C3:Transformations of graphs and the modulus function

C3:Transformations of graphs and the modulus function
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CHAPTER 2
C3:Transformations of
graphs and the modulus function
Learning objectives
After studying this chapter, you should be able to:
■ transform simple graphs to produce other graphs
■ understand the effect of composite transformations on equations of curves and describe them
geometrically
■ understand what is meant by a modulus function
■ sketch graphs of functions involving modulus functions
■ solve equations and inequalities involving modulus functions.
2.1 Review of simple transformations
of graphs
Some simple transformations of graphs were introduced in
chapter 5 of C2. The basic results are reviewed below. For
instance, the graph of y x2 can be transformed into the graph
3
.
of y (x 3)2 4 by a translation of
4
In general, a translation of
a
transforms the graph of
b
Although you may use a graphics
calculator to draw graphs, it is
important to see how the graph
of one curve can be obtained
from the graph of a simpler
curve using a sequence of
transformations.
y
y 2 x3
y f(x) into the graph of y f(x a) b.
You learnt how to find the equations of new curves after a
reflection in one of the coordinate axes.
For example, the graph of y 2 x3 is sketched opposite.
After reflection in the x-axis, the new curve will have equation
y 2 x3. The general result is given below.
The graph of y f(x) is transformed into the graph of
y f(x) by a reflection in the line y 0 (the x-axis).
When the curve y 2 x3 is reflected in the y-axis, the new
curve has equation y 2 x3.
The graph of y f(x) is transformed into the graph of
y f(x) by a reflection in the line x 0 (the y-axis).
2
O
x
2
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C3: Transformations of graphs and the modulus function
The graph of y sin x is transformed into the graph of y 3 sin x
by a stretch of scale factor 3 in the y-direction.
y
y
3
1
x
1
x
3
The general result is:
The graph of y f(x) is transformed into the graph of
y d f(x) by a stretch of scale factor d in the y-direction.
A stretch of scale factor 2 in the x-direction transforms the
x
graph of y sin x into the graph with equation y sin .
2
y
y
1
1
x
1
1
x
The graph of y f(x) is transformed into the graph of
x
y f by a stretch of scale factor c in the x-direction.
c
Worked example 2.1
Find the equation of the resulting curve when the curve
y 2 tan x is transformed by:
(a) a reflection in the y-axis,
(b) a translation of
3
5
,
(c) a stretch of scale factor 0.5 in the x-direction.
You are transforming the original
curve in each case. Successive
transformations will be discussed
in the next section.
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C3: Transformations of graphs and the modulus function
25
Solution
(a) The graph of y f(x) is transformed into the graph of
y f(x) by a reflection in the y-axis.
Hence the new curve has equation y 2 tan (x).
However, since tan(x) tan x, the equation of the new
curve can be written as y 2 tan x.
a
transforms the graph of
(b) Recall that a translation of
b
y f(x) into the graph of y f(x a) b.
2
After translation through the vector
3
5
, the curve
y 2 tan x has equation y 2 tan x 5
3
or
y 7 tan x .
3
(c) The graph of y f(x) is transformed into the graph of
x
y f by a stretch of scale factor c in the x-direction.
c
x
Hence y 2 tan x is transformed into y 2 tan 0.5
or y 2 tan 2x.
Worked example 2.2
Describe geometrically how the first curve is transformed into
the second curve in each of the following cases:
(a) y x2,
y (x 3)2 2
(b) y cos x, y cos 3x
(c) y 2x,
y 2x
Solution
(a) The first curve has been translated through the vector
to give the second curve.
3
2
(b) A one-way stretch in the x-direction has taken place. The
scale factor is 13.
(c) The curve has been reflected in the y-axis (or x 0).
EXERCISE 2A
1 Find the equation of the resulting curve after each of the
2
following has been translated through
.
1
(b) y x2 4x 5
(a) y x3
(c) y tan x
(d) y 3x
Notice that x has been divided by
1
to give 3x.
