Summer Transition Pack

2019

GCSE to A level Mathematics

Expanding brackets
and simplifying expressions
A LEVEL LINKS
Scheme of work: 1a. Algebraic expressions – basic algebraic manipulation, indices and surds

Key points
 When you expand one set of brackets you must multiply everything inside the bracket by
what is outside.
 When you expand two linear expressions, each with two terms of the form ax + b, where
a ≠ 0 and b ≠ 0, you create four terms. Two of these can usually be simplified by collecting
like terms.

Examples
Example 1 Expand 4(3x − 2)

4(3x − 2) = 12x − 8 Multiply everything inside the bracket

by the 4 outside the bracket

Example 2 Expand and simplify 3(x + 5) − 4(2x + 3)

3(x + 5) − 4(2x + 3) 1 Expand each set of brackets

= 3x + 15 − 8x – 12 separately by multiplying (x + 5) by
3 and (2x + 3) by −4
= 3 − 5x 2 Simplify by collecting like terms:
3x − 8x = −5x and 15 − 12 = 3

Example 3 Expand and simplify (x + 3)(x + 2)

(x + 3)(x + 2) 1 Expand the brackets by multiplying

= x(x + 2) + 3(x + 2) (x + 2) by x and (x + 2) by 3
= x2 + 2x + 3x + 6
= x2 + 5x + 6 2 Simplify by collecting like terms:
2x + 3x = 5x

Example 4 Expand and simplify (x − 5)(2x + 3)

(x − 5)(2x + 3) 1 Expand the brackets by multiplying

= x(2x + 3) − 5(2x + 3) (2x + 3) by x and (2x + 3) by −5
= 2x2 + 3x − 10x − 15
= 2x2 − 7x − 15 2 Simplify by collecting like terms:
3x − 10x = −7x
Practice
1 Expand. Watch out!
a 3(2x − 1) b −2(5pq + 4q2)
When multiplying (or
c −(3xy − 2y2)
dividing) positive and
2 Expand and simplify. negative numbers, if
the signs are the same
a 7(3x + 5) + 6(2x – 8) b 8(5p – 2) – 3(4p + 9)
the answer is ‘+’; if the
c 9(3s + 1) –5(6s – 10) d 2(4x – 3) – (3x + 5) signs are different the
answer is ‘–’.
3 Expand.
a 3x(4x + 8) b 4k(5k2 – 12)
c –2h(6h2 + 11h – 5) d –3s(4s2 – 7s + 2)

4 Expand and simplify.

a 3(y2 – 8) – 4(y2 – 5) b 2x(x + 5) + 3x(x – 7)
c 4p(2p – 1) – 3p(5p – 2) d 3b(4b – 3) – b(6b – 9)

5 Expand 12 (2y – 8)

6 Expand and simplify.

a 13 – 2(m + 7) b 5p(p2 + 6p) – 9p(2p – 3)

7 The diagram shows a rectangle.

Write down an expression, in terms of x, for the area of
the rectangle.
Show that the area of the rectangle can be written as
21x2 – 35x

8 Expand and simplify.

a (x + 4)(x + 5) b (x + 7)(x + 3)
c (x + 7)(x – 2) d (x + 5)(x – 5)
e (2x + 3)(x – 1) f (3x – 2)(2x + 1)
g (5x – 3)(2x – 5) h (3x – 2)(7 + 4x)
i (3x + 4y)(5y + 6x) j (x + 5)2
k (2x − 7)2 l (4x − 3y)2

Extend
9 Expand and simplify (x + 3)² + (x − 4)²

10 Expand and simplify.

2
 1  2  1
a  x   x   b x 
 x  x  x
Answers
1 a 6x – 3 b –10pq – 8q2
c –3xy + 2y2

2 a 21x + 35 + 12x – 48 = 33x – 13

b 40p – 16 – 12p – 27 = 28p – 43
c 27s + 9 – 30s + 50 = –3s + 59 = 59 – 3s
d 8x – 6 – 3x – 5 = 5x – 11

3 a 12x2 + 24x b 20k3 – 48k

c 10h – 12h3 – 22h2 d 21s2 – 21s3 – 6s

4 a –y2 – 4 b 5x2 – 11x

c 2p – 7p2 d 6b2

5 y–4

6 a –1 – 2m b 5p3 + 12p2 + 27p

7 7x(3x – 5) = 21x2 – 35x

8 a x2 + 9x + 20 b x2 + 10x + 21
c x2 + 5x – 14 d x2 – 25
e 2x2 + x – 3 f 6x2 – x – 2
g 10x2 – 31x + 15 h 12x2 + 13x – 14
i 18x2 + 39xy + 20y2 j x2 + 10x + 25
k 4x2 − 28x + 49 l 16x2 − 24xy + 9y2

9 2x2 − 2x + 25

2 1
10 a x2 1  b x2  2 
x2 x2
Completing the square
A LEVEL LINKS
Scheme of work: 1b. Quadratic functions – factorising, solving, graphs and the discriminants

Key points
 Completing the square for a quadratic rearranges ax2 + bx + c into the form p(x + q)2 + r
 If a ≠ 1, then factorise using a as a common factor.

Examples
Example 1 Complete the square for the quadratic expression x2 + 6x − 2

x2 + 6x − 2 1 Write x2 + bx + c in the form

2 2
 b b
= (x + 3) − 9 − 2  x     c
2

 2  2
= (x + 3)2 − 11 2 Simplify

Example 2 Write 2x2 − 5x + 1 in the form p(x + q)2 + r

2x2 − 5x + 1 1 Before completing the square write

ax2 + bx + c in the form
 b 
a  x2  x   c
 a 
 5 
= 2  x2  x   1 2 Now complete the square by writing
 2  5
x 2  x in the form
2
 5 5 
2 2
2 2
= 2  x      1
  b b
x   
 4   4  
 2 2

2
 5  25 3 Expand the square brackets – don’t
= 2 x    1
 4 8 5
2
forget to multiply   by the
4
factor of 2
2
 5  17
= 2 x    4 Simplify
 4 8
Practice
1 Write the following quadratic expressions in the form (x + p)2 + q
a x2 + 4x + 3 b x2 – 10x – 3
c x2 – 8x d x2 + 6x
e x2 – 2x + 7 f x2 + 3x – 2

2 Write the following quadratic expressions in the form p(x + q)2 + r

a 2x2 – 8x – 16 b 4x2 – 8x – 16
c 3x2 + 12x – 9 d 2x2 + 6x – 8

3 Complete the square.

a 2x2 + 3x + 6 b 3x2 – 2x
c 5x2 + 3x d 3x2 + 5x + 3

Extend
4 Write (25x2 + 30x + 12) in the form (ax + b)2 + c.
Answers
1 a (x + 2)2 – 1 b (x – 5)2 – 28

c (x – 4)2 – 16 d (x + 3)2 – 9

2
 3  17
e (x – 1)2 + 6 f x  
 2 4

2 a 2(x – 2)2 – 24 b 4(x – 1)2 – 20

2
 3  25
c 3(x + 2)2 – 21 d 2 x   
 2 2

2 2
 3  39  1 1
3 a 2 x    b 3 x   
 4 8  3 3

2 2
 3 9  5  11
c 5 x    d 3 x   
 10  20  6  12

4 (5x + 3)2 + 3
Rules of indices
A LEVEL LINKS
Scheme of work: 1a. Algebraic expressions – basic algebraic manipulation, indices and surds

Key points
 am × an = am + n
am
 n
 a mn
a
 (am)n = amn
 a0 = 1
1

 a n  n a i.e. the nth root of a

 a
m
m
 a n  n am  n

1
 am 
am
 The square root of a number produces two solutions, e.g. 16  4 .

