Mini-Lesson

Algebra and functions

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

Mission 1Why does $8^{2/3}$ 8^(2/3) come out as exactly 4?
Mission 2A ball's height is $h = 20 t - 5 t^{2}$ h = 20t − 5t². Does it ever reach 25 m?
Mission 3When does a line just touch a curve, and what one number tells you?
Mission 4 $x^{3} - 6 x^{2} + 11 x - 6 = 0$ x³−6x²+11x−6=0 has an easy root. How do the other two follow?
Mission 5 $y = f (2 x)$ y=f(2x) is narrower, but $y = 2 f (x)$ y=2f(x) is taller. Why opposite?
Mission 6Why must $y = x^{2}$ y=x² lose half its domain before it has an inverse?
Mission 7Why is the sum of $\frac{1}{n ( n + 1 )}$ 1/(n(n+1)) from $n = 1$ n=1 to $99$ 99 so neat?

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

Mission 1

Why does $8^{2/3}$ 8^(2/3) come out as exactly 4?

Make a guess: Why does $8^{2/3}$ 8^(2/3) come out as exactly 4?

Algebra · indices and surds

The laws of indices

Multiply, add the powers: $a^{m} \times a^{n} = a^{m + n}$ a^m × aⁿ = a^(m+n). Divide, subtract: $a^{m} \div a^{n} = a^{m - n}$ a^m ÷ aⁿ = a^(m−n).
Power of a power, multiply: $(a^{m})^{n} = a^{mn}$ (a^m)ⁿ = a^mn.
Zero and negative indices: $a^{0} = 1$ a⁰ = 1 and $a^{- n} = \frac{1}{a ^{n}}$ a⁻ⁿ = 1/aⁿ, e.g. $2^{- 3} = \frac{1}{8}$ 2⁻³ = 1/8.
The laws need the same base: $2^{3} \times 3^{2}$ 2³ × 3² cannot be combined into one power.

Algebra · indices and surds

Fractional indices

$(a^{1/ n})^{n} = a^{\frac{1}{n} \times n} = a^{1}$ (a^(1/n))ⁿ = a^(1/n × n) = a¹, so $a^{1/ n}$ a^(1/n) is the $n$ nth root: $a^{\frac{1}{n}} = n a$ a^(1/n) = n√a.
Then $a^{\frac{m}{n}} = (n a)^{m} = n a^{m}$ a^(m/n) = (n√a)^m = n√(a^m). The denominator is the root, the numerator is the power.
Root first keeps numbers small: $2 7^{2/3} = (327)^{2} = 3^{2} = 9$ 27^(2/3) = (∛27)² = 3² = 9.
Index form is how A-level writes roots: $x = x^{\frac{1}{2}}$ √x = x^(1/2) and $\frac{1}{x} = x^{- \frac{1}{2}}$ 1/(√x) = x^(−1/2).

Worked example Algebra · indices and surds

A negative fractional index

Question

Without a calculator, evaluate

(\frac{16}{81})^{- \frac{3}{4}}

.

1
Deal with the minus sign by taking the reciprocal: $(\frac{81}{16})^{\frac{3}{4}}$ (81/16)^(3/4)
2
Take the fourth root of top and bottom: $\frac{4 81}{4 16} = \frac{3}{2}$ (∜81)/(∜16) = 3/2
3
Cube: $(\frac{3}{2})^{3} = \frac{27}{8}$ (3/2)³ = 27/8

Answer

\frac{27}{8}

⚠️ Watch out

a negative index means "reciprocal", not "negative number".

(\frac{16}{81})^{- 3/4}

is positive.

Calculate

?Without a calculator, evaluate

3 2^{- \frac{2}{5}}

. Give your answer as a decimal.

Quick check

?Write

\frac{6 x ^{2} + x}{2 x}

in the form

a x^{p} + b x^{q}

.

Algebra · indices and surds

Simplifying surds

A surd is an irrational root such as $2$ √2. Keep it exact.
$ab = a b$ √ab = √a√b and $\frac{a}{b} = \frac{a}{b}$ √(a/b) = (√a)/(√b). Take out the largest square factor: $72 = 362 = 62$ √72 = √36√2 = 6√2.
$(a)^{2} = a$ (√a)² = a and $(x + y) (x - y) = x - y$ (√x + √y)(√x − √y) = x − y.
Never split a sum: $a + b \neq = a + b$ √(a + b) ≠ √a + √b.

Worked example Algebra · indices and surds

Rationalising the denominator

Question

Express

\frac{5}{3 - 2}

in the form

\frac{a + b 2}{c}

, where

a

,

b

and

c

are integers.

1
Multiply top and bottom by the conjugate $3 + 2$ 3 + √2: $\frac{5 ( 3 + 2 )}{( 3 - 2 ) ( 3 + 2 )}$ (5(3 + √2))/((3 − √2)(3 + √2))
2
Difference of two squares on the bottom: $9 - 2 = 7$ 9 − 2 = 7
3
Expand the top: $15 + 52$ 15 + 5√2

Answer

\frac{15 + 5 2}{7}

⚠️ Watch out

multiplying by

3 - 2

gives

(3 - 2)^{2} = 11 - 62

on the bottom, which is still irrational. Change the sign.

Calculate

?Without a calculator, write

72 + 50 - 8

in the form

k 2

. Find

k

.

Calculate

?Without a calculator, write

\frac{4 + 3}{2 - 3}

in the form

a + b 3

, where

a

and

b

are integers. Enter the value of

a

.

Mission check

?Why does

8^{2/3}

come out as exactly 4?