3
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C3: Transformations of graphs and the modulus function
2 Find the equation of the resulting curve after each of the
following has been stretched in the x-direction by scale factor 12.
(a) y x3
(b) y x2 4x 5
(c) y tan x
(d) y 3x
3 Describe geometrically how the curve y 1 sin x is
transformed into the following curves:
(a) y sin x
(b) y 1 sin x
(c) y 5 sin (x 2)
(d) y 4 4 sin x
(e) y 1 sin x
(f) y 1 sin 5x
4 Describe geometrically how the curve y 2x is transformed
into the following curves:
x
(a) y 25x
(b) y 2x 3
(c) y 23
(d) y 2x 7
(e) y 2x
5 Describe a geometrical transformation which maps the graph
of y 3x onto:
5x
(a) y 3x
(b) y 3x 4
(c) y 3 2
(d) y 2 3x
(e) y 4 3x
(f) y 9x
2.2 Composite transformations
You can perform each of the transformations described above in
sequence so as to produce a composite transformation.
Worked example 2.3
Find the equation of the resulting curve when the following
transformations are performed in sequence on the curve with
equation y x3:
(a) a stretch by a factor 2 in the y-direction,
1
,
(b) a translation through
3
(c) a reflection in the x-axis.
Solution
(a) After a stretch by a factor 2 in the y-direction, the curve
y x3 becomes the curve y 2x3.
(b) Applying a translation of
1
3 to the curve y 2x
the new curve y 2(x 1)3 3.
3
gives
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C3: Transformations of graphs and the modulus function
(c) Finally the effect of a reflection in the x-axis is to transform
y f(x) into the curve y f(x).
The curve y 2(x 1)3 3 therefore becomes the curve
with equation y 3 2(x 1)3.
It is important to perform the transformations in the given
order or you will not obtain the correct final equation.
2
Worked example 2.4
Describe geometrically a sequence of transformations that
transforms y sin x into y 2 5 sin 3x.
Solution
The curve y sin x is stretched by scale factor 13 in the x-direction
to give the curve y sin 3x.
Next the curve y sin 3x is stretched by scale factor 5 in the
y-direction to give y 5 sin 3x.
0
to give the final
Finally y 5 sin 3x is translated through
2
curve with equation y 2 5 sin 3x.
EXERCISE 2B
1 Describe a sequence of transformations that would map the
graph of y x3 onto the graph of y 3(x 5)3.
2 Find the resulting curve when the following transformations
are applied in sequence to the curve y cos x.
(a) a reflection in the x-axis,
0
(b) a translation through
,
2
(c) a stretch by a scale factor 3 in the x-direction.
3 Find the equation of the resulting graph when the graph of
1
then reflected in the y-axis.
y 3x is translated through
5
4 Find the equation of the new curve when the y sin x is
transformed by the following sequence of transformations:
,
(a) a translation through
0
(b) a reflection in the x-axis,
(c) a stretch by a scale factor 4 in the y-direction.
5 Express 2x2 12x 19 in the form 2(x a)2 b. Hence
decribe geometrically how the graph of y x2 can be
transformed into the graph of y 2x2 12x 19.
27
The two stretches could have
been done in any order to map
y sin x onto y 5 sin 3x.
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C3: Transformations of graphs and the modulus function
6 Describe geometrically how the first curve can be
transformed into the second curve by a sequence of
transformations:
(a) y x2, y 4(x 2)2
(b) y x2, y 4 3(x 1)2
(c) y x3, y (2x 1)3
(d) y x3, y (x 3)3
5
x
(e) y x4, y (3x 5)4
(f) y x5, y 4 2
3
7 Describe geometrically how the curve y 3x2 5 can be
transformed into the curve y x2 by a sequence of
transformations.
8 Describe geometrically how the curve y 5 sin(x 3) can be
transformed into the curve y sin(x 1) by a sequence of
transformations.