Examples
Example 1 Evaluate 100

100 = 1 Any value raised to the power of zero is

equal to 1

1
Example 2 Evaluate 9 2

1 1
92  9 Use the rule a n  n a
=3

2
Example 3 Evaluate 27 3

   a
2 m
2 m
27 3  3
27 1 Use the rule a n  n

= 32 2 Use 3
27  3
=9
Example 4 Evaluate 42

1 1
4 2  1 Use the rule a  m 
42 am
1
 2 Use 42  16
16

6 x5
Example 5 Simplify
2x2

6 x5 am
= 3x3 6 ÷ 2 = 3 and use the rule  a m  n to
2 x2 a n

x5
give 2
 x5  2  x3
x

x3  x5
Example 6 Simplify
x4

x3  x5 x35 x8 1 Use the rule a m  a n  a m  n

 4  4
x4 x x
am
= x8 − 4 = x4 2 Use the rule n
 a mn
a

1
Example 7 Write as a single power of x
3x

1 1 1 1
 x Use the rule m
 a  m , note that the
3x 3 a
1
fraction remains unchanged
3

4
Example 8 Write as a single power of x
x

4 4 1
 1 1 Use the rule a n  n a
x x 2


1 1
 4x 2 2 Use the rule m
 am
a
Practice
1 Evaluate.
a 140 b 30 c 50 d x0

2 Evaluate.
1 1 1 1
a 49 2 b 64 3 c 125 3 d 16 4

3 Evaluate.
3 5 3 3
a 25 2 b 83 c 49 2 d 16 4

4 Evaluate.

a 5–2 b 4–3 c 2–5 d 6–2

5 Simplify.
3x 2  x3 10 x5
a b
2 x2 2 x2  x
3x  2 x3 7 x3 y 2
c d Watch out!
2 x3 14 x 5 y
1 Remember that
y2 c2
e 1 f 3
any value raised to
y2  y c2  c 2 the power of zero

 2x  2 3 1 3
is 1. This is the
x2  x2 rule a0 = 1.
g h
4 x0 x 2  x 3

6 Evaluate.
1 2 1
  
a 4 2 b 27 3
c 9 2  23
1 2
1  
3  9  2  27  3
d 16 4 2 e   f  
 16   64 

7 Write the following as a single power of x.

1 1 4
a b c x
x x7
5 1 1
d x2 e 3
f
x 3
x2
8 Write the following without negative or fractional powers.
1
3 0
a x b x c x5
2 1 3
 
d x5 e x 2 f x 4

9 Write the following in the form axn.

2 1
a 5 x b c
x3 3x 4
2 4
d e 3
f 3
x x

Extend

10 Write as sums of powers of x.

x5  1  1  1 
a b x2  x   c x 4  x 2  3 
x2  x  x 
Answers
1 a 1 b 1 c 1 d 1

2 a 7 b 4 c 5 d 2

3 a 125 b 32 c 343 d 8

1 1 1 1
4 a b c d
25 64 32 36

3x3
5 a b 5x2
2
y
c 3x d
2x2
1
e y2 f c–3
g 2x6 h x

1 1 8
6 a b c
2 9 3
1 4 16
d e f
4 3 9

1
7 a x–1 b x–7 c x4
2 1 2
 
x 5 x 3 x 3
d e f

1 5
8 a b 1 c x
x3
5 1 1
d x2 e f
x 4
x3

1
1 4
9 a 5x 2 b 2x–3 c x
3
1 1
 
2x 2 4x 3
d e f 3x0

10 a x3  x 2 b x3  x c x 2  x 7
Surds and rationalising the denominator
A LEVEL LINKS
Scheme of work: 1a. Algebraic expressions – basic algebraic manipulation, indices and surds

Key points
 A surd is the square root of a number that is not a square number,
for example 2, 3, 5, etc.
 Surds can be used to give the exact value for an answer.
 ab  a  b
a a
 
b b
 To rationalise the denominator means to remove the surd from the denominator of a fraction.
a
 To rationalise you multiply the numerator and denominator by the surd b
b
a
 To rationalise you multiply the numerator and denominator by b  c
b c

Examples
Example 1 Simplify 50

50  25  2 1 Choose two numbers that are

factors of 50. One of the factors
must be a square number
 25  2 2 Use the rule ab  a  b
 5 2 3 Use 25  5
5 2

Example 2 Simplify 147  2 12

147  2 12 1 Simplify 147 and 2 12 . Choose

 49  3  2 4  3 two numbers that are factors of 147
and two numbers that are factors of
12. One of each pair of factors must
be a square number
 49  3  2 4  3 2 Use the rule ab  a  b
 7 3  2 2 3 3 Use 49  7 and 4 2
7 3 4 3
3 3 4 Collect like terms
Example 3 Simplify  7 2  7 2 
 7 2  7 2  1 Expand the brackets. A common
 7
2
mistake here is to write  49
= 49  7 2  2 7  4

=7–2 2 Collect like terms:

=5  7 2 2 7
 7 2 7 2 0

1
Example 4 Rationalise
3

1 1 3 1 Multiply the numerator and

= 
3 3 3 denominator by 3

1 3
= 2 Use 9 3
9

3
=
3

2
Example 5 Rationalise and simplify
12

2 2 12 1 Multiply the numerator and

= 
12 12 12 denominator by 12

2  43 2 Simplify 12 in the numerator.

=
12 Choose two numbers that are factors
of 12. One of the factors must be a
square number

3 Use the rule ab  a  b

2 2 3
= 4 Use 4 2
12
5 Simplify the fraction:
2 3 2 1
= simplifies to
6 12 6
 3
Example 6 Rationalise and simplify
2 5

3 3 2 5 1 Multiply the numerator and

= 
2 5 2 5 2 5 denominator by 2  5


3 2 5 
 2  5  2  5 
=
2 Expand the brackets

63 5
=
4 2 5 2 5 5 3 Simplify the fraction

63 5
=
1 4 Divide the numerator by −1
Remember to change the sign of all
= 3 5 6 terms when dividing by −1

Practice
1 Simplify. Hint
a 45 b 125 One of the two
c 48 d 175 numbers you
choose at the start
e 300 f 28 must be a square
g 72 h 162 number.

2 Simplify. Watch out!

a 72  162 b 45  2 5 Check you have
c 50  8 d 75  48 chosen the highest
square number at
e 2 28  28 f 2 12  12  27 the start.