Mission 2

A ball's height is $h = 20 t - 5 t^{2}$ h = 20t − 5t². Does it ever reach 25 m?

Make a guess: A ball's height is $h = 20 t - 5 t^{2}$ h = 20t − 5t². Does it ever reach 25 m?

Algebra · quadratics and the discriminant

Completing the square

Write $a x^{2} + b x + c$ ax² + bx + c as $a (x + p)^{2} + q$ a(x + p)² + q. For $a = 1$ a = 1: halve the $x$ x-coefficient, then correct the constant: $x^{2} + 6 x + 2 = (x + 3)^{2} - 9 + 2 = (x + 3)^{2} - 7$ x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7.
If $a \neq = 1$ a ≠ 1, take $a$ a out of the $x^{2}$ x² and $x$ x terms first.
The turning point of $y = a (x + p)^{2} + q$ y = a(x + p)² + q is $(- p, q)$ (−p, q), on the line of symmetry $x = - p$ x = −p.
$a > 0$ a > 0: a minimum, value $q$ q. $a < 0$ a < 0: a maximum, value $q$ q.

Worked example Algebra · quadratics and the discriminant

Turning point from the completed square

Question

Write

2 x^{2} - 12 x + 23

in the form

a (x + p)^{2} + q

. Hence state the turning point of

y = 2 x^{2} - 12 x + 23

and explain why the equation

2 x^{2} - 12 x + 23 = 0

has no real roots.

1
Take out the 2 from the first two terms: $2 (x^{2} - 6 x) + 23$ 2(x² − 6x) + 23
2
Complete the square inside: $2 [(x - 3)^{2} - 9] + 23$ 2[(x − 3)² − 9] + 23
3
Expand the bracket's constant: $2 (x - 3)^{2} - 18 + 23 = 2 (x - 3)^{2} + 5$ 2(x − 3)² − 18 + 23 = 2(x − 3)² + 5
4
Read off: minimum point $(3, 5)$ (3, 5). Since $2 (x - 3)^{2} \geq 0$ 2(x − 3)² ≥ 0, $y \geq 5 > 0$ y ≥ 5 > 0 for every $x$ x.

Answer

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

; turning point

(3, 5)

, a minimum; the curve never reaches

y = 0

, so there are no real roots.

⚠️ Watch out

the

- 9

inside the square bracket must be multiplied by 2. Writing

2 (x - 3)^{2} - 9 + 23

gives the wrong

q

.

Calculate

?By completing the square, find the maximum value of

y = 3 + 8 x - 2 x^{2}

.

Algebra · quadratics and the discriminant

The discriminant

For $a x^{2} + b x + c = 0$ ax² + bx + c = 0 the discriminant is $b^{2} - 4 a c$ b² − 4ac, the part under the root in $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ x = (−b ± √(b² − 4ac))/2a.
$b^{2} - 4 a c > 0$ b² − 4ac > 0: two distinct real roots; the graph crosses the $x$ x-axis twice.
$b^{2} - 4 a c = 0$ b² − 4ac = 0: one repeated root; the graph touches the $x$ x-axis.
$b^{2} - 4 a c < 0$ b² − 4ac < 0: no real roots; the graph misses the $x$ x-axis.
"Real roots" means $b^{2} - 4 a c \geq 0$ b² − 4ac ≥ 0.

Quick check

?Find the discriminant of

3 x^{2} + 5 x + 4

. What does it tell you about

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

?

Worked example Algebra · quadratics and the discriminant

Finding $k$ k for equal roots

Question

The equation

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

has equal roots. Find the possible values of

k

.

1
Identify the coefficients: $a = 1$ a = 1, $b = k$ b = k, $c = k + 3$ c = k + 3
2
Equal roots, so set the discriminant to zero: $k^{2} - 4 (k + 3) = 0$ k² − 4(k + 3) = 0
3
Simplify and factorise: $k^{2} - 4 k - 12 = 0$ k² − 4k − 12 = 0, so $(k - 6) (k + 2) = 0$ (k − 6)(k + 2) = 0
4
Check: $k = 6$ k = 6 gives $(x + 3)^{2} = 0$ (x + 3)² = 0; $k = - 2$ k = −2 gives $(x - 1)^{2} = 0$ (x − 1)² = 0.

Answer

k = 6

or

k = - 2

⚠️ Watch out

c

is the whole constant term

k + 3

, not just 3. Using

c = 3

gives

k^{2} - 12 = 0

.

Calculate

?The equation

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

, where

k

is a positive constant, has a repeated root. Find

k

.

Algebra · quadratics and the discriminant

Quadratics in disguise

An equation is a quadratic in a function of $x$ x if it fits $a (\dots)^{2} + b (\dots) + c = 0$ a(…)² + b(…) + c = 0.
$x^{4} - 5 x^{2} + 4 = 0$ x⁴ − 5x² + 4 = 0: let $u = x^{2}$ u = x², so $u^{2} - 5 u + 4 = 0$ u² − 5u + 4 = 0, $u = 1$ u = 1 or $4$ 4, and $x = \pm 1$ x = ± 1 or $\pm 2$ ± 2.
$x - 5 x + 6 = 0$ x − 5√x + 6 = 0: let $u = x$ u = √x, since $x = (x)^{2}$ x = (√x)².
Solve for $u$ u, then convert back to $x$ x. Reject impossible values: $x^{2}$ x² and $x$ √x cannot be negative.
Later you will meet the same idea with $sin x$ sin x, $e^{x}$ e^x and $ln x$ ln x.