9 Describe geometrically how the curve y 3 cos 2x can be
transformed into the curve y 5 cos x by a sequence of
transformations.
10 Describe geometrically how the curve y 4x 3 can be
2x
transformed into the curve y by a sequence of
5
stretches.
11 (a) Describe the geometrical transformation that
transforms the graph of y 3x into the graph of 3y x.
(b) Find the equations of the new graphs after each of the
folowing has been reflected in the line y x:
(i) y 3x 2,
(ii) y x2, (iii) (x 1)2 y2 4.
12 The graph of y f(x) is reflected in the x-axis and then the
y-axis to produce the graph with equation y g(x).
(a) Find g(x) in terms of f and x.
(b) Describe geometrically the single transformation that
maps the graph of y f(x) onto the graph of y g(x).
13 (a) Find the new equation resulting from reflecting the
curve y x2 in the line y 1.
(b) Describe a sequence of simple geometrical transformations
that will map y x2 onto your answer to (a).
14 (a) Find the new equation resulting from reflecting the
curve y 2x in the line x 5.
(b) Describe a sequence of simple geometrical transformations
that will map y 2x onto your answer to (a).
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29
15 (a) Describe a sequence of geometrical transformations
that will transform the graph of y f(x) into the graph
of y 6 f(x).
(b) Describe a single geometrical transformation that will
transform the graph of y f(x) into the graph of
y 6 f(x).
2
(c) Describe a single geometrical transformation that will
transform the graph of y f(x) into the graph of
y 2p f(x).
16 (a) Describe a sequence of geometrical transformations
that will transform the graph of y f(x) into the graph
of y f(4 x).
(b) Describe a single geometrical transformation that will
transform the graph of y f(x) into the graph of
y f(4 x).
(c) Describe a single geometrical transformation that will
transform the graph of y f(x) into the graph of
y f(2q x).
2.3 Modulus function
The modulus function finds the absolute value of a number. Any
negative sign in front of a number is disregarded and a positive
answer is returned. Consider the function box below.
input
Take the modulus
Often the key on a calculator
which finds the modulus is
denoted by ABS since it finds the
absolute value of a number.
output
When the value 3 is input then the output is 3, whereas when 7
is input the output is 7. An input of zero gives an output of zero.
y
The modulus of x is written as |x| and is usually read as ‘mod x’.
By taking a set of values of x it is easy to see that the graph of
y |x| would have the appearance of the V shape below.
y
3
2
1
3
2
1
O
1
2
3
x
O
x
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C3: Transformations of graphs and the modulus function
The modulus function is actually defined as follows:
 x when x 0
|x| 
x when x 0
Worked example 2.5
Sketch the graph of y |x 3|.
Solution
Method 1
When x 3, |x 3| x 3 since x 3 0.
y
3
O
x
graph of
yx3
3
So the graph for x 3 is the same as the graph of y x 3.
However when x 3, x 3 is negative and so |x 3| (x 3).
y
3
graph of
y3x
O
3
x
This means that when x 3, the graph of y |x 3| is the same
as the graph of y (x 3) 3 x.
The graph of y |x 3| is therefore as drawn below.
y
4
3
2
1
3 2 1 O
1
2
3
4
5 x
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Method 2
An alternative approach is to draw the graph of y x 3 and
then to reflect the section of the graph that lies below the x-axis
in the x-axis.
y
2
4
3
2
1
3 2 1 O
1
1
2
3
4
5 x
2
3
4
Worked example 2.6
Sketch the graphs of
y
(a) y |x 4| for 3 x 3,
8
2
(b) y |sin x| for 0 x 2.
4
Solution
2
(a) Draw the graph of y x2 4.
O
Now reflect in the x-axis all the parts of the graph which lie
below the x-axis.
2
x
4
The resulting graph is that of y |x2 4|.
y
There are now some ‘sharp’
corners on the graph called
cusps. Do not be tempted to
smooth these out.