3 Expand and simplify.

a ( 2  3)( 2  3) b (3  3)(5  12)
c (4  5)( 45  2) d (5  2)(6  8)
4 Rationalise and simplify, if possible.
1 1
a b
5 11
2 2
c d
7 8
2 5
e f
2 5
8 5
g h
24 45

5 Rationalise and simplify.

1 2 6
a b c
3 5 4 3 5 2

Extend

6 Expand and simplify  x y  x y 

7 Rationalise and simplify, if possible.
1 1
a b
9 8 x y
Answers

1 a 3 5 b 5 5
c 4 3 d 5 7
e 10 3 f 2 7
g 6 2 h 9 2

2 a 15 2 b 5
c 3 2 d 3
e 6 7 f 5 3

3 a −1 b 9 3
c 10 5  7 d 26  4 2

5 11
4 a b
5 11
2 7 2
c d
7 2
e 2 f 5
3 1
g h
3 3

3 5 2(4  3) 6(5  2)
5 a b c
4 13 23

6 x−y

x y
7 a 3 2 2 b
x y
Factorising expressions
A LEVEL LINKS
Scheme of work: 1b. Quadratic functions – factorising, solving, graphs and the discriminants

Key points
 Factorising an expression is the opposite of expanding the brackets.
 A quadratic expression is in the form ax2 + bx + c, where a ≠ 0.
 To factorise a quadratic equation find two numbers whose sum is b and whose product is ac.
 An expression in the form x2 – y2 is called the difference of two squares. It factorises to
(x – y)(x + y).

Examples
Example 1 Factorise 15x2y3 + 9x4y

15x2y3 + 9x4y = 3x2y(5y2 + 3x2) The highest common factor is 3x2y.

So take 3x2y outside the brackets and
then divide each term by 3x2y to find
the terms in the brackets

Example 2 Factorise 4x2 – 25y2

4x2 – 25y2 = (2x + 5y)(2x − 5y) This is the difference of two squares as
the two terms can be written as
(2x)2 and (5y)2

Example 3 Factorise x2 + 3x – 10

b = 3, ac = −10 1 Work out the two factors of

ac = −10 which add to give b = 3
(5 and −2)
So x2 + 3x – 10 = x2 + 5x – 2x – 10 2 Rewrite the b term (3x) using these
two factors
= x(x + 5) – 2(x + 5) 3 Factorise the first two terms and the
last two terms
= (x + 5)(x – 2) 4 (x + 5) is a factor of both terms
Example 4 Factorise 6x2 − 11x − 10

b = −11, ac = −60 1 Work out the two factors of

ac = −60 which add to give b = −11
So (−15 and 4)
6x2 − 11x – 10 = 6x2 − 15x + 4x – 10 2 Rewrite the b term (−11x) using
these two factors
= 3x(2x − 5) + 2(2x − 5) 3 Factorise the first two terms and the
last two terms
= (2x – 5)(3x + 2) 4 (2x − 5) is a factor of both terms

x 2  4 x  21
Example 5 Simplify
2x2  9x  9

x 2  4 x  21 1 Factorise the numerator and the

denominator
2 x2  9 x  9

For the numerator: 2 Work out the two factors of

b = −4, ac = −21 ac = −21 which add to give b = −4
(−7 and 3)
So
x2 − 4x – 21 = x2 − 7x + 3x – 21 3 Rewrite the b term (−4x) using these
two factors
= x(x − 7) + 3(x − 7) 4 Factorise the first two terms and the
last two terms
= (x – 7)(x + 3) 5 (x − 7) is a factor of both terms

For the denominator: 6 Work out the two factors of

b = 9, ac = 18 ac = 18 which add to give b = 9
(6 and 3)
So
2x2 + 9x + 9 = 2x2 + 6x + 3x + 9 7 Rewrite the b term (9x) using these
two factors
= 2x(x + 3) + 3(x + 3) 8 Factorise the first two terms and the
last two terms
= (x + 3)(2x + 3) 9 (x + 3) is a factor of both terms
So
x 2  4 x  21 ( x  7)( x  3) 10 (x + 3) is a factor of both the

2 x 2  9 x  9 ( x  3)(2 x  3) numerator and denominator so
x7 cancels out as a value divided by
= itself is 1
2x  3
Practice
1 Factorise.
Hint
a 6x4y3 – 10x3y4 b 21a3b5 + 35a5b2
c 25x2y2 – 10x3y2 + 15x2y3 Take the highest
common factor
2 Factorise outside the bracket.
a x2 + 7x + 12 b x2 + 5x – 14
c x2 – 11x + 30 d x2 – 5x – 24
e x2 – 7x – 18 f x2 + x –20
g x2 – 3x – 40 h x2 + 3x – 28

3 Factorise
a 36x2 – 49y2 b 4x2 – 81y2
c 18a2 – 200b2c2

4 Factorise
a 2x2 + x –3 b 6x2 + 17x + 5
c 2x2 + 7x + 3 d 9x2 – 15x + 4
e 10x2 + 21x + 9 f 12x2 – 38x + 20

5 Simplify the algebraic fractions.

2x2  4x x 2  3x
a b
x2  x x2  2 x  3
x2  2 x  8 x2  5x
c d
x2  4x x 2  25
x 2  x  12 2 x 2  14 x
e f
x2  4 x 2 x 2  4 x  70

6 Simplify
9 x 2  16 2 x 2  7 x  15
a b
3x 2  17 x  28 3x 2  17 x  10
4  25 x 2 6x2  x  1
c d
10 x 2  11x  6 2 x2  7 x  4

Extend

7 Simplify x 2  10 x  25

( x  2)2  3( x  2)2
8 Simplify
x2  4
Answers
1 a 2x3y3(3x – 5y) b 7a3b2(3b3 + 5a2)
c 5x2y2(5 – 2x + 3y)

2 a (x + 3)(x + 4) b (x + 7)(x – 2)
c (x – 5)(x – 6) d (x – 8)(x + 3)
e (x – 9)(x + 2) f (x + 5)(x – 4)
g (x – 8)(x + 5) h (x + 7)(x – 4)

3 a (6x – 7y)(6x + 7y) b (2x – 9y)(2x + 9y)

c 2(3a – 10bc)(3a + 10bc)

4 a (x – 1)(2x + 3) b (3x + 1)(2x + 5)

c (2x + 1)(x + 3) d (3x – 1)(3x – 4)
e (5x + 3)(2x +3) f 2(3x – 2)(2x –5)

2( x  2) x
5 a b
x 1 x 1
x2 x
c d
x x5
x3 x
e f
x x5

3x  4 2x  3
6 a b
x7 3x  2
2  5x 3x  1
c d
2x  3 x4

7 (x + 5)

4( x  2)
8
x2
Solving quadratic equations by
factorisation
A LEVEL LINKS
Scheme of work: 1b. Quadratic functions – factorising, solving, graphs and the discriminants

Key points
 A quadratic equation is an equation in the form ax2 + bx + c = 0 where a ≠ 0.
 To factorise a quadratic equation find two numbers whose sum is b and whose products is ac.
 When the product of two numbers is 0, then at least one of the numbers must be 0.
 If a quadratic can be solved it will have two solutions (these may be equal).

Examples
Example 1 Solve 5x2 = 15x

5x2 = 15x 1 Rearrange the equation so that all of

the terms are on one side of the
5x2 − 15x = 0 equation and it is equal to zero.
Do not divide both sides by x as this
would lose the solution x = 0.
5x(x − 3) = 0 2 Factorise the quadratic equation.
5x is a common factor.
So 5x = 0 or (x − 3) = 0 3 When two values multiply to make
zero, at least one of the values must
be zero.
Therefore x = 0 or x = 3 4 Solve these two equations.