Calculate

?Solve

x - 2 x - 8 = 0

.

Mission check

?A ball's height is

h = 20 t - 5 t^{2}

. Does it ever reach 25 m?

Mission 3

When does a line just touch a curve, and what one number tells you?

Make a guess: When does a line just touch a curve, and what one number tells you?

Algebra · simultaneous equations and inequalities

One linear, one quadratic

Two linear equations: eliminate a variable by adding or subtracting multiples, or substitute.
One linear, one quadratic: use substitution. Make $x$ x or $y$ y the subject of the linear equation; substitute into the quadratic.
Solve the quadratic, then put each value into the linear equation for its partner. Give answers in pairs: each is a meeting point.
The substituted quadratic's discriminant counts the meeting points: $> 0$ > 0 two, $= 0$ = 0 one (the line is a tangent), $< 0$ < 0 none.

Worked example Algebra · simultaneous equations and inequalities

Where a line meets a parabola

Question

Solve the simultaneous equations

y = 2 x + 3

and

y = x^{2} - 4 x + 8

.

1
Substitute for $y$ y: $2 x + 3 = x^{2} - 4 x + 8$ 2x + 3 = x² − 4x + 8
2
Rearrange to zero: $x^{2} - 6 x + 5 = 0$ x² − 6x + 5 = 0
3
Factorise: $(x - 1) (x - 5) = 0$ (x − 1)(x − 5) = 0, so $x = 1$ x = 1 or $x = 5$ x = 5
4
Find each $y$ y from the linear equation: $x = 1$ x = 1 gives $y = 5$ y = 5; $x = 5$ x = 5 gives $y = 13$ y = 13

Answer

x = 1, y = 5

and

x = 5, y = 13

⚠️ Watch out

listing

x = 1

,

x = 5

,

y = 5

,

y = 13

without pairing them loses the final mark.

Calculate

?The line

y = 4 x + c

is a tangent to the curve

y = x^{2} + 2 x + 5

. Find the value of the constant

c

.

Algebra · simultaneous equations and inequalities

Solving inequalities

Linear: solve like an equation, but reverse the sign when you multiply or divide by a negative number.
Quadratic: find the critical values (solve $= 0$ = 0), sketch the parabola, then read off where it is above ( $> 0$ > 0) or below ( $< 0$ < 0) the $x$ x-axis.
$p x^{2} + q x + r < a x + b$ px² + qx + r < ax + b means the curve is below the line: rearrange to one side first.
With $x$ x in a denominator, multiply by $x^{2}$ x² (positive, since $x \neq = 0$ x ≠ 0), not $x$ x (its sign is unknown): $\frac{6}{x} < 2$ 6/x < 2 gives $6 x < 2 x^{2}$ 6x < 2x², so $x < 0$ x < 0 or $x > 3$ x > 3.

Worked example Algebra · simultaneous equations and inequalities

A quadratic inequality

Question

Solve

x^{2} - 2 x - 15 > 0

. Give your answer in set notation.

1
Find the critical values: $(x + 3) (x - 5) = 0$ (x + 3)(x − 5) = 0, so $x = - 3$ x = −3 or $x = 5$ x = 5
2
Sketch: a $\cup$ ∪-shaped parabola crossing the $x$ x-axis at $- 3$ −3 and $5$ 5.
3
Read off where the curve is above the axis: to the left of $- 3$ −3 and to the right of 5.
4
Write it as two separate parts joined by "or": $x < - 3$ x < −3 or $x > 5$ x > 5

Answer

{x : x < - 3} \cup {x : x > 5}

⚠️ Watch out

writing

- 3 > x > 5

is wrong: no number is less than

- 3

and greater than 5 at once.

Algebra · simultaneous equations and inequalities

Set notation and interval notation

${x : x > 3}$ {x : x > 3} means "the set of $x$ x such that $x > 3$ x > 3".
"or" joins two parts with $\cup$ ∪: ${x : x < - 3} \cup {x : x > 5}$ {x : x < −3} ∪ {x : x > 5}.
"and" overlaps two conditions, $\cap$ ∩: ${x : x > - 3} \cap {x : x \leq 2} = {x : - 3 < x \leq 2}$ {x : x > −3} ∩ {x : x ≤ 2} = {x : −3 < x ≤ 2}.
$\emptyset$ ∅ is the empty set: no values work.
Interval notation: $(a, b)$ (a, b) means $a < x < b$ a < x < b; $[a, b]$ [a, b] means $a \leq x \leq b$ a ≤ x ≤ b; $[a, b)$ [a, b) means $a \leq x < b$ a ≤ x < b.

Quick check

?Which is the solution of

x^{2} + 4 x - 21 \leq 0

?

Calculate

?How many integers

x

satisfy both

3 x - 1 > 5

and

x^{2} - 5 x - 14 < 0

?

Algebra · simultaneous equations and inequalities

Regions on a graph

To show $y > x - 1$ y > x − 1, draw the boundary $y = x - 1$ y = x − 1 and identify the side where the inequality holds.
A strict inequality ( $<$ <, $>$ >) has a dashed boundary; $\leq$ ≤ or $\geq$ ≥ has a solid boundary.
Test a point not on the boundary, such as $(0, 0)$ (0, 0): $0 > 0 - 1$ 0 > 0 − 1 is true, so $(0, 0)$ (0, 0) is in the region.
Shade as the question instructs, and label the required region $R$ R. Region $R$ R in the diagram: $y > x - 1$ y > x − 1 and $y \leq 5 - x^{2}$ y ≤ 5 − x².

Mission check

?When does a line just touch a curve, and what one number tells you?

