CAIE AS & A Level Mathematics (9709)
Pure Mathematics 1 (P1)
Chapter 1.2: Functions
Comprehensive, Exam-Focused Lecture Notes & Study Guide
Contents
1 Key Definitions
1.1 Types of Mappings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
2 Domain and Range (Mastering Bounded Outputs)
2.1 The 3-Step Range Calculator Checklist . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Standard Syllabus Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Case 1: f (x) = x1 for x ≥ 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 Case 2: g(x) = x2 + 1 for x ∈ R . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Exam-Realistic Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
3
3
3
3
5
3 Composite Functions
3.1 Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
8
4 Inverse Functions
4.1 Key Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Syllabus Case Study: Finding h−1 (x) for Quadratic Curves . . . . . . . . . . . . . . .
4.3 Further Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
11
11
13
5 Graphical Relationship Between f and f −1
5.1 Annotated Graphs of f and f −1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 How to Sketch the Inverse Graph in the Exam . . . . . . . . . . . . . . . . . . . . . .
15
15
15
6 Transformations of Graphs (Unlocking the Mathematical Logic)
6.1 The Vertical World vs. The Horizontal World Paradigm . . . . . . . . . . . . . . . . .
6.2 Combined Transformations: Why Order Matters . . . . . . . . . . . . . . . . . . . .
6.2.1 Vertical Combinations: y = af (x) + b . . . . . . . . . . . . . . . . . . . . . . .
6.2.2 Horizontal Combinations: y = f (ax + b) . . . . . . . . . . . . . . . . . . . . . .
6.3 Exam-Realistic Worked Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4 Visualizing Transformations with TikZ . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.1 Set 1: Translations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.4.2 Set 2: Stretches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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7 Common Exam Pitfalls and Tips
21
8 Summary Checklist
22
1
1
Key Definitions
To excel in CAIE 9709 Functions questions, you must grasp the formal language used by examiners. A clear understanding of the difference between a relation and a function is crucial.
Fundamental Definitions
• Mapping (or Relation): Any rule that associates elements of an input set (Set X)
with elements of an output set (Set Y ).
• Function: A special type of mapping in which every input element x in the domain
maps to exactly one output element y in the codomain.
• Domain: The complete set of possible input values (usually x-values) for which the
function is defined.
• Range: The complete set of actual output values (usually f (x) or y-values) generated
by applying the function to every element in its domain.
1.1
Types of Mappings
There are four logical classifications of mappings, but only two qualify as functions:
1. One-to-One (1-1) Function: Each input maps to a unique output, and no two inputs map
to the same output. If x1 6= x2 , then f (x1 ) 6= f (x2 ).
2. Many-to-One Function: At least two different inputs map to the same output. For example, f (x) = x2 is many-to-one because both x = −2 and x = 2 map to y = 4.
3. One-to-Many Mapping (Not a function): One input maps to multiple outputs (e.g., y 2 =
√
x =⇒ y = ± x).
4. Many-to-Many Mapping (Not a function): Multiple inputs map to multiple outputs.
Domain X
Range Y
Domain X
a
1
a
Range Y
1
b
2
b
3
c
2
c
Many-to-One Function
One-to-One Function
Figure 1: Mapping diagrams showing the distinction between One-to-One and Many-to-One
functions.
2
2
Domain and Range (Mastering Bounded Outputs)
Many students struggle to visualize domain and range. Think of them through a physical projection analogy:
• Domain: The horizontal launchpad along the x-axis. It is the exact interval of x-values
where the curve is defined.
• Range: The vertical height shadow projected on the y-axis. If you were to squash the
entire curve horizontally against the y-axis, the covered region is the range.
2.1
The 3-Step Range Calculator Checklist
To find the range of any continuous algebraic function without failing, run this simple sequence:
1. Identify the Function Family: Is it a straight line, a quadratic cup/cap, a rational reciprocal, or a trig wave?
2. Check for turning points or asymptotes INSIDE the domain boundaries:
• For quadratics, complete the square to locate the vertex (h, k). Crucial Check: Is x = h
inside your domain? If yes, y = k is an absolute extreme boundary of your range!
• For rational functions, locate the vertical asymptote (where the denominator is 0).