Example 2 Solve x2 + 7x + 12 = 0

x2 + 7x + 12 = 0 1 Factorise the quadratic equation.

Work out the two factors of ac = 12
b = 7, ac = 12 which add to give you b = 7.
(4 and 3)
x2 + 4x + 3x + 12 = 0 2 Rewrite the b term (7x) using these
two factors.
x(x + 4) + 3(x + 4) = 0 3 Factorise the first two terms and the
last two terms.
(x + 4)(x + 3) = 0 4 (x + 4) is a factor of both terms.
So (x + 4) = 0 or (x + 3) = 0 5 When two values multiply to make
zero, at least one of the values must
be zero.
Therefore x = −4 or x = −3 6 Solve these two equations.
Example 3 Solve 9x2 − 16 = 0

9x2 − 16 = 0 1 Factorise the quadratic equation.

(3x + 4)(3x – 4) = 0 This is the difference of two squares
as the two terms are (3x)2 and (4)2.
So (3x + 4) = 0 or (3x – 4) = 0 2 When two values multiply to make
zero, at least one of the values must
4 4 be zero.
x or x  3 Solve these two equations.
3 3

Example 4 Solve 2x2 − 5x − 12 = 0

b = −5, ac = −24 1 Factorise the quadratic equation.

Work out the two factors of ac = −24
which add to give you b = −5.
(−8 and 3)
So 2x2 − 8x + 3x – 12 = 0 2 Rewrite the b term (−5x) using these
two factors.
2x(x − 4) + 3(x − 4) = 0 3 Factorise the first two terms and the
last two terms.
(x – 4)(2x + 3) = 0 4 (x − 4) is a factor of both terms.
So (x – 4) = 0 or (2x +3) = 0 5 When two values multiply to make
zero, at least one of the values must
3 be zero.
x  4 or x   6 Solve these two equations.
2

Practice
1 Solve
a 6x2 + 4x = 0 b 28x2 – 21x = 0
c x2 + 7x + 10 = 0 d x2 – 5x + 6 = 0
e x2 – 3x – 4 = 0 f x2 + 3x – 10 = 0
g x2 – 10x + 24 = 0 h x2 – 36 = 0
i x2 + 3x – 28 = 0 j x2 – 6x + 9 = 0
k 2x2 – 7x – 4 = 0 l 3x2 – 13x – 10 = 0

2 Solve
a x2 – 3x = 10 b x2 – 3 = 2x Hint
c x2 + 5x = 24 d x2 – 42 = x
Get all terms
e x(x + 2) = 2x + 25 f x2 – 30 = 3x – 2
onto one side
g x(3x + 1) = x2 + 15 h 3x(x – 1) = 2(x + 1)
of the equation.
Solving quadratic equations by
completing the square
A LEVEL LINKS
Scheme of work: 1b. Quadratic functions – factorising, solving, graphs and the discriminants

Key points
 Completing the square lets you write a quadratic equation in the form p(x + q)2 + r = 0.

Examples
Example 5 Solve x2 + 6x + 4 = 0. Give your solutions in surd form.

x2 + 6x + 4 = 0 1 Write x2 + bx + c = 0 in the form

2 2
 b b
(x + 3)2 − 9 + 4 = 0 x    c  0
 2 2
(x + 3)2 − 5 = 0 2 Simplify.
(x + 3)2 = 5 3 Rearrange the equation to work out
x. First, add 5 to both sides.
x+3=  5 4 Square root both sides.
Remember that the square root of a
value gives two answers.
x =  5 3
5 Subtract 3 from both sides to solve
the equation.
So x =  5  3 or x = 5 3 6 Write down both solutions.

Example 6 Solve 2x2 − 7x + 4 = 0. Give your solutions in surd form.

2x2 − 7x + 4 = 0 1 Before completing the square write

ax2 + bx + c in the form
 7   b 
2  x2  x   4 = 0 a  x2  x   c
 2   a 

 7 7 
2 2 2 Now complete the square by writing
2  x        4 = 0 7
 4   4   x 2  x in the form
2
2 2
 b   b 
x   
 2a   2a 
2
 7  49
2 x    4 =0 3 Expand the square brackets.
 4 8
2
 7  17
2 x    =0 4 Simplify.
 4 8
(continued on next page)
  7  17
2 5 Rearrange the equation to work out
2 x    17
 4 8 x. First, add to both sides.
8
2
 7  17 6 Divide both sides by 2.
x  
 4  16

7 Square root both sides. Remember

7 17
x  that the square root of a value gives
4 4 two answers.
17 7 7
x  8 Add to both sides.
4 4 4

7 17 7 17 9 Write down both the solutions.

So x   or x  
4 4 4 4

Practice
3 Solve by completing the square.
a x2 – 4x – 3 = 0 b x2 – 10x + 4 = 0
c x2 + 8x – 5 = 0 d x2 – 2x – 6 = 0
e 2x2 + 8x – 5 = 0 f 5x2 + 3x – 4 = 0

4 Solve by completing the square.

Hint
a (x – 4)(x + 2) = 5
b 2x2 + 6x – 7 = 0 Get all terms
c x2 – 5x + 3 = 0 onto one side
of the equation.
Solving quadratic equations by using the
formula
A LEVEL LINKS
Scheme of work: 1b. Quadratic functions – factorising, solving, graphs and the discriminants

Key points
 Any quadratic equation of the form ax2 + bx + c = 0 can be solved using the formula
b  b 2  4ac
x
2a
 If b2 – 4ac is negative then the quadratic equation does not have any real solutions.
 It is useful to write down the formula before substituting the values for a, b and c.

Examples
Example 7 Solve x2 + 6x + 4 = 0. Give your solutions in surd form.

a = 1, b = 6, c = 4 1 Identify a, b and c and write down

b  b 2  4ac the formula.
x
2a Remember that b  b 2  4ac is
all over 2a, not just part of it.

2 Substitute a = 1, b = 6, c = 4 into the

6  62  4(1)(4)
x formula.
2(1)
6  20 3 Simplify. The denominator is 2, but
x this is only because a = 1. The
2
denominator will not always be 2.
6  2 5 4 Simplify 20 .
x
2 20  4  5  4  5  2 5
x  3  5 5 Simplify by dividing numerator and
denominator by 2.
So x  3  5 or x  5  3 6 Write down both the solutions.
Example 8 Solve 3x2 − 7x − 2 = 0. Give your solutions in surd form.

a = 3, b = −7, c = −2 1 Identify a, b and c, making sure you

b  b  4ac
2 get the signs right and write down
x the formula.
2a
Remember that b  b 2  4ac is
all over 2a, not just part of it.
(7)  (7)2  4(3)(2) 2 Substitute a = 3, b = −7, c = −2 into
x
2(3) the formula.

7  73 3 Simplify. The denominator is 6

x when a = 3. A common mistake is
6 to always write a denominator of 2.
7  73 7  73
So x  or x  4 Write down both the solutions.
6 6

Practice
5 Solve, giving your solutions in surd form.
a 3x2 + 6x + 2 = 0 b 2x2 – 4x – 7 = 0

6 Solve the equation x2 – 7x + 2 = 0

a b
Give your solutions in the form , where a, b and c are integers.
c

Hint
7 Solve 10x2 + 3x + 3 = 5
Give your solution in surd form. Get all terms onto one
side of the equation.

