Cambridge International Catalogue 2026 FINAL - Flipbook - Page 55
Cambridge IGCSE™ and O Level
Additional Mathematics (0606/4037)
Student’s Book, Second Edition
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Using graphs to solve cubic inequalities
4 EQUATIONS, INEQUALITIES AND GRAPHS
Using graphs to solve cubic
inequalities
Solve the inequality 3(x + 2)(x − 1)(x − 7) −100 graphically.
Cubic graphs have distinctive shapes determined by the coefficient of x³.
Solution
Worked example
Because you are solving the inequality graphically, you will need to draw
the curve as accurately as possible on graph paper, so start by drawing up a
table of values.
Negative x3 term
Positive x3 term
y = 3(x + 2)(x − 1)(x − 7)
x
The centre part of each of these curves may not have two distinct
turning points like those shown above, but may instead ‘flatten out’
to give a point of inflection. When the modulus of a cubic function is
required, any part of the curve below the x-axis is reflected in that axis.
−3
−2
View sample material
from our Student’s
Books
1
1
2
2
3
3
4
4
5
5
8
8
9
−2
−1
0
5
6
(x − 7)
−10
−9
−8
−7
−6
−5
−4
−3
−2
−1
0
1
x
−120
0
48
42
0
−60
−120
−162
−168
−120
0
210
2
7
7
0
−3
1
6
6
−1
−4
Worked example
Look inside
0
−1
(x + 2)
(x − 1)
3
4
10
7
The solution is given by the values of x that correspond to the parts of the
curve on or below the line y = −100.
a Sketch the graph of y = 3(x + 2)(x − 1)(x − 7). Identify the points where the
You are asked for
curve cuts the axes.
a sketch graph,
b Sketch the graph of y = |3(x + 2)(x − 1)(x − 7)|.
so although it
must show the
Solution
main features,
a The curve crosses the x-axis at −2, 1 and 7. Notice that the distance
it does not need
between consecutive points is 3 and 6 units, respectively, so the y-axis is
to be absolutely
between the points −2 and 1 on the x-axis, but closer to the 1.
accurate. You
The curve crosses the y-axis when x = 0, i.e. when y = 3(2)(−1)(−7) = 42.
may find it easier
y
to draw the curve
42
first, with the
positive x³ term
–2
1
7 x
determining the
y = 3(x + 2) (x – 1) (x – 7)
shape of the curve,
and then position
the x-axis so that
b To obtain a sketch of the modulus curve, reflect any part of the curve that
the distance
is below the x-axis in the x-axis.
between the
y
first and second
intersections is
y = | 3(x + 2) (x – 1) (x – 7) |
about half that
between the
42
second and third,
since these are
–2
1
7 x
3 and 6 units,
4 EQUATIONS, INEQUALITIES AND GRAPHS
respectively.
4
y
50
–2.9
y = – 100
4
2
3
6.2
4
5
6
–150
–200
From the graph, the solution is x −2.9 or 2.6 x 6.2.
Exercise 4.3
Remember: ( x )
means the positive
square root of x.
1 Where possible, use the substitution x = u ² to solve the following
equations:
b x+2 x =8
a x − 4 x = −4
c x − 2 x = 15
d x + 6 x = −5
2
4
3
c
2
b x −2 x +1= 0
1
Using graphs to solve cubic inequalities
x 3 + 3 x 3 = 10
5
Past-paper questions
1 (i) Sketch the graph of y = |(2x + 3)(2x − 7)|.
(ii) How many values of x satisfy the equation
|(2x + 3)(2x − 7)| = 2x?
2
[4]
[2]
Cambridge O Level Additional Mathematics 4037
Paper 23 Q6 November 2011
2 (i)
Cambridge IGCSE Additional Mathematics 0606
Paper 23 Q6 November 2011
On a grid like the one below, sketch the graph of
y = |(x − 2)(x + 3)| for −5 x 4, and state the coordinates of
the points where the curve meets the coordinate axes.
[4]
y
b
y
2
3 Use the substitution x = u 2 to solve the equation x 3 − 10 x 3 = −9.
4 Using a suitable substitution, solve the following equations:
8 Identify the following cubic graphs:
y
4
4
3
3
2
2
1
O
–3 –2 –1
–1
1
2 Use the substitution x = u 3 to solve the equation x 3 + 3 x 3 = 4 .
a x − 7 x = −12
ii x 3 + 5 x 3 + 4 = 0
i x 3 − 5x 3 + 4 = 0
6 Sketch the following graphs, indicating the points where they cross
the x-axis:
a y = x(x – 2)(x + 2)
b y = |x(x – 2)(x + 2)|
c y = 3(2 x – 1)(x + 1)(x + 3)
d y = |3(2 x – 1)(x + 1)(x + 3)|
7 Solve the following equations graphically. You will need to use graph
paper.
a x(x + 2)(x − 3) 1
b x(x + 2)(x − 3) −1
c (x + 2)(x − 1)(x − 3) > 2
d (x + 2)(x − 1)(x − 3) < −2
a
8 x
7
–100
y = 3(x + 2) (x – 1) (x – 7)
1
2
1
–50
5 a Use the substitution x = u 2 to solve the equation x 4 − 5 x 2 + 4 = 0.
b Using the same substitution, show that the equation x 4 + 5 x 2 + 4 = 0
has no solution.
c Solve where possible:
4
2.6
–1 O
–7 –6 –5 –4 –3 –2
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solutions for assessment
exercises, ESL support
1
1
2
3
x
–2
–2
–1
O
–1
1
x
–5
–4
–3
–2
–1
O
1
2
3
4
x
–2
–3
–3
9 Identify these graphs. (They are the moduli of cubic graphs.)
a
b
y
c
y
y
12
12
12
11
11
11
10
10
10
9
9
9
8
8
8
7
7
7
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
O
–3 –2 –1
–1
1
2
3
4
x
O
–5 –4 –3 –2 –1
–1
1
2
3
4
x –3 –2 –1 O
–1
–2
–2
–2
–3
–3
–3
(ii) Find the coordinates of the stationary point on the curve
y = |(x − 2)(x + 3)|.
[2]
(iii) Given that k is a positive constant, state the set of values of
k for which |(x − 2)(x + 3)| = k has 2 solutions only.
[1]
Cambridge O Level Additional Mathematics 4037
Paper 12 Q8 November 2013
Cambridge IGCSE Additional Mathematics 0606
Paper 12 Q8 November 2013
3 Solve the inequality 9x 2 + 2x − 1 < (x + 1)2 .
[3]
Cambridge O Level Additional Mathematics 4037
Paper 22 Q2 November 2014
Cambridge IGCSE Additional Mathematics 0606
Paper 22 Q2 November 2014
1
2
3
4
x
6
7
This series is endorsed for the Cambridge Pathway to support the syllabuses for
examination from 2025.
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