4
2
O
2
x
(b) Draw the graph of y sin x for 0 x 2.
Reflect in the x-axis the portion of the graph
where y is negative so as to produce the
graph of y |sin x|.
y
1
O
π
2π
x
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C3: Transformations of graphs and the modulus function
Worked example 2.7
Sketch the graphs of:
(a) y 3 |x|,
(b) y |x 1| |x 2|.
Solution
(a) For x 0 , the graph is identical to that of y 3 x.
When x 0, y 3.
For x 0 , the graph is identical to that of y 3 (x) 3 x.
Hence we can sketch the graph of y 3 |x|.
Alternatively, the graph of
y 3 | x | can be obtained from
the graph of y | x | by a
reflection in the x-axis followed
0
by a translation of
.
3
y
3
3
O
3
x
(b) It is necessary to consider three separate intervals in order
to sketch y |x 1| |x 2|.
Firstly, when x 1,
|x 1| (x 1) x 1 and also
|x 2| (x 2) x 2.
Therefore y |x 1| |x 2| x 1 (x 2) 3 2x.
The graph is identical to y 3 2x for x 1.
Next, when 1 x 2, |x 1| x 1.
But
|x 2| (x 2) x 2.
Therefore y |x 1| |x 2| x 1 (x 2) 1.
The graph is identical to y 1 for 1 x 2.
Finally, when x 2
|x 1| x 1 and also
|x 2| x 2.
Therefore y |x 1| |x 2| x 1 x 2 2x 3.
The graph is identical to y 2x 3 for x 2.
The graph of y |x 1| |x 2| can now be sketched.
The end points of these intervals
were not included but serve as
useful check points.
y
3
When x 1, y 0 1 1.
2
When x 2, y 1 0 1.
1
1
O
1
2
3
x
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33
EXERCISE 2C
Sketch the graph of each of the following, showing the values of
any intercepts on the axes.
1 y |3x|
2 y |x 4|
3 y |3x 5|
4 y |5 2x|
5 y |x| 5
6 y 4 |x|
2
7 y |x 1|
8 y |(x 5)(x 2)|
9 y |x| |x 3|
10 y |x| |x 3|
11 y |(x 1)(x 2)(x 3)| 12 y |x2 5| 4
13 y |x| |x 1| |x 2| 14 y sin|x|, 2 x 2
15 y |cos x|, 2 x 2
16 y |tan x|, 0 x 17 y |cos 3x|, x 18 y 1 |sin 2x|, x 2
2.4 Equations involving modulus
functions
Often the graphical approach is best since you can see the
approximate solutions and how many solutions to expect.
Worked example 2.8
Solve the equation |3 2x| x 1.
Solution
y
Method 1
Firstly, sketch the graph of y |3 2x|. This is the
V-shaped graph. The sections have gradients 2.
5
Now add the straight line y x 1, which has
gradient 1. Hence there are two points of intersection.
3
This means the equation |3 2x| x 1 will have
two solutions, and from the graph one of these is less
than 1.5 and the other is greater than 1.5.
3
When x , |3 2x| 3 2x.
2
One solution is given by 3 2x x 1
4
⇒ 3 1 2x x ⇒ x .
3
3
Similarly, when x , |3 2x| (3 2x).
2
The second solution is given by (3 2x) x 1
⇒ 3 2x x 1 ⇒ 2x x 3 1 ⇒ x 2.
4
The two solutions are x 2, .
3
4
2
1
1 O
1
2
3
4
5
6
x
1
3
This solution is valid since x .
2
3
This solution is valid since x .
2
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Method 2
Square both sides of the equation |3 2x| x 1.
(|3 2x|)2 (x 1)2 ⇒ 9 12x 4x2 x2 2x 1.
⇒ 3x2 10x 8 0 ⇒ (3x 4)(x 2) 0.
4
The two solutions are x 2, .