Extend
8 Choose an appropriate method to solve each quadratic equation, giving your answer in surd form
when necessary.
a 4x(x – 1) = 3x – 2
b 10 = (x + 1)2
c x(3x – 1) = 10
Answers
2 3
1 a x = 0 or x =  b x = 0 or x =
3 4
c x = –5 or x = –2 d x = 2 or x = 3
e x = –1 or x = 4 f x = –5 or x = 2
g x = 4 or x = 6 h x = –6 or x = 6
i x = –7 or x = 4 j x=3
1 2
k x=  or x = 4 l x=  or x = 5
2 3

2 a x = –2 or x = 5 b x = –1 or x = 3
c x = –8 or x = 3 d x = –6 or x = 7
e x = –5 or x = 5 f x = –4 or x = 7
1 1
g x = –3 or x = 2 h x =  or x = 2
2 3

3 a x=2+ 7 or x = 2 – 7 b x=5+ 21 or x = 5 – 21
c x = –4 + 21 or x = –4 – 21 d x=1+ 7 or x = 1 – 7
3  89 3  89
e x = –2 + 6.5 or x = –2 – 6.5 f x= or x =
10 10

3  23 3  23
4 a x=1+ 14 or x = 1 – 14 b x= or x =
2 2
5  13 5  13
c x= or x =
2 2

3 3 3 2 3 2
5 a x = –1 + or x = –1 – b x=1+ or x = 1 –
3 3 2 2

7  41 7  41
6 x= or x =
2 2

3  89 3  89
7 x = or x =
20 20

7  17 7  17
8 a x= or x =
8 8
b x = –1 + 10 or x = –1 – 10
2
c x = –1 or x = 2
3
Sketching quadratic graphs
A LEVEL LINKS
Scheme of work: 1b. Quadratic functions – factorising, solving, graphs and the discriminants

Key points
 The graph of the quadratic function
y = ax2 + bx + c, where a ≠ 0, is a curve
called a parabola.
 Parabolas have a line of symmetry and for a > 0 for a < 0
a shape as shown.
 To sketch the graph of a function, find the points where the graph intersects the axes.
 To find where the curve intersects the y-axis substitute x = 0 into the function.
 To find where the curve intersects the x-axis substitute y = 0 into the function.
 At the turning points of a graph the gradient of the curve is 0 and any tangents to the curve at
these points are horizontal.
 To find the coordinates of the maximum or minimum point (turning points) of a quadratic
curve (parabola) you can use the completed square form of the function.

Examples
Example 1 Sketch the graph of y = x2.

The graph of y = x2 is a parabola.

When x = 0, y = 0.

a = 1 which is greater
than zero, so the graph
has the shape:

Example 2 Sketch the graph of y = x2 − x − 6.

When x = 0, y = 02 − 0 − 6 = −6 1 Find where the graph intersects the

So the graph intersects the y-axis at y-axis by substituting x = 0.
(0, −6)
When y = 0, x2 − x − 6 = 0 2 Find where the graph intersects the
x-axis by substituting y = 0.
(x + 2)(x − 3) = 0 3 Solve the equation by factorising.

x = −2 or x = 3 4 Solve (x + 2) = 0 and (x − 3) = 0.

So, 5 a = 1 which is greater

the graph intersects the x-axis at (−2, 0) than zero, so the graph
and (3, 0) has the shape:

(continued on next page)

 2
 1 1 6 To find the turning point, complete
x2 − x − 6 =  x     6
 2 4 the square.
2
 1  25
= x  
 2 4
2
 1 1
When  x    0 , x  and 7 The turning point is the minimum
 2 2 value for this expression and occurs
25 when the term in the bracket is
y , so the turning point is at the
4 equal to zero.
 1 25 
point  ,  
2 4 

Practice
1 Sketch the graph of y = −x2.

2 Sketch each graph, labelling where the curve crosses the axes.
a y = (x + 2)(x − 1) b y = x(x − 3) c y = (x + 1)(x + 5)

3 Sketch each graph, labelling where the curve crosses the axes.
a y = x2 − x − 6 b y = x2 − 5x + 4 c y = x2 – 4
d y = x2 + 4x e y = 9 − x2 f y = x2 + 2x − 3

4 Sketch the graph of y = 2x2 + 5x − 3, labelling where the curve crosses the axes.

Extend
5 Sketch each graph. Label where the curve crosses the axes and write down the coordinates of the
turning point.
a y = x2 − 5x + 6 b y = −x2 + 7x − 12 c y = −x2 + 4x

6 Sketch the graph of y = x2 + 2x + 1. Label where the curve crosses the axes and write down the
equation of the line of symmetry.
Answers
1

2 a b c

3 a b c

d e f
4

5 a b c

Line of symmetry at x = −1.

Solving linear simultaneous equations
using the elimination method
A LEVEL LINKS
Scheme of work: 1c. Equations – quadratic/linear simultaneous

Key points
 Two equations are simultaneous when they are both true at the same time.
 Solving simultaneous linear equations in two unknowns involves finding the value of each
unknown which works for both equations.
 Make sure that the coefficient of one of the unknowns is the same in both equations.
 Eliminate this equal unknown by either subtracting or adding the two equations.

Examples
Example 1 Solve the simultaneous equations 3x + y = 5 and x + y = 1

3x + y = 5 1 Subtract the second equation from

– x+y=1 the first equation to eliminate the y
2x =4 term.
So x = 2

Using x + y = 1 2 To find the value of y, substitute

2+y=1 x = 2 into one of the original
So y = −1 equations.

Check: 3 Substitute the values of x and y into

equation 1: 3 × 2 + (−1) = 5 YES both equations to check your
equation 2: 2 + (−1) = 1 YES answers.

Example 2 Solve x + 2y = 13 and 5x − 2y = 5 simultaneously.

x + 2y = 13 1 Add the two equations together to

+ 5x − 2y = 5 eliminate the y term.
   6x = 18
So x = 3

Using x + 2y = 13 2 To find the value of y, substitute

3 + 2y = 13 x = 3 into one of the original
So y = 5 equations.

Check: 3 Substitute the values of x and y into

equation 1: 3 + 2 × 5 = 13 YES both equations to check your
equation 2: 5 × 3 − 2 × 5 = 5 YES answers.
Example 3 Solve 2x + 3y = 2 and 5x + 4y = 12 simultaneously.

(2x + 3y = 2) × 4  8x + 12y = 8 1 Multiply the first equation by 4 and

(5x + 4y = 12) × 3  15x + 12y = 36 the second equation by 3 to make
7x = 28 the coefficient of y the same for
both equations. Then subtract the
So x = 4 first equation from the second
equation to eliminate the y term.

Using 2x + 3y = 2 2 To find the value of y, substitute

2 × 4 + 3y = 2 x = 4 into one of the original
So y = −2 equations.

Check: 3 Substitute the values of x and y into

equation 1: 2 × 4 + 3 × (−2) = 2 YES both equations to check your
equation 2: 5 × 4 + 4 × (−2) = 12 YES answers.

Practice
Solve these simultaneous equations.