Mission 4

$x^{3} - 6 x^{2} + 11 x - 6 = 0$ x³−6x²+11x−6=0 has an easy root. How do the other two follow?

Make a guess: $x^{3} - 6 x^{2} + 11 x - 6 = 0$ x³−6x²+11x−6=0 has an easy root. How do the other two follow?

Algebra · polynomials

The factor theorem

Factor theorem: $f (a) = 0 \Leftrightarrow (x - a)$ f(a)=0 Leftrightarrow (x−a) is a factor of $f (x)$ f(x).
More generally, $f (\frac{b}{a}) = 0 \Leftrightarrow (a x - b)$ f(b/a)=0 Leftrightarrow (ax−b) is a factor.
A root of $f (x) = 0$ f(x)=0 is an $x$ x-intercept of $y = f (x)$ y=f(x).
To find a first root of a cubic, try the factors of the constant term: $\pm 1, \pm 2, \pm 3, \dots$ ±1, ±2, ±3, … If none work, try fractions $\frac{b}{a}$ b/a, with $b$ b a factor of the constant term and $a$ a a factor of the leading coefficient.

Worked example Algebra · polynomials

Show that, hence factorise

Question

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

. Show that

(x - 1)

is a factor of

f (x)

and hence factorise

f (x)

completely.

1
Substitute $x = 1$ x=1: $f (1) = 1 + 3 - 4 = 0$ f(1)=1+3−4=0, so $(x - 1)$ (x−1) is a factor by the factor theorem.
2
Divide by $(x - 1)$ (x−1), writing the missing $x$ x term as $0 x$ 0x: $x^{3} + 3 x^{2} + 0 x - 4 = (x - 1) (x^{2} + 4 x + 4)$ x³+3x²+0x−4 = (x−1)(x²+4x+4)
3
Factorise the quadratic: $x^{2} + 4 x + 4 = (x + 2)^{2}$ x²+4x+4=(x+2)²

Answer

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

, so the graph crosses the

x

-axis at 1 and touches it at

- 2

.

⚠️ Watch out

forgetting the

0 x

placeholder makes the columns of the division slip.

Algebra · polynomials

Algebraic division by a linear factor

Divide $f (x)$ f(x) by $(x - a)$ (x−a) with long division, or compare coefficients in $f (x) \equiv (x - a) (p x^{2} + q x + r)$ f(x) ≡ (x−a)(px²+qx+r).
Match $x^{3}$ x³ terms for $p$ p, constants for $r$ r, then $x^{2}$ x² terms for $q$ q.
Dividing a cubic by a linear factor leaves a quadratic quotient.
A-level only: to simplify a rational expression, factorise top and bottom fully, then cancel common factors, never single terms.

Quick check

?

(x - 3)

is a factor of

2 x^{3} - 3 x^{2} - 11 x + 6

. What is the quotient when you divide by

(x - 3)

?

Calculate

?

f (x) = 2 x^{3} + p x^{2} - 5 x + 6

, where

p

is a constant. Given that

(x - 2)

is a factor of

f (x)

, find the value of

p

.

Worked example Algebra · polynomials

A cubic with a fractional root

Question

Solve

6 x^{3} + 11 x^{2} - x - 6 = 0

.

1
Try factors of $- 6$ −6: $f (1) = 10$ f(1)=10, but $f (- 1) = - 6 + 11 + 1 - 6 = 0$ f(−1)=−6+11+1−6=0, so $(x + 1)$ (x+1) is a factor.
2
Divide: $6 x^{3} + 11 x^{2} - x - 6 = (x + 1) (6 x^{2} + 5 x - 6)$ 6x³+11x²−x−6=(x+1)(6x²+5x−6)
3
Factorise the quadratic: $6 x^{2} + 5 x - 6 = (2 x + 3) (3 x - 2)$ 6x²+5x−6=(2x+3)(3x−2)
4
Set each factor to zero: $x = - 1$ x=−1, $x = - \frac{3}{2}$ x=−3/2, $x = \frac{2}{3}$ x=2/3

Answer

x = - 1

,

x = - \frac{3}{2}

or

x = \frac{2}{3}

(check:

f (\frac{2}{3}) = 0

, so

(3 x - 2)

is a factor).

⚠️ Watch out

a cubic can have roots that are fractions; trying only whole numbers can miss them.

Calculate

?

f (x) = 2 x^{3} + x^{2} - 13 x + 6

. Use the factor theorem to show that

(x - 2)

is a factor, then solve

f (x) = 0

. One root lies between 0 and 1. Give that root as a decimal.

Quick check

?(A-level only) Simplify

\frac{x ^{2} + x - 6}{x ^{2} - 4}

fully.

Mission check

?

x^{3} - 6 x^{2} + 11 x - 6 = 0

has the easy root

x = 1

. How do the other two roots follow?

Mission 5

$y = f (2 x)$ y=f(2x) is narrower, but $y = 2 f (x)$ y=2f(x) is taller. Why opposite?

Make a guess: $y = f (2 x)$ y=f(2x) is narrower, but $y = 2 f (x)$ y=2f(x) is taller. Why opposite?

Algebra · graphs and transformations

Sketching cubics and quartics

From factorised form, each factor gives an $x$ x-intercept; put $x = 0$ x=0 for the $y$ y-intercept.
A single factor: the curve crosses the axis. A repeated root (squared factor): the curve touches the axis and turns.
The sign of the leading term fixes the ends: a positive cubic rises from bottom left to top right; a positive quartic goes up at both ends.

Quick check

?Which statement describes the graph of

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

?