3
It is always a little dangerous
when you square both sides of
an equation, since you may
produce spurious answers. You
should therefore check that
these solutions satisfy the
original equation. In this case the
solutions are valid.
Worked example 2.9
Solve the equation |x 2| 3.
Solution
An approach involving squaring both sides yields
(|x 2|)2 (3)2 ⇒ x2 4x 4 9 ⇒ x2 4x 5 0.
⇒ (x 5)(x 1) 0 ⇒ x 5, x 1.
Checking each of these values in the original equation shows
that neither of the answers are correct solutions.
When x 5, |x 2| |5 2| 3.
When x 1, |x 2| |(1)2| |3| 3.
y
4
3
2
1
2
1 O
1
2
3
4
5
x
1
2
3
The graph of y |x 2| is never negative and so the equation
|x 2| 3 has no solutions.
Can you spot the flaw in the following argument?
For solutions to |x 2| 3, solve x 2 3 ⇒ x 2 3 1.
Also solve (x 2) 3 ⇒ x 2 3 ⇒ x 5.
Hence the solutions are x 1 and x 5.
A sketch can easily reveal when
no solutions actually exist since
the graphs of y | x 2| and
y 3 do not intersect.
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Worked example 2.10
35
y
Solve the equation |x 1| 6x.
2
Solution
A sketch showing the graphs of y |x2 1| and y 6x enables
you to see that there are two solutions, one in the interval
0 x 1 and the other satisfying x 1.
For 0 x 1, you can write |x2 1| (x2 1) x2 1
Hence, solving
36 4
6 x2 1 6x ⇒ 0 x2 6x 1 ⇒ x 2
40
⇒ x 3 3 10
.
2
But the only valid solution in the interval 0 x 1 is
0.162… . You must reject the solution that is
x 3 10
negative.
For x 1, you can write |x2 1| x2 1
4
6 36
Hence, solving x2 1 6x ⇒ 0 x2 6x 1 ⇒ x 2
40
⇒ x 3 3 10
. Once again, rejecting the negative
2
6.162… .
solution gives the solution x 3 10
The two solutions are
x 3 10
0.162…
6.162… .
x 3 10
EXERCISE 2D
Solve the following equations. In some cases there are no solutions.
1 |x 3| 2
2 |4 x| 5
3 |x 2| 7
4 |x 1| x
5 |2 x| x 1
6 |2x 3| x 1
7 |2x 3| 2
8 |4 3x| x
9 |x2 2| 3x
10 |2x2 3x| 1
11 |4 3x2| x
12 |x2 2x| 3x 2
13 |x2 x| 2 0
14 |4x2 x| 3
15 |4x2 3| x
2.5 Inequalities involving modulus
functions
It is advisable to draw a sketch and then to find the critical points
when trying to solve inequalities involving modulus functions.
2
1
1
O
1
2
Using the quadratic equation
b ac
b2 4 formula x .
2a
x
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C3: Transformations of graphs and the modulus function
Worked example 2.11
Solve the inequality |x 2| 2x 3.
y
Solution
The graphs y |x 2| and y 2x 3 are drawn opposite and
there is a single point of intersection when x is negative.
3
2
Since x 2 for this point of intersection, you can write
1
|x 2| (x 2) 2 x.
2
Solving 2 x 2x 3 gives 1 3x. Hence x 13.
This is the critical point and by looking at the graph, the y-value
for the graph of y |x 2| is less than the y-value on the graph
of y 2x 3 whenever you are to the right-hand side of this
critical point.
1
O
1
2
x
You can check this solution
by taking a test value. For
example when x 0, | x 2 | 2
and 2x 3 3. Since 2 3, it
gives a check on the solution.
The solution is therefore x 13.
Worked example 2.12
Solve the inequality |x2 3x| 2.
Solution
The graph of y x2 3x x(x 3) is a parabola cutting
the x-axis at (0, 0) and (3, 0). The graph of y |x2 3x| is
sketched opposite together with the straight line y 2.
y
4
There are four points of intersection and hence four
critical points.