1 4x + y = 8 2 3x + y = 7
x + y = 5 3x + 2y = 5

3 4x + y = 3 4 3x + 4y = 7
3x – y = 11 x – 4y = 5

5 2x + y = 11 6 2x + 3y = 11
x – 3y = 9 3x + 2y = 4
Solving linear simultaneous equations
using the substitution method
A LEVEL LINKS
Scheme of work: 1c. Equations – quadratic/linear simultaneous
Textbook: Pure Year 1, 3.1 Linear simultaneous equations

Key points
 The subsitution method is the method most commonly used for A level. This is because it is
the method used to solve linear and quadratic simultaneous equations.

Examples
Example 4 Solve the simultaneous equations y = 2x + 1 and 5x + 3y = 14

5x + 3(2x + 1) = 14 1 Substitute 2x + 1 for y into the

second equation.
5x + 6x + 3 = 14 2 Expand the brackets and simplify.
11x + 3 = 14
11x = 11 3 Work out the value of x.
So x = 1

Using y = 2x + 1 4 To find the value of y, substitute

y=2×1+1 x = 1 into one of the original
So y = 3 equations.

Check: 5 Substitute the values of x and y into

equation 1: 3 = 2 × 1 + 1 YES both equations to check your
equation 2: 5 × 1 + 3 × 3 = 14 YES answers.

Example 5 Solve 2x − y = 16 and 4x + 3y = −3 simultaneously.

y = 2x − 16 1 Rearrange the first equation.

4x + 3(2x − 16) = −3 2 Substitute 2x − 16 for y into the
second equation.
4x + 6x − 48 = −3 3 Expand the brackets and simplify.
10x − 48 = −3
10x = 45 4 Work out the value of x.
So x = 4 12
Using y = 2x − 16 5 To find the value of y, substitute
y = 2 × 4 12 − 16 x = 4 12 into one of the original
So y = −7 equations.
Check: 6 Substitute the values of x and y into
equation 1: 2 × 4 12 – (–7) = 16 YES both equations to check your
equation 2: 4 ×  4 12  + 3 × (−7) = −3 YES answers.
Practice
Solve these simultaneous equations.

7 y=x–4 8 y = 2x – 3
2x + 5y = 43 5x – 3y = 11

9 2y = 4x + 5 10 2x = y – 2
9x + 5y = 22 8x – 5y = –11

11 3x + 4y = 8 12 3y = 4x – 7
2x – y = –13 2y = 3x – 4

13 3x = y – 1 14 3x + 2y + 1 = 0
2y – 2x = 3 4y = 8 – x

Extend

3( y  x)
15 Solve the simultaneous equations 3x + 5y − 20 = 0 and 2( x  y )  .
4
Answers
1 x = 1, y = 4

2 x = 3, y = –2

3 x = 2, y = –5

1
4 x = 3, y = –
2

5 x = 6, y = –1

6 x = –2, y = 5

7 x = 9, y = 5

8 x = –2, y = –7

1 1
9 x= ,y=3
2 2

1
10 x = ,y=3
2

11 x = –4, y = 5

12 x = –2, y = –5

1 3
13 x = ,y=1
4 4

1
14 x = –2, y = 2
2

1 1
15 x = –2 , y = 5
2 2
Solving linear and quadratic simultaneous
equations
A LEVEL LINKS
Scheme of work: 1c. Equations – quadratic/linear simultaneous

Key points
 Make one of the unknowns the subject of the linear equation (rearranging where necessary).
 Use the linear equation to substitute into the quadratic equation.
 There are usually two pairs of solutions.

Examples
Example 1 Solve the simultaneous equations y = x + 1 and x2 + y2 = 13

x2 + (x + 1)2 = 13 1 Substitute x + 1 for y into the second

equation.
x2 + x2 + x + x + 1 = 13 2 Expand the brackets and simplify.
2x2 + 2x + 1 = 13

2x2 + 2x − 12 = 0 3 Factorise the quadratic equation.

(2x − 4)(x + 3) = 0
So x = 2 or x = −3 4 Work out the values of x.

Using y = x + 1 5 To find the value of y, substitute

When x = 2, y = 2 + 1 = 3 both values of x into one of the
When x = −3, y = −3 + 1 = −2 original equations.

So the solutions are

x = 2, y = 3 and x = −3, y = −2

Check: 6 Substitute both pairs of values of x

equation 1: 3 = 2 + 1 YES and y into both equations to check
and −2 = −3 + 1 YES your answers.
equation 2: 22 + 32 = 13 YES
and (−3)2 + (−2)2 = 13 YES
Example 2 Solve 2x + 3y = 5 and 2y2 + xy = 12 simultaneously.

5  3y
x 1 Rearrange the first equation.
2
 5  3y  5  3y
2 y2    y  12 2 Substitute for x into the
 2  2
second equation. Notice how it is
5 y  3y2 easier to substitute for x than for y.
2 y2   12
2 3 Expand the brackets and simplify.
4 y 2  5 y  3 y 2  24
y 2  5 y  24  0
(y + 8)(y − 3) = 0 4 Factorise the quadratic equation.
So y = −8 or y = 3 5 Work out the values of y.

Using 2x + 3y = 5 6 To find the value of x, substitute

When y = −8, 2x + 3 × (−8) = 5, x = 14.5 both values of y into one of the
When y = 3, 2x + 3 × 3 = 5, x = −2 original equations.

So the solutions are

x = 14.5, y = −8 and x = −2, y = 3

Check: 7 Substitute both pairs of values of x

equation 1: 2 × 14.5 + 3 × (−8) = 5 YES and y into both equations to check
and 2 × (−2) + 3 × 3 = 5 YES your answers.
2
equation 2: 2×(−8) + 14.5×(−8) = 12 YES
and 2 × (3)2 + (−2) × 3 = 12 YES

Practice
Solve these simultaneous equations.
1 y = 2x + 1 2 y=6−x
x2 + y2 = 10 x2 + y2 = 20

3 y=x–3 4 y = 9 − 2x
x2 + y2 = 5 x2 + y2 = 17

5 y = 3x – 5 6 y=x−5
y = x2 − 2x + 1 y = x2 − 5x − 12

7 y=x+5 8 y = 2x – 1
x2 + y2 = 25 x2 + xy = 24

9 y = 2x 10 2x + y = 11
y2 – xy = 8 xy = 15

Extend
11 x – y = 1 12 y – x = 2
x2 + y2 = 3 x2 + xy = 3
Answers
1 x = 1, y = 3
9 13
x , y
5 5

2 x = 2, y = 4
x = 4, y = 2

3 x = 1, y = −2
x = 2, y = –1

4 x = 4, y = 1
16 13
x , y
5 5

5 x = 3, y = 4
x = 2, y = 1

6 x = 7, y = 2
x = −1, y = −6

7 x = 0, y = 5
x = –5, y = 0

8 19
8 x =  , y = 
3 3
x = 3, y = 5

9 x = –2, y = –4
x = 2, y = 4

5
10 x = ,y=6
2
x = 3, y = 5

1 5 1  5
11 x = ,y=
2 2
1 5 1  5
x= , y =
2 2

1  7 3 7
12 x = ,y=
2 2
1  7 3 7
x= ,y=
2 2
Solving simultaneous equations
graphically
A LEVEL LINKS
Scheme of work: 1c. Equations – quadratic/linear simultaneous

Key points
 You can solve any pair of simultaneous equations by drawing the graph of both equations and
finding the point/points of intersection.