Algebra · graphs and transformations

Reciprocal graphs, asymptotes and intersections

$y = \frac{a}{x}$ y=a/x (with $a > 0$ a>0): two branches, in the first and third quadrants.
$y = \frac{a}{x ^{2}}$ y=a/x² (with $a > 0$ a>0): both branches above the $x$ x-axis.
Both have asymptotes $x = 0$ x=0 and $y = 0$ y=0: lines the curve approaches but never meets.
The solutions of $f (x) = g (x)$ f(x)=g(x) are the $x$ x-coordinates of the intersection points of $y = f (x)$ y=f(x) and $y = g (x)$ y=g(x); count intersections to count solutions.
Proportion: $y \propto x$ y ∝ x means $y = k x$ y=kx (a line through the origin); $y \propto \frac{1}{x}$ y ∝ 1/x means $y = \frac{k}{x}$ y=k/x; $y \propto \frac{1}{x ^{2}}$ y ∝ 1/x² means $y = \frac{k}{x ^{2}}$ y=k/x².

Algebra · graphs and transformations

Transformations of $y = f (x)$ y=f(x)

Equation	Transformation	Point $(x, y)$ (x, y) goes to
$y = f (x) + a$ y=f(x)+a	translation $(0 a)$ (0; a)	$(x, y + a)$ (x, y+a)
$y = f (x + a)$ y=f(x+a)	translation $(- a 0)$ (−a; 0)	$(x - a, y)$ (x−a, y)
$y = a f (x)$ y=af(x)	stretch parallel to the $y$ y-axis, scale factor $a$ a	$(x, a y)$ (x, ay)
$y = f (a x)$ y=f(ax)	stretch parallel to the $x$ x-axis, scale factor $\frac{1}{a}$ 1/a	$(\frac{x}{a}, y)$ (x/a, y)

Changes outside the bracket act on $y$ y as you expect. Changes inside act on $x$ x and work in reverse.
$y = - f (x)$ y=−f(x) reflects in the $x$ x-axis; $y = f (- x)$ y=f(−x) reflects in the $y$ y-axis.

Worked example Algebra · graphs and transformations

Where does the maximum go?

Question

The curve

y = f (x)

has a maximum point at

(4, 6)

. Find the maximum point on each of

y = f (2 x)

,

y = 2 f (x)

,

y = f (x + 3)

and

y = f (x) - 5

.

1
$y = f (2 x)$ y=f(2x): stretch parallel to the $x$ x-axis, scale factor $\frac{1}{2}$ 1/2, so halve $x$ x: $(2, 6)$ (2, 6)
2
$y = 2 f (x)$ y=2f(x): stretch parallel to the $y$ y-axis, scale factor 2, so double $y$ y: $(4, 12)$ (4, 12)
3
$y = f (x + 3)$ y=f(x+3): translation $(- 3 0)$ (−3; 0): $(1, 6)$ (1, 6)
4
$y = f (x) - 5$ y=f(x)−5: translation $(0 - 5)$ (0; −5): $(4, 1)$ (4, 1)

Answer

(2, 6)

,

(4, 12)

,

(1, 6)

and

(4, 1)

⚠️ Watch out

f (x + 3)

moves the graph 3 to the left, not the right.

Calculate

?

f (x) = x^{2} - 8 x + 20

, which can be written

(x - 4)^{2} + 4

. The graph of

y = f (x + a)

has its minimum point on the

y

-axis. Find the value of

a

.

Calculate

?The graph of

y = f (x)

crosses the

x

-axis only at

(- 3, 0)

and

(6, 0)

. Find the distance between the two points where the graph of

y = f (\frac{1}{3} x)

crosses the

x

-axis.

Worked example Algebra · graphs and transformations

Combining transformations (A-level only)

Question

The point

P (2, 5)

lies on

y = f (x)

. Find the image of

P

on the curve

y = 3 f (x + 1) - 4

.

1
Inside the bracket: $x + 1$ x+1 translates by $(- 1 0)$ (−1; 0): $(1, 5)$ (1, 5)
2
Multiply the output by 3 (stretch parallel to the $y$ y-axis): $(1, 15)$ (1, 15)
3
Subtract 4 (translation $(0 - 4)$ (0; −4)): $(1, 11)$ (1, 11)

Answer

(1, 11)

⚠️ Watch out

the

y

-coordinate is

3 \times 5 - 4 = 11

. Subtracting first,

3 (5 - 4) = 3

, is the transformation

y = 3 (f (x + 1) - 4)

, a different curve.

Sort it

?Does each graph come from changing the

x

-coordinates or the

y

-coordinates of

y = f (x)

?

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

Mission check

?Why does the 2 make

y = f (2 x)

narrower but

y = 2 f (x)

taller?

Mission 6

Why must $y = x^{2}$ y=x² lose half its domain before it has an inverse?

Make a guess: Why must $y = x^{2}$ y=x² lose half its domain before it has an inverse?

Algebra · functions

Functions, domain and range (A-level only)

A function maps every input in its domain to exactly one output. The set of all outputs is the range.
One-one: each output comes from only one input. Many-one: some output comes from two or more inputs, e.g. $f (x) = x^{2}$ f(x)=x², $x \in R$ x ∈ ℝ, has $f (3) = f (- 3) = 9$ f(3)=f(−3)=9.
Notation: $f : x \mapsto x^{2}$ f: x mapsto x² or $f (x) = x^{2}$ f(x)=x². Range: $f (x) \geq 0$ f(x) ≥ 0.
To find a range, sketch the graph; for a quadratic, complete the square.
Modelling: the context sets a model's domain (e.g. time $t \geq 0$ t ≥ 0). Unrealistic outputs show a limitation: refine the model.