Ensure your domain does not cross this asymptote.
3. Evaluate the Domain Endpoints: Calculate the function values at the boundaries of your
domain and combine them with any internal turning points to state your final range.
2.2
Standard Syllabus Examples
2.2.1
Case 1: f (x) = x1 for x ≥ 1
• The curve is continuous and strictly decreasing for x ≥ 1.
• Left boundary evaluation: At x = 1, f (1) = 1.
• Right behavior evaluation: As x → ∞, f (x) → 0 but never reaches 0.
• Combining these: The outputs start just above 0 and go up to (and include) 1.
Range of f :
2.2.2
0 < f (x) ≤ 1
Case 2: g(x) = x2 + 1 for x ∈ R
• This is a quadratic with vertex at (0, 1).
• The vertex x = 0 is within the domain R.
• Since x2 ≥ 0 for all real numbers, the absolute minimum of the graph is at y = 1.
• There is no upper limit because x2 → ∞ as x → ±∞.
Range of g :
3
g(x) ≥ 1
The Most Common Range Error in CAIE Exams!
Never write your final range in terms of x.
• Incorrect: Range of f is x ≥ 1. (This refers to a domain).
• Correct: f (x) ≥ 1, or y ≥ 1.
Examiners strictly award 0 marks for range questions written using the variable x.
4
2.3
Exam-Realistic Worked Examples
Worked Example 2.1: The Quadratic turning Point Trap [4 Marks]
The function f is defined by f (x) = 2x2 − 12x + 11 for the domain:
(i) 0 ≤ x ≤ 5
(ii) 4 ≤ x ≤ 7
Find the range of f in each case, showing your complete working.
Solution for (i):
• Step 1: Complete the square to find the vertex.
f (x) = 2(x2 − 6x) + 11 = 2[(x − 3)2 − 9] + 11 = 2(x − 3)2 − 18 + 11 = 2(x − 3)2 − 7
This curve is a U-shaped quadratic cup with a minimum turning point (vertex) at
(3, −7).
• Step 2: Check if the vertex x = 3 lies in the domain [0, 5]. Since 3 ∈ [0, 5], the
absolute lowest value the function can ever output is the vertex value:
Minimum output = −7
• Step 3: Evaluate the endpoints to find the maximum output.
At x = 0 :
f (0) = 2(0 − 3)2 − 7 = 11
At x = 5 :
f (5) = 2(5 − 3)2 − 7 = 1
The highest point reached on this interval is 11.
• Final Range for (i): The outputs span from the minimum vertex height of −7 up to
the boundary peak of 11:
−7 ≤ f (x) ≤ 11
Solution for (ii):
• Step 1: Inspect the same vertex on the new domain [4, 7]. The vertex is at x = 3.
This is outside our domain because 3 ∈
/ [4, 7].
• Step 2: Evaluate boundaries directly. Since the vertex is to the left of our domain
and the curve is strictly increasing for x > 3:
At x = 4 :
f (4) = 2(4 − 3)2 − 7 = −5
(This is the absolute minimum on this interval)
At x = 7 :
f (7) = 2(7 − 3)2 − 7 = 25
(This is the absolute maximum on this interval)
• Final Range for (ii): The turning point is ignored because it lies outside the active
domain:
−5 ≤ f (x) ≤ 25
5
Worked Example 2.2: Bounded Rational Curve [4 Marks]
3
The function g is defined by g(x) = 5 − 2x−1
for x ≥ 2. Find the range of g.
Solution:
• Step 1: Check for asymptotes. The vertical asymptote is where 2x − 1 = 0 =⇒ x =
0.5. Since our domain is x ≥ 2, the vertical asymptote is completely avoided, and the
curve is continuous throughout.
• Step 2: Test the behavior of the reciprocal term. For x ≥ 2, let’s trace the values:
As x increases, the denominator (2x − 1) grows larger and larger. Thus, the fraction
3
2x−1 becomes smaller and smaller, shrinking toward zero.
• Step 3: Analyze boundary values.
– At the left boundary (x = 2):
g(2) = 5 −
3
3
=5− =4
2(2) − 1
3
– At the extreme right behavior (x → ∞):
lim g(x) = 5 − 0 = 5
x→∞
3
Since 2x−1
> 0 for all x ≥ 2, we are always subtracting a positive number from 5,
meaning g(x) will always be strictly less than 5.
