year 8 maths

Chapter summary
455
y
y = 2x − 1
−4
−3 (2, 3)
−2
−1
x
O 1 2 3
To graphically solve 2x − 1 = 3:
• Locate point with y = 3.
• x-coordinate is solution,
i.e. x = 2.
y
O
Point of
intersection
x
The point of intersection is the only
solution that satisfies both linear
equations.
Cartesian plane
(number plane)
y
(−1, 3) 3
(2, 3)
2
(−3, 0)
1
x
−3 −2 −1 O
−1
1 2 3
−2 origin
(−2, −2)
−3 (0, −3) (0, 0)
Linear
relationships 1
Rules, tables and graphs
y = −2x + 3
x −2 −1 0 1 2
y 7 5 3 1 −1
y
7
6
5
4
3
2
1
−3 −2 −1O
−1
1 2 3
−2
Finding the rule from
tables
x −2 −1 0 1 2
y −8 −5 −2 1 4
x
Chapter summary
Non-linear graphs
y = x 2 − 2
x −2 −1 0 1 2
y 2 −1 −2 −1 2
y
3
2
1
−2 −1 O
x
−1
1 2
−2 A parabola
y = mx + b
Gradient
y
y-intercept
5 (2, 5)
4
3
2
1
(0, 1)
x
−1 O
−1 1 2 3
4 m =
b = 1
y = 2x + 1
+3 +3 +3 +3
y = × x +
= 3x + (−2)
= 3x − 2
Applications
• Distance increases by
20 km per hour.
t 0 1 2 3
d 0 20 40 60
d = 20t
• Volume decreases from
1000 L by 200 L per minute.
t 0 1 2 3 4 5
V 1000 800 600 400 200 0
V = −200t + 1000
2 = 2 −
y = 2x + 6
0 = 2x + 6
−6 = 2x
−3 = x
x-intercept
y = 0
y
(0, 6)
(−3, 0)
O
x
Gradient
y
rise 8 =
Negative
run 6 (0, 8) gradient
4 = −
3
(6, 0)
x
O
O
y
Zero
gradient
x
Undefined
gradient
Cambridge Maths NSW
Stage 4 Year 8 Second edition
ISBN 978-1-108-46627-1 © Palmer et al. 2018
Cambridge University Press
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