2
Two of these are given by solving
x2 3x 2 ⇒ x2 3x 2 0.
This equation must be solved using the formula
3
1
1 O
b 2 4
8
b ac
3 9
x . Hence x .
2a
2
The approximate values of x are 3.56 … and 0.56… .
The other two critical points are given by solving
(x2 3x) 2 ⇒ x2 3x 2 0 ⇒ (x 1)(x 2) 0.
Therefore x 1, x 2.
By considering the graph, the intervals satisfying |x2 3x| 2 are
3 17
3 17
x , 1 x 2, x .
2
2
1
2
3
4
5
The four critical points in
ascending order are
3 17
,
2
1,
2,
3 17
.
2
x
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37
EXERCISE 2E
Solve the following inequalities:
1 |x 5| 2
2 |4 x| 1
3 |x 3| 7
4 |x 2| x
5 |3 x| x 4
6 |2x 3| x 2
7 |2x 5| 2 x
8 |2 5x| x
9 |x2 4| 5x
10 |2x2 x| 1
2
11 |7 3x2| 2x 1
12 |x2 2x| 7x 6
Worked examination question
The function f is defined for all real values of x by
f(x) |x| |x 3|.
(a) For values of x such that x 0 , show that f(x) 3 2x.
(b) Write down expressions for f(x) in a form not involving
modulus signs for each of the intervals:
(i) x 3;
(ii) 0 x 3.
(c) Sketch the graph of f and write down the equation of its
line of symmetry.
(d) State the range of f.
(e) Solve the equation f(x) 4.
(f) Explain whether it is possible to find an inverse of the
function f.
[A]
Solution
(a) When x 0 , |x| x and |x 3| (x 3) 3 x.
Hence f(x) x 3 x 3 2x.
(b) (i) When x 3, |x| x and |x 3| x 3.
Hence f(x) x x 3 2x 3.
(ii) When 0 x 3, |x| x and |x 3| (x 3) 3 x.
Hence f(x) x 3 x 3.
(c) The graph consists of the three sections defined above
and is sketched opposite.
3
The line of symmetry has equation x .
2
(d) Since the function has least value equal to 3, the range
is given by f(x) 3.
(e) From the graph, the equation f(x) 4 has two solutions.
One of these is when x 0 ⇒ 3 2x 4 ⇒ x 12.
The other is when
x 3 ⇒ 2x 3 4 ⇒ x 312.
(f) Since the graph of f is many-one, it does NOT have an
inverse.
y
5
4
3
2
1
1 O
1
2
3
4
5
x
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C3: Transformations of graphs and the modulus function
MIXED EXERCISE
1 The function f is defined for all real values of x by
f(x) |2x 3| 1.
(a) Sketch the graph of y f(x). Indicate the coordinates of
the points where the graph crosses the coordinate axes.
(b) State the range of y f(x).
(c) Find the values of x for which f(x) x.
[A]
2 Sketch the graphs of y |2x 3|and y |2x 5| on the
same axes. Hence solve the inequality |2x 3| |2x 5|.
3 The function f is defined by f(x) |x 3|, x .
(a) Sketch the graph of y f(x).
(b) Solve the inequality |x 3| 12x.
[A]
4 The functions f and g are defined for all real values of x by
f(x) x2 10 and g(x) |x 2|:
(a) Show that ff(x) x4 20x2 90.
Find all the values of x for which ff(x) 26.
(b) Show that gf(x) |x2 12|. Sketch the graph of
y gf(x). Hence or otherwise, solve the equation
gf(x) x.
[A]
5 (a) Determine the two values of x for which
|2x 3| |5 x|.
y
(b) The function f is defined for all real values of x. The graph
of y |f(x)| is sketched opposite. Sketch two possible
graphs of y f(x) on separate axes.
[A]
6 (a) Sketch the graphs of
(i)i y x2 6x 5;
(ii) y |x2 6x 5|.