• Final Range:
4 ≤ g(x) < 5
6
Worked Example 2.3: Bounded Trigonometric Wave [3 Marks]
The function h is defined by h(x) = 4 − 3 sin2 (2x) for 0 ≤ x ≤ π4 . Find the range of h.
Solution:
• Step 1: Track the angle limits. The input angle is 2x. Since 0 ≤ x ≤ π4 :
0 ≤ 2x ≤
π
2
• Step
π 2: Determine limits of the trigonometric component. Over the interval
0, 2 , the sine function behaves as:
0 ≤ sin(2x) ≤ 1
Squaring this inequality yields:
0 ≤ sin2 (2x) ≤ 1
• Step 3: Apply transformations algebraically to find the range. Multiply by −3
(which reverses the inequality signs):
−3 ≤ −3 sin2 (2x) ≤ 0
Add 4 to all parts of the inequality:
4 − 3 ≤ 4 − 3 sin2 (2x) ≤ 4 + 0 =⇒ 1 ≤ h(x) ≤ 4
• Final Range:
1 ≤ h(x) ≤ 4
7
3
Composite Functions
A composite function is formed when one function is applied to the output of another. The
notation gf (x) (or (g ◦ f )(x)) means that you apply the inner function f first, and then apply the
outer function g to the result:
gf (x) = g(f (x))
The Crucial Compatibility Condition
The composite function gf exists (or can be formed) if and only if:
Range of f ⊆ Domain of g
This means that every output produced by the inner function f must be a valid input that
the outer function g is allowed to receive.
3.1
Worked Examples
Worked Example 3.1: Verifying if Composition is Possible
Two functions are defined by:
f (x) = x2 − 4 for x < 2
√
g(x) = x for x ≥ 0
Determine, with a clear reason, whether the composite function gf (x) can be formed.
Solution: Step 1: Determine the Range of the inner function f . For f (x) = x2 − 4
defined on x < 2: As x → −∞, f (x) → ∞. At the boundary x = 2, f (2) = 22 − 4 = 0. Since
x < 2, the output value 0 is exclusive. Therefore, the range of f is f (x) > −4.
Step 2: Compare Range(f ) with Domain(g). The domain of g is x ≥ 0. The range of f
is (−4, ∞). For gf to exist, we require Range(f ) ⊆ Domain(g). However, values like −3
lie in the range of f but are outside the domain of g (you cannot take the square root of
a negative number in Pure 1). Specifically, (−4, ∞) 6⊆ [0, ∞). Therefore, the composite
function gf (x) cannot be formed.
8
Worked Example 3.2: Finding Domain and Range of a Composite Function
Let f (x) = 2x + 1 for x ≥ 1, and g(x) = x2 − 3 for x ≥ 0.
1. Explain why the composite function gf can be formed.
2. Find an expression for gf (x).
3. State the domain and find the range of gf (x).
Solution: 1. Verification:
• Range(f ): Since x ≥ 1 =⇒ 2x + 1 ≥ 3. Thus, Range(f ) = [3, ∞).
• Domain(g) = [0, ∞).
• Since [3, ∞) ⊆ [0, ∞), the range of f is a subset of the domain of g. Therefore, gf can
be formed.
2. Algebraic Expression:
gf (x) = g(f (x)) = (2x + 1)2 − 3 = 4x2 + 4x + 1 − 3 = 4x2 + 4x − 2
3. Domain and Range of gf :
• The domain of the composite function is always identical to the domain of the inner
function:
Domain of gf : x ≥ 1
• To find the range of gf (x), evaluate the outer function g(u) over the set of inputs u
provided by the range of the inner function f (which is u ≥ 3): For u ≥ 3: g(u) =
u2 − 3 ≥ 32 − 3 = 6. Hence:
Range of gf :
9
gf (x) ≥ 6
Worked Example 3.3: Setting a Domain to Allow Composition
1
Functions f and g are defined by f (x) = x2 + 2x for x ∈ R, and g(x) = x−8
for x > 8. Find
the set of values of x for which the composite function gf (x) can be formed.
