Solutions to Exercise 7D
1 a
i Write y0 = 2(x0 − 1)2 + 3
y0 − 3
∴
= (x0 − 1)2
2
y0 − 3
0
Choose x = x − 1 and y =
2
∴ x0 = x + 1 and y0 = 2y + 3
In summary,
A dilation of factor 2 from the
x-axis, then a translation of 1 unit
in the positive direction of the
x-axis and 3 units in the positive
direction of the y-axis
ii Write y0 = −(x0 + 1)2 + 2
∴ −y0 + 2 = (x0 + 1)2
Choose x = x0 + 1 and y = −y0 + 2
∴ x0 = x − 1 and y0 = −y + 2
In summary,
A reflection in the x-axis, then a
translation of 1 unit in the negative direction of the x-axis and
2 units in the positive direction of
the y-axis
iii Write y0 = (2x0 + 1)2 − 2
∴ y0 + 2 = (2x0 + 1)2
Choose x = 2x0 + 1 and y = y0 + 2
x−1 x 1
= − and
∴ x0 =
2
2 2
0
y =y−2
In summary,
A dilation of factor 21 from the
y-axis, then a translation of 12 unit
in the negative direction of the
x-axis and 2 units in the negative
direction of the y-axis
b
2
i Write y0 = 0
x +3
y0
1
∴ = 0
2
x +3
y0
Choose x = x0 + 3 and y =
2
∴ x0 = x − 3 and y0 = 2y
In summary,
A dilation of factor 2 from the
x-axis, then a translation of
3 units in the negative direction of
the x-axis
1
ii Write y0 = 0
+2
x +3
0
y −2 1
∴
= x0 + 3
Choose x = x0 + 3 and y = y0 − 2
∴ x0 = x − 3 and y0 = y + 2
In summary,
A translation of 3 units in the
negative direction of the x-axis
and 2 units in the positive
direction of the y-axis
1
−2
−3
1
∴ y0 + 2 = 0
x −3
Choose x = x0 − 3 and y = y0 + 2
∴ x0 = x + 3 and y0 = y − 2
In summary,
A translation of 3 units in the
positive direction of the x-axis
and 2 units in the negative
direction of the y-axis
√
c i Write y0 = √ x0 + 3 + 2
∴ y0 − 2 = x0 + 3
Choose x = x0 + 3 and y = y0 − 2
∴ x0 = x + 3 and y0 = y − 2
In summary,
A translation of 3 units in the
negative direction of the x-axis
and 2 units in the positive
direction of the y-axis
√
ii Write y0 = 2 3x0
iii Write y0 =
x0
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y0 √ 0
= 3x
2
y0
Choose x = 3x0 and y =
2
∴ x0 = 13 x and y0 = 2y
In summary,
A dilation of factor 31 from the
y-axis, then a dilation of factor 2
from the x-axis
y−3 √
= −(x − 4)
2
y−3
Choose −y0 + 6 =
and
2
√
0
x = −(x − 4)
15 − y ∴ (x, y) → −(x − 4),
2
∴
√
iii Write y0 = − x0 + 2
√
∴ −y0 + 2 = x0
Choose x = x0 and y = −y0 + 2
∴ x0 = x and y0 = −y + 2y
In summary,
A reflection in the x-axis, then
a translation of 2 units in the
positive direction of the y-axis
3 a y = 2x + 7 maps to y = 3x + 2
Rewrite as:
y − 7 = 2x and y0 − 2 = 3x0
Therefore we can write:
y0 − 2 = y − 7 and 3x0 = 2x
2
y0 = y − 5 and x0 = x
3
The rule is:
!
2
(x, y) → x, y − 5
3
1
maps to
(x − 2)2
1
y−4
=
3
(x − 5)2
Therefore we can write:
b y+1=
y+7
1
1
and
=
02
5
x
(x − 3)2
y
+
7
Choose y0 =
and x0 = x − 3
5 y + 7
∴ (x, y) → x − 3,
5
2 a Write y0 =
b Write y0 = (x0 )2 and y − 5 = (3x + 2)2
Choose y0 = y − 5 and x0 = 3x + 2
∴ (x, y) → (3x + 2, y − 5)
c Write y0 = (x0 )2 and
−y + 7 = 3(3x + 1)2
−y + 7
Choose y0 =
and x0 = 3x + 1
3
y − 7
∴ (x, y) → 3x + 1, −
3
c y − 4 = (x + 2)2 maps to
y + 5 = (3x − 2)2
Therefore we can write:
y0 + 5 = y − 4 and 3x0 − 2 = x + 2
1
y0 = y − 9 and x0 = (x + 4)
3
The rule is:
!
1
(x, y) → x + 4, y − 9
3
√
y √
x0 and = −(x − 4)
2 √
y
Choose y0 = and x0 = −(x − 4)
2
y
∴ (x, y) → −(x − 4),
2
d Write y0 =
y0 − 4
= y + 1 and x0 − 5 = x − 2
3
y0 = 3y + 7 and x0 = x + 3
The rule is:
(x, y) → (x + 3, 3y + 7)
√
e Write y0 = − x0 + 6 and
d
y √
y √
= 3 − x maps to = x − 6
2
5
Rewrite
as:
0
y y
= and x0 − 6 = 3 − x
2
5
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Therefore we can write:
5y
y0 =
and x0 = 9 − x
2
The rule is:
!
5y
(x, y) → 9 − x,
2
e
y−3 √
y−6 √
= 2 − x maps to
= x
2
−5
Rewrite
as:
y0 − 6 y − 3
=
and x0 = 2 − x
−5
2
Therefore we can write:
−5y + 27
y0 =
and x0 = 2 − x
2
The rule is:
−5y + 27
(x, y) → 2 − x,
2
!
4 a 2(x − a)3 + b
= 2(x3 + 3ax2 + 3a2 x − a3 ) + b
= 2x3 + 6ax2 + 6a2 x − 2a3 + b
= 2x3 − 12x2 + 24x − 13
Therefore a = 2 and b = 3
b The rule is:
(x, y) → (x + 2, y + 3)
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