(b) Calculate the four roots of the equation |x2 6x 5| 3,
expressing the irrational solutions in surd form.
(c) Using this result and the sketch to (a) (ii), or otherwise,
solve the inequality
|x2 6x 5| 3.
[A]
O
x
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C3: Transformations of graphs and the modulus function
39
7 The diagram shows a sketch of the curve with equation
y sin 2x for x .
2
2
y
2
1
π2
O
π
2
x
1
(a) Draw on the same diagram sketches of the graphs with
equations y |x| and y |sin 2x| for x .
2
2
(b) Hence state the number of times the graph of the curve
with equation y |sin 2x| |x| intersects the x-axis in
[A]
the interval x .
2
2
8 (a) Sketch the graph of y |2x 4|. Indicate the
coordinates of the points where the graph meets the
coordinate axes.
(b) (i) The line y x intersects the graph of y |2x 4| at
two points P and Q. Find the x-coordinates of the
points P and Q.
(ii) Hence solve the inequality |2x 4| x.
(c) The graph of y |2x 4| k touches the line y x at
only one point. Find the value of the constant k.
[A]
9 A function f is defined for all real values of x by
f(x) 3 |2x 1|.
(a) (i) Sketch the graph of y f(x). Indicate the
coordinates of the points where the graph crosses
the coordinate axes.
(ii) Hence show that the equation f(x) 4 has no real
roots.
(b) State the range of f.
(c) By finding the values of x for which f(x) x, solve the
inequality f(x) x.
[A]
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C3: Transformations of graphs and the modulus function
Key point summary
1 A translation of
b transforms the graph of y f(x)
a
p23
into the graph of y f(x a) b.
2 The graph of y f(x) is transformed into the graph of
y f(x) by a reflection in the line y 0 (the x-axis).
p23
3 The graph of y f(x) is transformed into the graph of
y f(x) by a reflection in the line x 0 (the y-axis).
p23
4 The graph of y f(x) is transformed into the graph of p24
y d f(x) by a stretch of scale factor d in the y-direction.
5 The graph of y f(x) is transformed into the graph of p24
x
y f by a stretch of scale factor c in the x-direction.
c
6 The modulus function |x| is defined by
x when x 0.
|x|
x when x 0.
p30
Test yourself
1 Describe geometrically how the curve:
What to review
Section 2.1
(a) y x5 is transformed into y (x 1)5 3,
(b) y tan x is transformed into y tan 4x.
2 Describe geometrically a sequence of transformations that
transforms y x2 into y 3(x 1)2 4.
Section 2.2
3 Describe geometrically a sequence of transformations that
transforms y sin x into y 3 2 sin 4x.
Section 2.2
4 The function f is defined for all values of x by
Section 2.3
f(x) |x 5|.
(a) Express f(x) in a form not involving modulus signs
when x 5.
(b) Sketch the graph of y f(x), indicating any values where
the graph meets the axes.
5 (a) Sketch the graph of y 4 x2.
Section 2.3
(b) Hence sketch the graph of y |4 x2|.
(c) State the number of roots of the equation |4 x2| 1.
6 Solve the equation |3x 5| 2 x.
Section 2.4
7 Solve the inequality|2x 7| 3 x.
Section 2.5
2
1 (a) Translation through
13;
(b) Stretch in x-direction with scale factor 41.
2 Translation of
01, followed by stretch in y-direction with scale factor 3,
followed by translation of
40.
3 Stretch in x-direction with scale factor 41 and stretch in y-direction with
scale factor 2, followed by translation of
30.
4 (a) f(x) 5 x, when x 5;
(b)
y
5
O
5 (a)
5
x
(b)
y
4
2
O
2
y
4
2
x
O
2
x
(c) 4 roots.
6 x 121, 143.
7 131 x 10.
Test yourself
ANSWERS
C3: Transformations of graphs and the modulus function
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41
Page 41
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