Solution: For gf (x) to be defined, the output of f (x) must lie within the domain of g(x).
• The domain of g is g(x) > 8.
• Thus, we require the outputs of f (x) to satisfy: f (x) > 8.
Solve the quadratic inequality:
x2 + 2x > 8 =⇒ x2 + 2x − 8 > 0
Factorize the quadratic expression:
(x + 4)(x − 2) > 0
The critical values are x = −4 and x = 2. Since we require the product to be positive, the
solution set is:
x < −4 or x > 2
These are the restricted domain values of f that allow gf to exist.
10
4
Inverse Functions
The inverse function f −1 (x) reverses the operation of f (x). Applying f followed by f −1 returns
you to the starting value: f −1 (f (x)) = x.
Existence of an Inverse Function
An inverse function f −1 exists if and only if the function f is a One-to-One function. If f
is Many-to-One, its inverse mapping would be One-to-Many, which is not a function.
4.1
Key Identities
The coordinates on the graph of f are swapped to obtain coordinates on the graph of f −1 .
Consequently, the boundaries of inputs and outputs swap:
Domain of f −1 = Range of f
Range of f −1 = Domain of f
4.2
Syllabus Case Study: Finding h−1 (x) for Quadratic Curves
This is a highly popular and technical exam question. Let’s study the syllabus case:
h(x) = (2x + 3)2 − 4
11
for x < −
3
2
Detailed Analysis of Syllabus Case
Find the inverse function h−1 (x) and state its domain.
Solution: Step 1: Let y = h(x) and rearrange to isolate x.
y = (2x + 3)2 − 4 =⇒ y + 4 = (2x + 3)2
Take the square root of both sides:
p
± y + 4 = 2x + 3
Step 2: Choose the correct sign for the root (CRITICAL STEP). Look at the domain of h:
x < − 32 . Multiply by 2 and add 3 to see the sign of (2x + 3):
2x < −3 =⇒ 2x + 3 < 0
Since the quantity (2x+3) must be strictly negative, we must choose the negative square
root:
p
− y + 4 = 2x + 3
Solve for x:
p
−3 −
2x = −3 − y + 4 =⇒ x =
√
y+4
2
Swap the variables to write the final inverse function:
√
−3 − x + 4
−1
h (x) =
2
Step 3: State the Domain of h−1 . The domain of h−1 is equal to the range of h. Let’s
find the range of h: For x < − 32 , we have 2x + 3 < 0 =⇒ (2x + 3)2 > 0. Therefore,
(2x + 3)2 − 4 > −4. The range of h is h(x) > −4. Hence, the domain of h−1 is:
x > −4
12
4.3
Further Worked Examples
Worked Example 4.1: Rational Function Inverse
Find the inverse of the function f (x) = 3x+1
x−2 for x > 2, and state its domain.
Solution: Let y = 3x+1
x−2 . Multiply through by (x − 2):
y(x − 2) = 3x + 1 =⇒ yx − 2y = 3x + 1
Group all terms containing x on one side:
yx − 3x = 2y + 1
Factorize out x:
x(y − 3) = 2y + 1 =⇒ x =
Thus, the inverse function is:
f −1 (x) =
2y + 1
y−3
2x + 1
x−3
To find the domain of f −1 , we need the range of f for x > 2. Rewrite f (x) by performing
algebraic division:
3(x − 2) + 7
7
f (x) =
=3+
x−2
x−2
7
Since x > 2, the term x − 2 > 0, meaning x−2
> 0. Thus, f (x) > 3 for all x > 2. The range
of f is f (x) > 3. Consequently, the domain of f −1 is:
x>3
13
Worked Example 4.2: Quadratic Requiring Completing the Square First
The function f is defined by f (x) = x2 − 8x + 7 for x ≥ k.
1. State the smallest value of k for which f has an inverse.
2. Using this value of k, find an expression for f −1 (x) and state its domain.
Solution: 1. Smallest value of k: For f to have an inverse, it must be a one-one function.
A quadratic is only one-one if its domain is restricted to one side of its line of symmetry.
