational symmetry of order 2 about the origin: $x^{3}$ x³, $sin x$ sin x, $tan x$ tan x.
$f^{- 1}$ f⁻¹ exists only where $f$ f is one-to-one, so you may need to restrict the domain: $x^{2}$ x² with $x \geq 0$ x ≥ 0 has inverse $x$ √x. The graph of $f^{- 1}$ f⁻¹ is the reflection of $f$ f in $y = x$ y = x.
Self-inverse: $f^{- 1} (x) = f (x)$ f⁻¹(x) = f(x), so $f (f (x)) = x$ f(f(x)) = x. Examples: $\frac{1}{x}$ 1/x, $5 - x$ 5 − x, $\frac{2 x + 1}{x - 2}$ (2x + 1)/(x − 2).

Calculate

?The function

f (x) = \frac{3 x + 5}{x - k}

,

x \neq = k

, is self-inverse. Find the value of the constant

k

.

Sort it

?Is each function odd, even or neither?

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

Worked example Functions · rational functions and transformations

Solving $g (x) \geq f (x)$ g(x) ≥ f(x): a rational inequality

Question

Solve

\frac{x + 1}{x - 2} \geq 2

.

1
Don't multiply by $x - 2$ x − 2 (it may be negative). Subtract 2 and combine: $\frac{x + 1 - 2 ( x - 2 )}{x - 2} = \frac{5 - x}{x - 2} \geq 0$ (x + 1 − 2(x − 2))/(x − 2) = (5 − x)/(x − 2) ≥ 0
2
Critical values, where the top or the bottom is zero: $x = 5$ x = 5 and $x = 2$ x = 2
3
Sign table (the same method works for a factorised cubic $\geq 0$ ≥ 0): negative for $x < 2$ x < 2, positive for $2 < x < 5$ 2 < x < 5, negative for $x > 5$ x > 5
4
Include $x = 5$ x = 5 (the fraction is 0), exclude $x = 2$ x = 2 (undefined)
5
Check with your GDC: $y = \frac{x + 1}{x - 2}$ y = (x + 1)/(x − 2) is on or above $y = 2$ y = 2 only for $2 < x \leq 5$ 2 < x ≤ 5

Answer

2 < x \leq 5

⚠️ Watch out

multiplying both sides by

x - 2

gives

x \leq 5

alone, which wrongly includes values below 2, where

x - 2

is negative.

Functions · rational functions and transformations

Modulus and other transformations

Graph	How to get it from $y = f (x)$ y = f(x)
$y = ∣ f (x)∣$ y = \| f(x) \|	reflect the parts below the $x$ x-axis in the $x$ x-axis
$y = f (∣ x ∣)$ y = f(\| x \|)	keep the part for $x \geq 0$ x ≥ 0; reflect it in the $y$ y-axis
$y = \frac{1}{f ( x )}$ y = 1/(f(x))	zeros become vertical asymptotes; $y$ y-values become reciprocals ( $y = \pm 1$ y = ± 1 stay put)
$y = f (a x + b)$ y = f(ax + b)	translate by $- b$ −b horizontally, then stretch horizontally by scale factor $\frac{1}{a}$ 1/a
$y = [f (x)]^{2}$ y = [f(x)]²	never negative; zeros become points where the graph touches

Modulus equations: $∣ f (x) ∣ = c$ |f(x)| = c means $f (x) = c$ f(x) = c or $f (x) = - c$ f(x) = −c.
Modulus inequalities: $∣ f (x) ∣ < c$ |f(x)| < c means $- c < f (x) < c$ −c < f(x) < c; use a graph or GDC when it gets harder.

Calculate

?How many integers

x

satisfy

∣ x^{2} - 4∣ < 3

?

Mission check

?A function is its own inverse. What must its graph look like?

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.
A root $α$ α of $P (x) = 0$ P(x) = 0 means a factor $(x - α)$ (x − α) and an $x$ x-intercept; single roots cross, double roots touch.
Remainder on dividing by $(a x - b)$ (ax − b) is $P (\frac{b}{a})$ P(b/a); $(x - a)$ (x − a) is a factor exactly when $P (a) = 0$ P(a) = 0.
Sum of roots $\frac{- a _{n - 1}}{a _{n}}$ (−aₙ₋₁)/aₙ, product $\frac{( - 1 ) ^{n} a _{0}}{a _{n}}$ ((−1)ⁿ a₀)/aₙ, counting complex and repeated roots.
Rational graphs: vertical asymptotes from the denominator; divide to find an oblique asymptote; mark all intercepts.
Even: $f (- x) = f (x)$ f(−x) = f(x); odd: $f (- x) = - f (x)$ f(−x) = −f(x); self-inverse: $f (f (x)) = x$ f(f(x)) = x, symmetric in $y = x$ y = x.
Inequalities: one side zero, critical values, sign table; never multiply by something that could be negative. $∣ f (x) ∣ < c$ |f(x)| < c means $- c < f (x) < c$ −c < f(x) < c.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Functions for IB Diploma Mathematics: Analysis & Approaches 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 fountain jet traces a parabola. How far out does it land?
Mission 4Why does a drug's concentration never quite reach zero?
Mission 5How can you add a quartic's four roots without finding any of them?
Mission 6A function is its own inverse. What must its graph look like?

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Functions

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

Gradient and the forms of a line

The line through two points

Parallel and perpendicular lines

Functions, domain, range and inverses

Key features of graphs, with technology

Where two graphs meet

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

Composite functions

Inverse functions

An inverse with a restricted domain

Transformations of graphs

Order in a composite transformation

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

Three forms of a quadratic

Changing between forms

Solving, and the discriminant

A quadratic inequality

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

Reciprocal and rational functions

Asymptotes and intercepts

Exponential and logarithmic functions

Solving equations analytically and graphically

A hidden quadratic

How can you add a quartic's four roots without finding any of them?

Polynomial graphs, zeros and factors

The remainder and factor theorems

Finding unknown coefficients

Factorising a cubic fully

Sum and product of the roots

A function is its own inverse. What must its graph look like?

Rational functions and their asymptotes

Finding an oblique asymptote

Odd, even and self-inverse functions

Solving g(x)≥f(x)g(x) ≥ f(x): a rational inequality

Modulus and other transformations

Big ideas

Mini-lesson complete!

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Solving $g (x) \geq f (x)$ g(x) ≥ f(x): a rational inequality