Examples
Example 1 Solve the simultaneous equations y = 5x + 2 and x + y = 5 graphically.

y=5–x 1 Rearrange the equation x + y = 5

to make y the subject.
y = 5 – x has gradient –1 and y-intercept 5. 2 Plot both graphs on the same grid
y = 5x + 2 has gradient 5 and y-intercept 2. using the gradients and
y-intercepts.

Lines intersect at 3 The solutions of the simultaneous

x = 0.5, y = 4.5 equations are the point of
intersection.
Check:
First equation y = 5x + 2: 4 Check your solutions by
4.5 = 5 × 0.5 + 2 YES substituting the values into both
Second equation x + y = 5: equations.
0.5 + 4.5 = 5 YES
Example 2 Solve the simultaneous equations y = x − 4 and y = x2 − 4x + 2 graphically.

1 Construct a table of values and

x 0 1 2 3 4 calculate the points for the quadratic
y 2 –1 –2 –1 2 equation.

2 Plot the graph.

3 Plot the linear graph on the same

grid using the gradient and
y-intercept.
y = x – 4 has gradient 1 and
y-intercept –4.

The line and curve intersect at 4 The solutions of the simultaneous

x = 3, y = −1 and x = 2, y = −2 equations are the points of
intersection.
Check:
First equation y = x − 4: 5 Check your solutions by substituting
−1 = 3 − 4 YES the values into both equations.
−2 = 2 − 4 YES
Second equation y = x2 − 4x + 2:
−1 = 32 − 4 × 3 + 2 YES
−2 = 22 − 4 × 2 + 2 YES

Practice
1 Solve these pairs of simultaneous equations graphically.
a y = 3x − 1 and y = x + 3
b y = x − 5 and y = 7 − 5x
c y = 3x + 4 and y = 2 − x

2 Solve these pairs of simultaneous equations graphically. Hint

a x + y = 0 and y = 2x + 6
Rearrange the
b 4x + 2y = 3 and y = 3x − 1
equation to make
c 2x + y + 4 = 0 and 2y = 3x − 1 y the subject.
3 Solve these pairs of simultaneous equations graphically.
a y = x − 1 and y = x2 − 4x + 3
b y = 1 − 3x and y = x2 − 3x − 3
c y = 3 − x and y = x2 + 2x + 5

4 Solve the simultaneous equations x + y = 1 and x2 + y2 = 25 graphically.

Extend
5 a Solve the simultaneous equations 2x + y = 3 and x2 + y = 4
i graphically
ii algebraically to 2 decimal places.
b Which method gives the more accurate solutions? Explain your answer.
Answers
1 a x = 2, y = 5
b x = 2, y = −3
c x = −0.5, y = 2.5

2 a x = −2, y = 2
b x = 0.5, y = 0.5
c x = −1, y = −2

3 a x = 1, y = 0 and x = 4, y = 3
b x = −2, y = 7 and x = 2, y = −5
c x = −2, y = 5 and x = −1, y = 4

4 x = −3, y = 4 and x = 4, y = −3

5 a i x = 2.5, y = −2 and x = −0.5, y = 4

ii x = 2.41, y = −1.83 and x = −0.41, y = 3.83
b Solving algebraically gives the more accurate solutions as the solutions from the graph are
only estimates, based on the accuracy of your graph.
Quadratic inequalities
A LEVEL LINKS
Scheme of work: 1d. Inequalities – linear and quadratic (including graphical solutions)

Key points
 First replace the inequality sign by = and solve the quadratic equation.
 Sketch the graph of the quadratic function.
 Use the graph to find the values which satisfy the quadratic inequality.

Examples
Example 1 Find the set of values of x which satisfy x2 + 5x + 6 > 0

x2 + 5x + 6 = 0 1 Solve the quadratic equation by

(x + 3)(x + 2) = 0 factorising.
x = −3 or x = −2

2 Sketch the graph of

y = (x + 3)(x + 2)

3 Identify on the graph where

x2 + 5x + 6 > 0, i.e. where y > 0

x < −3 or x > −2 4 Write down the values which satisfy

the inequality x2 + 5x + 6 > 0

Example 2 Find the set of values of x which satisfy x2 − 5x ≤ 0

x2 − 5x = 0 1 Solve the quadratic equation by

x(x − 5) = 0 factorising.
x = 0 or x = 5
2 Sketch the graph of y = x(x − 5)

3 Identify on the graph where

x2 − 5x ≤ 0, i.e. where y ≤ 0

0≤x≤5 4 Write down the values which satisfy

the inequality x2 − 5x ≤ 0
Example 3 Find the set of values of x which satisfy −x2 − 3x + 10 ≥ 0

−x2 − 3x + 10 = 0 1 Solve the quadratic equation by

(−x + 2)(x + 5) = 0 factorising.
x = 2 or x = −5
y
2 Sketch the graph of
y = (−x + 2)(x + 5) = 0

3 Identify on the graph where

−x2 − 3x + 10 ≥ 0, i.e. where y ≥ 0

–5 O 2 x

−5 ≤ x ≤ 2 3 Write down the values which satisfy

the inequality −x2 − 3x + 10 ≥ 0

Practice
1 Find the set of values of x for which (x + 7)(x – 4) ≤ 0

2 Find the set of values of x for which x2 – 4x – 12 ≥ 0

3 Find the set of values of x for which 2x2 –7x + 3 < 0

4 Find the set of values of x for which 4x2 + 4x – 3 > 0

5 Find the set of values of x for which 12 + x – x2 ≥ 0

Extend
Find the set of values which satisfy the following inequalities.

6 x2 + x ≤ 6

7 x(2x – 9) < –10

8 6x2 ≥ 15 + x
Answers
1 –7 ≤ x ≤ 4

2 x ≤ –2 or x ≥ 6

1
3  x3
2

3 1
4 x<  or x >
2 2

5 –3 ≤ x ≤ 4

6 –3 ≤ x ≤ 2

1
7 2<x<2
2
3 5
8 x or x 
2 3
Sketching cubic and reciprocal graphs
A LEVEL LINKS
Scheme of work: 1e. Graphs – cubic, quartic and reciprocal

Key points
 The graph of a cubic function,
which can be written in the
form y = ax3 + bx2 + cx + d,
where a ≠ 0, has one of the
shapes shown here.

 The graph of a reciprocal

a
function of the form y  has
x
one of the shapes shown here.

 To sketch the graph of a function, find the points where the graph intersects the axes.
 To find where the curve intersects the y-axis substitute x = 0 into the function.
 To find where the curve intersects the x-axis substitute y = 0 into the function.
 Where appropriate, mark and label the asymptotes on the graph.
 Asymptotes are lines (usually horizontal or vertical) which the curve gets closer to but never
touches or crosses. Asymptotes usually occur with reciprocal functions. For example, the
a
asymptotes for the graph of y  are the two axes (the lines y = 0 and x = 0).
x
 At the turning points of a graph the gradient of the curve is 0 and any tangents to the curve at
these points are horizontal.
 A double root is when two of the solutions are equal. For example (x – 3)2(x + 2) has a
double root at x = 3.
 When there is a double root, this is one of the turning points of a cubic function.
Examples
Example 1 Sketch the graph of y = (x − 3)(x − 1)(x + 2)

To sketch a cubic curve find intersects with both axes and use the key points above
for the correct shape.