Complete the square:
f (x) = (x − 4)2 − 16 + 7 = (x − 4)2 − 9
The axis of symmetry is at x = 4. For the domain x ≥ k to ensure a one-one function, the
domain must start at or to the right of the vertex. Thus, the smallest possible value of k
is:
k=4
2. Find f −1 (x) using k = 4: Set y = (x − 4)2 − 9 for x ≥ 4.
p
y + 9 = (x − 4)2 =⇒ ± y + 9 = x − 4
Since x ≥ 4 =⇒ x − 4 ≥ 0, we must choose the positive square root:
p
√
x = 4 + y + 9 =⇒ f −1 (x) = 4 + x + 9
The domain of f −1 is the range of f . Since the vertex is (4, −9), for x ≥ 4, the range is
f (x) ≥ −9. Therefore, the domain of f −1 is:
x ≥ −9
14
Graphical Relationship Between f and f −1
5
There is a fundamental geometric relationship between a function and its inverse.
Geometric Property of Inverses
The graph of y = f −1 (x) is a reflection of the graph of y = f (x) in the line y = x. This is
because the operation of finding an inverse swaps every coordinate (a, b) on f to (b, a) on
f −1 .
5.1
Annotated Graphs of f and f −1
Let us plot a concrete example to visualize this transformation. Let f (x) = x2 for x ≥ 0. The
√
corresponding inverse function is f −1 (x) = x for x ≥ 0.
4
y
(2, 4)
f (x) = x2
√
−1
f (x) = x
y=x
3
(4, 2)
2
(1, 1)
1
x
1
2
3
4
Figure 2: Visual representation of the reflection of f (x) = x2 onto f −1 (x) =
y = x.
5.2
√
x across the line
How to Sketch the Inverse Graph in the Exam
If you are asked to sketch the graph of f −1 on the same axes as f :
1. Draw the line of symmetry y = x as a dashed line first.
2. Identify key points on the graph of y = f (x), such as y-intercepts, x-intercepts, turning
points, or boundary endpoints.
3. Swap the coordinates of these points. For example, (0, c) → (c, 0) and (a, b) → (b, a). Plot
these new points on your grid.
4. Sketch the reflected curve through the new points, ensuring it forms a perfect mirror
image across the y = x line. Note that any intersection point between y = f (x) and y =
f −1 (x) must lie exactly on the line y = x.
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6
Transformations of Graphs (Unlocking the Mathematical Logic)
Many students find graph transformations difficult because they try to memorize rules without
understanding the underlying math. The simplest way to master transformations is to divide
them into two independent ”worlds.”
6.1
The Vertical World vs. The Horizontal World Paradigm
Table 1: The Honest Vertical World vs. The Opposite Horizontal World
The Vertical World (Outside the
Brackets)
The Horizontal World (Inside the
Brackets)
Affects: y-coordinates
Affects: x-coordinates
Strict Rule: It is Honest. It behaves
exactly as you would expect.
Strict Rule: It is the Opposite. It does
the complete reverse of what it looks like.
Addition (+a): Moves the curve up by a.
Addition (+a): Moves the curve left by a.
Multiplication (×a): Stretches the curve
vertically by scale factor a.
Multiplication (×a):
Compresses/stretches horizontally by
scale factor a1 .
Unlocking the Logic of the ”Opposite” Horizontal World
Why does y = f (2x) compress the curve by factor 12 instead of stretching it by 2?
• Let’s think: To get the same output height y, the inner input has to equal the original
input.
• If the original function reached a height of y = 10 at x = 6, the new function y = f (2x)
only needs 2x = 6 =⇒ x = 3 to reach that exact same height!
• Because the process runs twice as fast, the curve needs only half the horizontal
space to complete its shape. Thus, it is compressed by a factor of 12 .
6.2
Combined Transformations: Why Order Matters
When multiple transformations are combined, we must be incredibly careful about the order.
6.2.1
Vertical Combinations: y = af (x) + b
Since vertical transformations are ”honest,” they follow standard order of operations (BIDMAS):
1. Multiply first: Stretch vertically by scale factor a parallel to the y-axis.
0
2. Add second: Translate vertically by vector
.
b
6.2.2
Horizontal Combinations: y = f (ax + b)
Because horizontal transformations do the opposite of what they look like, their order of operations is also reversed. To get from f (x) to f (ax + b), there are two standard pathways. **Choose
ONE and use it consistently:**
16
Two Pathways to Describe f (ax + b) in Exams
Pathway A: The Translation-First Method (Highly Recommended)
−b
1. Translate horizontally by vector
first. (Replacing x with x + b).