Quick check

?

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

,

x \in R

. What is the range of

f

?

Algebra · functions

Composite and inverse functions

$f g (x) = f (g (x))$ fg(x)=f(g(x)): do $g$ g first, then $f$ f. Usually $f g (x) \neq = g f (x)$ fg(x) ≠ gf(x).
The inverse $f^{- 1}$ f⁻¹ undoes $f$ f: $f^{- 1} f (x) = x$ f⁻¹f(x)=x. Only one-one functions have an inverse.
To find $f^{- 1} (x)$ f⁻¹(x): write $y = f (x)$ y=f(x), rearrange to make $x$ x the subject, then swap $x$ x and $y$ y.
Domain of $f^{- 1}$ f⁻¹ = range of $f$ f; range of $f^{- 1}$ f⁻¹ = domain of $f$ f.
The graph of $y = f^{- 1} (x)$ y=f⁻¹(x) is the reflection of $y = f (x)$ y=f(x) in the line $y = x$ y=x.

Worked example Algebra · functions

An inverse with a restricted domain

Question

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

,

x \geq 2

. Find

f^{- 1} (x)

and state its domain.

1
Range of $f$ f: the least value is $f (2) = 1$ f(2)=1, so $f (x) \geq 1$ f(x) ≥ 1.
2
Write $y = (x - 2)^{2} + 1$ y=(x−2)²+1 and rearrange: $(x - 2)^{2} = y - 1$ (x−2)²=y−1
3
Since $x \geq 2$ x ≥ 2, take the positive root: $x = 2 + y - 1$ x=2+√(y−1)
4
Swap $x$ x and $y$ y: $f^{- 1} (x) = 2 + x - 1$ f⁻¹(x)=2+√(x−1)

Answer

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

, with domain

x \geq 1

(the range of

f

).

⚠️ Watch out

keeping

\pm

gives two outputs; the domain

x \geq 2

tells you to use the

+

root.

Calculate

?

f (x) = 3 x - 1

and

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

, both for

x \in R

. Find the positive value of

x

for which

f g (x) = 32

.

Calculate

?A model for the number of fish in a lake

t

years after it is stocked is

N = f (t) = \frac{800 t + 200}{t + 4}

,

t \geq 0

. Find

f^{- 1} (500)

, the time at which the model predicts 500 fish.

years

Algebra · functions

The modulus function

$∣ x ∣$ |x| is the size of $x$ x without its sign: $∣ - 4 ∣ = 4$ |−4|=4.
Sketch $y = ∣ a x + b ∣$ y=|ax+b|: draw $y = a x + b$ y=ax+b, then reflect the part below the $x$ x-axis in the $x$ x-axis. You get a V with vertex $(- \frac{b}{a}, 0)$ (−b/a, 0).
Solve $∣ f (x) ∣ = g (x)$ |f(x)|=g(x): solve $f (x) = g (x)$ f(x)=g(x) and $- f (x) = g (x)$ −f(x)=g(x), then check each answer in the original. A sketch shows how many solutions to expect.
$y = ∣ f (x) ∣$ y=|f(x)|: reflect any part of $y = f (x)$ y=f(x) below the $x$ x-axis in the $x$ x-axis. For $y = ∣ f (- x) ∣$ y=|f(−x)|, reflect in the $y$ y-axis first.

Worked example Algebra · functions

A modulus equation with a false root

Question

Solve

∣ x - 3∣ = 2 x + 1

.

1
Case $x - 3 = 2 x + 1$ x−3=2x+1: $x = - 4$ x=−4. Check: $2 (- 4) + 1 = - 7$ 2(−4)+1=−7, but a modulus is never negative, so reject it.
2
Case $- (x - 3) = 2 x + 1$ −(x−3)=2x+1: $3 - x = 2 x + 1$ 3−x=2x+1, so $x = \frac{2}{3}$ x=2/3.
3
Check: $\frac{2}{3} - 3 = \frac{7}{3}$ |2/3−3|=7/3 and $2 (\frac{2}{3}) + 1 = \frac{7}{3}$ 2(2/3)+1=7/3. It works.

Answer

x = \frac{2}{3}

only; the graphs meet once.

⚠️ Watch out

solving both cases is not enough. Always check, because one case can give a false root.

Quick check

?Solve the inequality

∣2 x - 7∣ > x + 1

.

Mission check

?Why must

y = x^{2}

lose half its domain before it has an inverse?

Mission 7

Why is the sum of $\frac{1}{n ( n + 1 )}$ 1/(n(n+1)) from $n = 1$ n=1 to $99$ 99 so neat?

Make a guess: Why is the sum of $\frac{1}{n ( n + 1 )}$ 1/(n(n+1)) from $n = 1$ n=1 to $99$ 99 so neat?

Algebra · partial fractions

Splitting a fraction (A-level only)

Adding fractions makes one denominator: $\frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n ( n + 1 )}$ 1/n−1/(n+1)=1/(n(n+1)). Partial fractions reverse this.
Distinct linear factors: $\frac{p x + q}{( a x + b ) ( c x + d )} \equiv \frac{A}{a x + b} + \frac{B}{c x + d}$ (px+q)/((ax+b)(cx+d)) ≡ A/(ax+b)+B/(cx+d), and similarly with three factors.
Multiply through by the denominator to get an identity, then substitute the value of $x$ x that makes each factor zero (or compare coefficients).
Here the numerator is constant or linear: lower degree than the denominator.