When x = 0, y = (0 − 3)(0 − 1)(0 + 2) 1 Find where the graph intersects the

= (−3) × (−1) × 2 = 6 axes by substituting x = 0 and y = 0.
The graph intersects the y-axis at (0, 6) Make sure you get the coordinates
the right way around, (x, y).
When y = 0, (x − 3)(x − 1)(x + 2) = 0 2 Solve the equation by solving
So x = 3, x = 1 or x = −2 x − 3 = 0, x − 1 = 0 and x + 2 = 0
The graph intersects the x-axis at
(−2, 0), (1, 0) and (3, 0)

3 Sketch the graph.

a = 1 > 0 so the graph has the shape:

Example 2 Sketch the graph of y = (x + 2)2(x − 1)

To sketch a cubic curve find intersects with both axes and use the key points above
for the correct shape.

When x = 0, y = (0 + 2)2(0 − 1) 1 Find where the graph intersects the

= 22 × (−1) = −4 axes by substituting x = 0 and y = 0.
The graph intersects the y-axis at (0, −4)

When y = 0, (x + 2)2(x − 1) = 0 2 Solve the equation by solving

So x = −2 or x =1 x + 2 = 0 and x − 1 = 0

(−2, 0) is a turning point as x = −2 is a

double root.
The graph crosses the x-axis at (1, 0)

3 a = 1 > 0 so the graph has the shape:

Practice
1 Here are six equations.
Hint
5
A y B y = x + 3x – 10
2
C 3
y = x + 3x 2
x Find where each
D y = 1 – 3x2 – x3 E y = x – 3x – 1
3 2
F x+y=5 of the cubic
equations cross
the y-axis.
Here are six graphs.
i ii iii

iv v vi

a Match each graph to its equation.

b Copy the graphs ii, iv and vi and draw the tangent and normal each at point P.

Sketch the following graphs

2 y = 2x3 3 y = x(x – 2)(x + 2)

4 y = (x + 1)(x + 4)(x – 3) 5 y = (x + 1)(x – 2)(1 – x)

6 y = (x – 3)2(x + 1) 7 y = (x – 1)2(x – 2)

3 2
8 y = a 9 y = 
x x
Hint: Look at the shape of y = x
in the second key point.

Extend
1 1
10 Sketch the graph of y  11 Sketch the graph of y 
x2 x 1
Answers
1 a i–C
ii – E
iii – B
iv – A
v–F
vi – D

b ii iv

vi

2 3

4 5
6 7

8 9 y

O x

10 11
Translating graphs
A LEVEL LINKS
Scheme of work: 1f. Transformations – transforming graphs – f(x) notation

Key points
 The transformation y = f(x) ± a is a
translation of y = f(x) parallel to the y-axis;
it is a vertical translation.

As shown on the graph,

o y = f(x) + a translates y = f(x) up
o y = f(x) – a translates y = f(x) down.

 The transformation y = f(x ± a) is a

translation of y = f(x) parallel to the x-axis;
it is a horizontal translation.

As shown on the graph,

o y = f(x + a) translates y = f(x) to the left
o y = f(x – a) translates y = f(x) to the right.

Examples
Example 1 The graph shows the function y = f(x).
Sketch the graph of y = f(x) + 2.

For the function y = f(x) + 2 translate

the function y = f(x) 2 units up.
Example 2 The graph shows the function y = f(x).

Sketch the graph of y = f(x − 3).

For the function y = f(x − 3) translate

the function y = f(x) 3 units right.

Practice
1 The graph shows the function y = f(x).
Copy the graph and on the same axes sketch and
label the graphs of y = f(x) + 4 and y = f(x + 2).

2 The graph shows the function y = f(x).

Copy the graph and on the same axes sketch and
label the graphs of y = f(x + 3) and y = f(x) – 3.

3 The graph shows the function y = f(x).

Copy the graph and on the same axes sketch the
graph of y = f(x – 5).
4 The graph shows the function y = f(x) and two
transformations of y = f(x), labelled C1 and C2.
Write down the equations of the translated curves
C1 and C2 in function form.

5 The graph shows the function y = f(x) and two

transformations of y = f(x), labelled C1 and C2.
Write down the equations of the translated curves
C1 and C2 in function form.

6 The graph shows the function y = f(x).

a Sketch the graph of y = f(x) + 2

b Sketch the graph of y = f(x + 2)
Stretching graphs
A LEVEL LINKS
Scheme of work: 1f. Transformations – transforming graphs – f(x) notation
Textbook: Pure Year 1, 4.6 Stretching graphs

Key points

 The transformation y = f(ax) is a horizontal

1
stretch of y = f(x) with scale factor
a
parallel to the x-axis.

 The transformation y = f(–ax) is a

horizontal stretch of y = f(x) with scale
1
factor parallel to the x-axis and then a
a
reflection in the y-axis.

 The transformation y = af(x) is a vertical

stretch of y = f(x) with scale factor a
parallel to the y-axis.

 The transformation y = –af(x) is a vertical

stretch of y = f(x) with scale factor a
parallel to the y-axis and then a reflection
in the x-axis.
Examples
Example 3 The graph shows the function y = f(x).
Sketch and label the graphs of
y = 2f(x) and y = –f(x).

The function y = 2f(x) is a vertical

stretch of y = f(x) with scale
factor 2 parallel to the y-axis.
The function y = −f(x) is a
reflection of y = f(x) in the
x-axis.

Example 4 The graph shows the function y = f(x).

Sketch and label the graphs of

y = f(2x) and y = f(–x).

The function y = f(2x) is a horizontal

stretch of y = f(x) with scale factor
1
2 parallel to the x-axis.

The function y = f(−x) is a reflection

of y = f(x) in the y-axis.
Practice
7 The graph shows the function y = f(x).
a Copy the graph and on the same axes
sketch and label the graph of y = 3f(x).
b Make another copy of the graph and on
the same axes sketch and label the graph
of y = f(2x).

8 The graph shows the function y = f(x).

Copy the graph and on the same axes
sketch and label the graphs of
y = –2f(x) and y = f(3x).

9 The graph shows the function y = f(x).

Copy the graph and, on the same axes,
sketch and label the graphs of
y = –f(x) and y = f  12 x  .

10 The graph shows the function y = f(x).

Copy the graph and, on the same axes,
sketch the graph of y = –f(2x).

11 The graph shows the function y = f(x) and a

transformation, labelled C.
Write down the equation of the translated
curve C in function form.
12 The graph shows the function y = f(x) and a
transformation labelled C.
Write down the equation of the translated
curve C in function form.

13 The graph shows the function y = f(x).

a Sketch the graph of y = −f(x).
b Sketch the graph of y = 2f(x).

Extend
14 a Sketch and label the graph of y = f(x), where f(x) = (x – 1)(x + 1).
b On the same axes, sketch and label the graphs of y = f(x) – 2 and y = f(x + 2).

15 a Sketch and label the graph of y = f(x), where f(x) = –(x + 1)(x – 2).
b On the same axes, sketch and label the graph of y = f   12 x  .
Answers
1 2

4 C1: y = f(x – 90°)

C2: y = f(x) – 2

5 C1: y = f(x – 5)
C2: y = f(x) – 3

6 a b
7 a b

8 9

10

11 y = f(2x)

12 y = –2f(2x) or y = 2f(–2x)

13 a b
14

15