0
2. Stretch horizontally by scale factor a1 parallel to the x-axis second. (Replacing x with
ax).
translate
stretch
3. Result: f (x) −−−−−→ f (x + b) −−−−→ f (ax + b).
Pathway B: The Stretch-First Method (Requires Factoring)
1. Factor out the horizontal coefficient: f (ax + b) =⇒ f a x + ab
.
2. Stretch horizontally by scale factor a1 parallel to the x-axis first. (Replacing x with ax).
−b/a
3. Translate horizontally by vector
second. (Replacing x with x + ab ).
0
stretch
translate
4. Result: f (x) −−−−→ f (ax) −−−−−→ f a x + ab
17
= f (ax + b).
6.3
Exam-Realistic Worked Examples
Worked Example 6.1: Describing Complex Multi-Step Transformations [6 Marks]
√
√
The graph of y = x is transformed to the graph of y = 3 2x − 8 + 5.
√
(i) State a sequence of four geometric transformations that maps y = x onto the
transformed curve.
√
(ii) A point P (4, 2) lies on the original curve y = x. Find the coordinates of the corresponding point P 0 on the transformed curve.
√
Solution for (i): Let’s break down the transformation of the base function f (x) = x into
the vertical and horizontal components of y = 3f (2x − 8) + 5:
Horizontal World (Inside): To map f (x) → f (2x − 8), we use **Pathway A (TranslationFirst Method)**:
8
1. Horizontal Translation: Translate horizontally by vector
(or 8 units to the right).
0
2. Horizontal Stretch: Stretch horizontally by scale factor 12 parallel to the x-axis.
Vertical World (Outside): To map f (2x − 8) → 3f (2x − 8) + 5, we follow standard BIDMAS:
3. Vertical Stretch: Stretch vertically by scale factor 3 parallel to the y-axis.
0
4. Vertical Translation: Translate vertically by vector
(or 5 units up).
5
*(Note: Any valid order that separates horizontal/vertical components correctly will be
awarded full marks in the exam.)*
Solution for (ii): Let’s track the coordinate conversions step-by-step for the point P (4, 2):
1. Translate right by 8: (x, y) → (x + 8, y)
(4, 2) → (12, 2)
2. Horizontal Stretch (SF 12 ): (x, y) →
1
2 x, y
(12, 2) → (6, 2)
3. Vertical Stretch (SF 3): (x, y) → (x, 3y)
(6, 2) → (6, 6)
4. Translate up by 5: (x, y) → (x, y + 5)
(6, 6) → (6, 11)
Thus, the final coordinates are:
P 0 = (6, 11)
18
Worked Example 6.2: Algebraic Modeling of Transformations [5 Marks]
A curve has the equation y = x2 − 4x. The curve undergoes three consecutive transformations:
1. A vertical stretch with scale factor 3.
−2
2. A horizontal translation by vector
.
0
0
3. A vertical translation by vector
.
1
Find the equation of the resulting curve in the form y = px2 + qx + r, where p, q and r are
integers.
Solution: Let’s perform the algebraic operations step-by-step on the starting equation
y = f (x) = x2 − 4x:
• Step 1: Apply vertical stretch with scale factor 3. This multiplies the entire function by 3:
y1 = 3 · f (x) = 3(x2 − 4x) = 3x2 − 12x
−2
• Step 2: Apply horizontal translation by vector
. This shifts the curve left by
0
2. We replace every occurrence of x with (x + 2):
y2 = 3(x + 2)2 − 12(x + 2)
= 3(x2 + 4x + 4) − 12x − 24
= 3x2 + 12x + 12 − 12x − 24
= 3x2 − 12
0
• Step 3: Apply vertical translation by vector
. This shifts the curve up by 1. We
1
add 1 to the entire equation:
y3 = (3x2 − 12) + 1
= 3x2 − 11
Thus, the final transformed equation of the curve is:
y = 3x2 − 11
*(Here, p = 3, q = 0, and r = −11.)*
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6.4
Visualizing Transformations with TikZ
Here we examine two sets of graph transformations to visualize the physical shifts and stretches.