Worked example Algebra · partial fractions

Two linear factors

Question

Express

\frac{5 x + 1}{( x - 1 ) ( x + 2 )}

in partial fractions.

1
Write the form: $\frac{5 x + 1}{( x - 1 ) ( x + 2 )} \equiv \frac{A}{x - 1} + \frac{B}{x + 2}$ (5x+1)/((x−1)(x+2)) ≡ A/(x−1)+B/(x+2)
2
Multiply by $(x - 1) (x + 2)$ (x−1)(x+2): $5 x + 1 \equiv A (x + 2) + B (x - 1)$ 5x+1 ≡ A(x+2)+B(x−1)
3
Put $x = 1$ x=1: $6 = 3 A$ 6=3A, so $A = 2$ A=2
4
Put $x = - 2$ x=−2: $- 9 = - 3 B$ −9=−3B, so $B = 3$ B=3

Answer

\frac{2}{x - 1} + \frac{3}{x + 2}

⚠️ Watch out

the identity holds for every

x

, so you may substitute

x = 1

even though the original fraction is undefined there.

Calculate

?

\frac{7 x - 7}{( 2 x + 1 ) ( x - 3 )} \equiv \frac{A}{2 x + 1} + \frac{B}{x - 3}

. Find the value of

A

.

Algebra · partial fractions

Repeated linear factors

A squared factor needs two fractions: $\frac{p x + q}{( a x + b ) ( c x + d ) ^{2}} \equiv \frac{A}{a x + b} + \frac{B}{c x + d} + \frac{C}{( c x + d ) ^{2}}$ (px+q)/((ax+b)(cx+d)²) ≡ A/(ax+b)+B/(cx+d)+C/((cx+d)²)
Multiply through by the whole denominator; the $B$ B term keeps one factor of $(c x + d)$ (cx+d).
Substituting the two roots gives $A$ A and $C$ C. Find $B$ B by comparing coefficients of $x^{2}$ x², or by substituting another value such as $x = 0$ x=0.

Worked example Algebra · partial fractions

A repeated factor

Question

Express

\frac{4}{( x + 1 ) ( x - 1 ) ^{2}}

in partial fractions.

1
Form: $\frac{4}{( x + 1 ) ( x - 1 ) ^{2}} \equiv \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{( x - 1 ) ^{2}}$ 4/((x+1)(x−1)²) ≡ A/(x+1)+B/(x−1)+C/((x−1)²)
2
Multiply through: $4 \equiv A (x - 1)^{2} + B (x + 1) (x - 1) + C (x + 1)$ 4 ≡ A(x−1)²+B(x+1)(x−1)+C(x+1)
3
Put $x = 1$ x=1: $4 = 2 C$ 4=2C, so $C = 2$ C=2. Put $x = - 1$ x=−1: $4 = 4 A$ 4=4A, so $A = 1$ A=1.
4
Compare $x^{2}$ x² coefficients: $0 = A + B$ 0=A+B, so $B = - 1$ B=−1

Answer

\frac{1}{x + 1} - \frac{1}{x - 1} + \frac{2}{( x - 1 ) ^{2}}

⚠️ Watch out

there is no

x^{2}

term on the left, so its coefficient is 0, not 4.

Quick check

?Which form is correct for splitting

\frac{3 x + 2}{x ( x + 2 ) ^{2}}

into partial fractions?

Calculate

?

\frac{x + 11}{( x + 3 ) ( x - 1 ) ^{2}} \equiv \frac{A}{x + 3} + \frac{B}{x - 1} + \frac{C}{( x - 1 ) ^{2}}

. Find the value of

B

.

Give your answer as a decimal.

Algebra · partial fractions

Why split fractions?

Integration: each simple fraction integrates to a logarithm, e.g. $\frac{2}{x - 1} + \frac{3}{x + 2}$ 2/(x−1)+3/(x+2).
Series: expand each simple fraction with the binomial expansion, then add.
Sums that cancel: $\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}$ 1/(n(n+1))=1/n−1/(n+1). Write each term of a sum this way and neighbouring pieces cancel, leaving only the first and last.

Calculate

?Use

\frac{1}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1}

to find

\frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{9 \times 10}

.

Give your answer as a decimal.

Mission check

?Why is

\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{99 \times 100}

so neat?

