Chapter review 4

1 a y = x2(x − 2) 6
0 = x2(x − 2) 2 b =1+x
x
So x = 0 or x = 2 6 = x + x2
The curve crosses the x-axis at (2, 0) 0 = x2 + x − 6
and touches it at (0, 0). 0 = (x + 3)(x − 2)
x → ∞, y → ∞ So x = 2 or x = −3
x → − ∞, y → − ∞ Using y = 1 + x:
y = 2x − x2 when x = 2, y = 1 + 2 = 3
= x(2 − x) when x = −3, y = 1 − 3 = −2
As a = −1 is negative, the graph has a So A is (−3, −2) and B is (2, 3).
shape and a maximum point.
0 = x(2 − x) c Substituting the points A and B into
So x = 0 or x = 2 y = x2 + px + q:
The curve crosses the x-axis at (0, 0) A: −2 = 9 − 3p + q (1)
and (2, 0). B: 3 = 4 + 2p + q (2)
(1) − (2):
  −5 = 5 − 5p
p=2
Substituting in (1):
−2 = 9 − 6 + q
q = −5

d y = x2 + 2x − 5
As a = 1 is positive, the graph has a
shape and a minimum point.
b x2(x − 2) = x(2 − x) y = (x + 1)2 − 6
x2(x − 2) − x(2 − x) = 0 So the minimum is at (−1, −6).
x2(x − 2) + x(x − 2) = 0
x(x − 2)(x + 1) = 0 3 a f(2x) is a stretch with scale factor 1
2
So x = 0, x = 2 or x = −1 in the x-direction.
Using y = x(2 − x):
when x = 0, y = 0 × 2 = 0
when x = 2, y = 2 × 0 = 0
when x = −1, y = (−1) × 3 = −3
The points of intersection are (0, 0),
(2, 0) and (−1, −3).

6 1
2 a y= is like y = .
x x
y = 1 + x is a straight line.
 3 , 4 , B(0, 0)
A 2
The asymptote is y = 2.

1 1
b 2f(x) is a stretch with scale factor 2

in the y-direction.

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3 b e A(6, 4), B(3, 0)
The asymptote is y = 2.

0
f f(x) + 1 is a translation by   ,
1
or one unit up.

A(3, 2), B(0, 0)

The asymptote is y = 1.

 0
c f(x) − 2 is a translation by   ,
 2 
or two units down.
A(3, 5), B(0, 1)
The asymptote is y = 3.

4 2 = 5 + 2x − x2
2
x − 2x − 3 = 0
(x − 3)(x + 1) = 0
So x = −1 or x = 3
The points of intersection
are A(−1, 2), B(3, 2).
A(3, 2), B(0, −2)
The asymptote is y = 0.
5 a f(−x) is a reflection in the y-axis.
 3 
d f(x + 3) is a translation by   ,
0
or three units to the left.

A(0, 4), B(−3, 0) b −f(x) is a reflection in the x-axis.

The asymptote is y = 2.

 3
e f(x − 3) is a translation by   ,
0
or three units to the right.

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6 a Let y = a(x − p)(x − q). 7 c f(x) = (x − 1)(x − 2)(x + 1)
Since (1, 0) and (3, 0) are on the curve then 0 = (x − 1)(x − 2)(x + 1)
p = 1 and q = 3. So x = 1, x = 2 or x = −1
So y = a(x − 1)(x − 3) The curve crosses the x-axis at (1, 0),
Using (2, −1): (2, 0) and (−1, 0).
−1 = a(1)(−1) y = f(x + b) is a translation b units
a=1 to the left.
So y = (x − 1)(x − 3) = x2 − 4x + 3 For the point (0, 0) to lie on the translated
curve, either the point (1, 0), (2, 0) or
b i f(x + 2) = (x + 1)(x − 1), or a translation (−1, 0) has translated to the point (1, 0).
 2  For the coordinate (1, 0) to be translated to
by   , or two units to the left. (0, 0), b = 1.
 0 For the coordinate (2, 0) to be translated to
(0, 0), b = 2.
For the coordinate (−1, 0) to be translated to
(0, 0), b = −1.
b = −1, b = 1 or b = 2

8 a i y = f(3x) is a stretch with scale factor 1

3

in the x-direction. Find 13 of the

x-coordinate.
P is transformed to  43 ,3 .

ii f(2x) = (2x − 1)(2x − 3), or a stretch with ii 1

2 y = f(x)
scale factor 12 in the x-direction. y = 2f(x), which is a stretch with scale
factor 2 in the y-direction.
P is transformed to (4, 6).