6.4.1
Set 1: Translations
Below we visualize the base function y = x2 transformed via a translation of
2
to become
1
y = (x − 2)2 + 1.
5 y
Original: y = x2
Translated:
y = (x − 2)2 + 1
4
3
2
(2, 1)
1
+1
(0, 0)
−2
+2
1
−1
2
x
3
4
5
2
Figure 3: Translation of y = f (x) to y = f (x − 2) + 1 via vector
.
1
6.4.2
Set 2: Stretches
Here we show a trigonometric wave f (x) = sin(x) stretched vertically by factor 2 to become
y = 2 sin(x), alongside a horizontal compression y = sin(2x).
2 y
Base: y = sin(x)
Vertical Stretch: y = 2 sin(x)
Horizontal Comp.: y = sin(2x)
1
x
π
2
π
3π
2
2π
−1
−2
Figure 4: Comparison of a vertical stretch (scale factor 2) and a horizontal stretch (scale factor
1
2 ).
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7
Common Exam Pitfalls and Tips
To maximize your marks, pay close attention to these common errors identified in past Examiner
Reports.
Common Mistakes & How to Avoid Them
1. Stating Domain of f −1 using the wrong variable:
• Error: Writing the domain of f −1 as f −1 (x) ≥ 3 or y ≥ 3.
• Correction: Even though the domain of f −1 is calculated from the range of f , it
is a set of inputs and must be written in terms of x (e.g., x ≥ 3).
2. Incorrect composite function ordering:
• Error: Computing f g(x) when asked for gf (x).
• Correction: Always work from the inside out. For gf (x), take the expression for
f (x) and substitute it wherever there is an x inside the formula for g(x).
3. Forgetting to check the One-to-One condition:
• Error: Trying to find the algebraic inverse of a quadratic function across its entire
domain R.
• Correction: A quadratic function over R is many-one and does not have an inverse. You must first restrict the domain to one side of the axis of symmetry
b
b
(where x ≥ − 2a
or x ≤ − 2a
).
4. Losing the correct sign during inverse rearrangement:
√
• Error: Writing x = 2 ± y as the final step.
• Correction: You must inspect the original restricted domain of f to select either
the positive or negative square root. If the domain is x ≤ c, choose the negative
root. If the domain is x ≥ c, choose the positive root.
5. Using coordinate geometry terminology for graph transformations:
• Error: Describing f (x + 3) as “moving left by 3”.
• Correction:Useformal mathematical transformations. State it as a Translation
−3
by vector
or parallel to the x-axis.
0
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8
Summary Checklist
Use this end-of-topic checklist to rate your understanding of each syllabus point before the
exam.
Syllabus Competency Checklist
Core Terms: I can state and use the definitions of function, domain, range, one-one
function, inverse function, and composition of functions.
Mapping Diagrams: I can identify functions from mapping diagrams and classify
them as one-one or many-one.
Domain & Range Identification: I can find the range of a given function (including
linear, quadratic, rational, and trigonometric functions) over restricted domains.
Composition Feasibility: I can determine whether a composite function gf can be
formed by verifying if Range(f ) ⊆ Domain(g).
Composite Expressions: I can algebraically form gf (x) and state its corresponding
domain and range.
One-to-One Test: I can determine if an inverse exists by checking whether a function is one-to-one (using algebra or sketches).
Finding the Inverse: I can find f −1 (x) for linear, rational, and quadratic functions
(by completing the square).
Inverse Domain/Range Identities: I can apply the rule that Domain(f −1 ) =
Range(f ) and Range(f −1 ) = Domain(f ).
Graphical Reflection: I can sketch the graph of f −1 on the same axes as f using a
reflection in the mirror line y = x.
Single Transformations: I can describe and apply translations (f (x) + a, f (x + a)),
reflections (−f(x), f (−x)), and stretches (af (x), f (ax)).
Combined Transformations: I can perform and describe combinations of up to
three geometric transformations in the correct sequence.
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