Recap

Big ideas

$a^{m} \times a^{n} = a^{m + n}$ a^m × aⁿ = a^(m+n), $a^{m} \div a^{n} = a^{m - n}$ a^m ÷ aⁿ = a^(m−n), $(a^{m})^{n} = a^{mn}$ (a^m)ⁿ = a^mn, $a^{0} = 1$ a⁰ = 1, $a^{- n} = \frac{1}{a ^{n}}$ a⁻ⁿ = 1/aⁿ.
$a^{\frac{m}{n}} = (n a)^{m}$ a^(m/n) = (n√a)^m: denominator root, numerator power; negative index means reciprocal.
Simplify surds with $ab = a b$ √ab = √a√b; rationalise $\frac{1}{p - q}$ 1/(p − √q) by multiplying by $p + q$ p + √q.
Completing the square, $a (x + p)^{2} + q$ a(x + p)² + q, gives the turning point $(- p, q)$ (−p, q) and the greatest or least value.
The discriminant $b^{2} - 4 a c$ b² − 4ac decides the roots: $> 0$ > 0 two distinct, $= 0$ = 0 repeated, $< 0$ < 0 none.
Spot quadratics in $x^{2}$ x² or $x$ √x, substitute $u$ u, and reject impossible values.
Linear and quadratic together: substitute, solve the quadratic, pair each $x$ x with its $y$ y; the discriminant counts the meeting points.
Quadratic inequalities: find the critical values, sketch, read off; "or" is $\cup$ ∪, "and" is $\cap$ ∩.
Regions: dashed boundary for $<$ < or $>$ >, solid for $\leq$ ≤ or $\geq$ ≥; test a point to pick the side.
Factor theorem: $f (a) = 0 \Leftrightarrow (x - a)$ f(a)=0 Leftrightarrow (x−a) is a factor, and $f (\frac{b}{a}) = 0 \Leftrightarrow (a x - b)$ f(b/a)=0 Leftrightarrow (ax−b) is a factor.
Find one root by trial, divide (or compare coefficients) to get a quadratic, then factorise or solve it.
Factorised polynomials: single roots cross, repeated roots touch; $y = \frac{a}{x}$ y=a/x and $y = \frac{a}{x ^{2}}$ y=a/x² have asymptotes $x = 0$ x=0 and $y = 0$ y=0.
Outside the bracket acts on $y$ y as written; inside acts on $x$ x in reverse: $f (x + a)$ f(x+a) moves left, $f (a x)$ f(ax) has scale factor $\frac{1}{a}$ 1/a.
Intersections of two graphs give the solutions of $f (x) = g (x)$ f(x)=g(x).
A function gives one output per input; only one-one functions have inverses, so a many-one function needs its domain restricted first.
$f g (x)$ fg(x) means do $g$ g first; $y = f^{- 1} (x)$ y=f⁻¹(x) is the reflection of $y = f (x)$ y=f(x) in $y = x$ y=x, and their domain and range swap.
For $∣ f (x) ∣ = g (x)$ |f(x)|=g(x), solve both cases and check each answer.
Distinct linear factors give one fraction each; a squared factor $(c x + d)^{2}$ (cx+d)² needs both $\frac{B}{c x + d}$ B/(cx+d) and $\frac{C}{( c x + d ) ^{2}}$ C/((cx+d)²).
Multiply through to an identity, substitute the roots, then compare coefficients for any constant left.
Partial fractions make integrals, binomial expansions and some sums easy.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Algebra and functions for Edexcel A-level Mathematics (9MA0). 🎉

Mission 1Why does $8^{2/3}$ 8^(2/3) come out as exactly 4?
Mission 2A ball's height is $h = 20 t - 5 t^{2}$ h = 20t − 5t². Does it ever reach 25 m?
Mission 3When does a line just touch a curve, and what one number tells you?
Mission 4 $x^{3} - 6 x^{2} + 11 x - 6 = 0$ x³−6x²+11x−6=0 has an easy root. How do the other two follow?
Mission 5 $y = f (2 x)$ y=f(2x) is narrower, but $y = 2 f (x)$ y=2f(x) is taller. Why opposite?
Mission 6Why must $y = x^{2}$ y=x² lose half its domain before it has an inverse?
Mission 7Why is the sum of $\frac{1}{n ( n + 1 )}$ 1/(n(n+1)) from $n = 1$ n=1 to $99$ 99 so neat?

Your stars: 0 / 0

Next: test yourself in the Verify stage.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to A-Level Mathematics · Edexcel

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

Algebra and functions

Why does 82/38^(2/3) come out as exactly 4?

The laws of indices

Fractional indices

A negative fractional index

Simplifying surds

Rationalising the denominator

A ball's height is h=20t−5t2h = 20t − 5t². Does it ever reach 25 m?

Completing the square

Turning point from the completed square

The discriminant

Finding kk for equal roots

Quadratics in disguise

When does a line just touch a curve, and what one number tells you?

One linear, one quadratic

Where a line meets a parabola

Solving inequalities

A quadratic inequality

Set notation and interval notation

Regions on a graph

x3−6x2+11x−6=0x³−6x²+11x−6=0 has an easy root. How do the other two follow?

The factor theorem

Show that, hence factorise

Algebraic division by a linear factor

A cubic with a fractional root

y=f(2x)y=f(2x) is narrower, but y=2f(x)y=2f(x) is taller. Why opposite?

Sketching cubics and quartics

Reciprocal graphs, asymptotes and intersections

Transformations of y=f(x)y=f(x)

Where does the maximum go?

Combining transformations (A-level only)

Why must y=x2y=x² lose half its domain before it has an inverse?

Functions, domain and range (A-level only)

Composite and inverse functions

An inverse with a restricted domain

The modulus function

A modulus equation with a false root

Why is the sum of n(n+1)1​1/(n(n+1)) from n=1n=1 to 9999 so neat?

Splitting a fraction (A-level only)

Two linear factors

Repeated linear factors

A repeated factor

Why split fractions?

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Why does $8^{2/3}$ 8^(2/3) come out as exactly 4?

A ball's height is $h = 20 t - 5 t^{2}$ h = 20t − 5t². Does it ever reach 25 m?

Finding $k$ k for equal roots

$x^{3} - 6 x^{2} + 11 x - 6 = 0$ x³−6x²+11x−6=0 has an easy root. How do the other two follow?

$y = f (2 x)$ y=f(2x) is narrower, but $y = 2 f (x)$ y=2f(x) is taller. Why opposite?

Transformations of $y = f (x)$ y=f(x)

Why must $y = x^{2}$ y=x² lose half its domain before it has an inverse?

Why is the sum of $\frac{1}{n ( n + 1 )}$ 1/(n(n+1)) from $n = 1$ n=1 to $99$ 99 so neat?