5
iii y = f(x − 5) is a translation by   ,
0
or five units to the right.
P is transformed to (9, 3).

iv −y = f(x)
y = −f(x), which is a reflection of the
curve in the x-axis.
(4, −3)
7 a f(x) = (x − 1)(x − 2)(x + 1)
When x = 0, y = (−1) × (−2) × 1 = 2 v 2(y + 2) = f(x)
So the curve crosses the y-axis at (0, 2).
y = 12 f(x) − 2, which is a stretch with
1
b y = af(x) is a stretch with scale factor a scale factor 2 in the y-direction and
in the y-direction.  0
The y-coordinate has multiplied by −2, then a translation by   , or two
therefore y = −2f(x).  2 
a = −2 units down.
P is transformed to  4,  12  .

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8 b P(4, 3) is transformed to (2, 3). c y = (x − k)3 − 6(x − k)2 +9(x − k) is a
Either the x-coordinate has halved, which translation of the curve y = x3 − 6x2 + 9x by
is a stretch with scale factor 12 in the k 
x-direction, or it has had 2 subtracted from   , or k units to the right.
0
 2 
it, which is a translation by   , or two
 0 For the point (−4, 0) to lie on the translated
units to the left. curve, either the point (0, 0) or (3, 0) has
translated to the point (−4, 0).
So the transformation is y = f(2x) or For the coordinate (0, 0) to be translated to
y = f(x + 2). (−4, 0), k = −4.
For the coordinate (3, 0) to be translated to
c i P(4, 3) is translated to the point (8, 6). (−4, 0), k = −7.
The x-coordinate of P has 4 added to it k = −4 or k = −7
and the y-coordinate has 3 added to it.
y = f(x − 4) + 3 10 a y = x(x − 2)2
0 = x(x − 2)2
ii P(4, 3) is stretched to the point (8, 6). So x = 0 or x = 2
The x-coordinate of P has doubled and The curve crosses the x-axis at (0, 0)
the y-coordinate has doubled. and touches it at (2, 0).
x → ∞, y → ∞
y = 2f  12 x 
x → − ∞, y → − ∞
   

9 a x3 − 6x2 + 9x
= x(x2 − 6x + 9)
= x(x − 3)2

b y = x(x − 3)2
0 = x(x − 3)2
So x = 0 or x = 3
The curve crosses the x-axis at (0, 0)
and touches it at (3, 0).
x → ∞, y → ∞
x → − ∞, y → − ∞
   

 3 
b y = f(x + 3) is a translation by vector  
0
of y = f(x), or three units to the left.
So the curve crosses the x-axis at (−3, 0)
and touches it at (−1, 0).
When x = 3, f(x) = 3(3 − 2)2 = 3
So f(x + 3) crosses the y-axis at (0, 3).

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  0 Challenge
11 a y = f(x) − 2 is a translation by   , or
 2  R(6, −4)
two units down.
 c 
y = f(x + c) − d is a translation by   ,
0
or c units to the left and a translation by
 0 
  , or d units down.
 d 
So R is transformed to (6 − c, −4 − d).

The horizontal asymptote is y = −2.

The vertical asymptote is x = 0.

b From the sketch, the curve crosses

the x-axis.

y = f(x) − 2
1
= 2
x
1
0 = 2
x
x= 2 1

So the curve cuts the x-axis at  12 ,0 .

 3 
c y = f(x + 3) is a translation by   ,
0
or three units to the left.

d The horizontal asymptote is y = 0.

The vertical asymptote is x = −3.
y = f(x + 3)
1
=
x3
When x = 0, y = 13
So the curve cuts the y-axis at  0, 13  .

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