DP Maths AA HL - Functions - Transformations of Functions

DP Maths AA HL - Functions - Transformations of Functions - Full notes

RevisionPrep

DP Maths AA HL | Transformations of Functions

Transformations of Functions

This topic contains 1 note:

2.11 Transformations of Graphs

DP MATHEMATICS AA · TOPIC 2: FUNCTIONS

2.11

Transformations of Graphs

Full notes for HL · Syllabus SL 2.11

Contents

1	Translations
2	Reflections
3	Vertical stretches
4	Horizontal stretches
5	Composite transformations and order
6	Transforming points, asymptotes and key features
7	Finding equations of transformed graphs
8	Transformations of standard functions
9	Translations in practice
10	Vertical and horizontal stretches in practice
11	Reflections in practice
12	Order of composite transformations
13	Images of points and key features
14	Transforming asymptotes
15	Sketching transformed graphs
16	Transformations of trigonometric graphs
17	Transformations and the equation of a model
18	Modulus transformations (HL)
19	Mixed review: worked problems
20	Additional exam-style practice
21	Transformations with technology
22	Exam-style questions with markscheme
23	Common misconceptions and links
24	Check your understanding
25	Summary and topic map
26	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the Quick check.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe and apply translations y = f(x) + b and y = f(x − a)
describe and apply reflections y = −f(x) and y = f(−x)
describe and apply vertical stretches y = pf(x) and horizontal stretches y = f(qx)
apply composite transformations in the correct order
find the image of a point, asymptote or key feature under a transformation
deduce a transformed equation from a description or graph
apply transformations to quadratic, reciprocal, exponential, logarithmic and trigonometric functions

This booklet covers SL 2.11, which is common to SL and HL. HL students extend this topic in booklet 2.15.

Guiding questions

How do changes to a function's equation move or reshape its graph? Why do changes inside the bracket act 'the opposite way'?

How this topic is examined

Transformations are tested on Paper 1 (describe a sequence of transformations; find the image of a point; sketch a transformed graph) and support many Paper 2 modelling questions, especially with trigonometric and exponential functions.

Use precise language: 'translation by the vector (a, b)', 'vertical stretch with scale factor p', 'horizontal stretch with scale factor 1/q', 'reflection in the x-axis'.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently on Paper 2
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 Translations

Shifting without changing shape.

Think about it

The graph of y = x2 has its vertex at the origin. Where is the vertex of y = (x − 3)2 + 2? Why does the '− 3' move the graph to the right, not the left?

y = f(x) + b is a vertical translation by b units: up if b > 0, down if b < 0. Every point (x, y) moves to (x, y + b).

y = f(x − a) is a horizontal translation by a units: right if a > 0, left if a < 0. Every point (x, y) moves to (x + a, y).

Why the opposite sign? The new function reaches the value f(0) when x − a = 0, that is, when x = a. Everything happens a units later, so the graph moves a units to the right.

Combined, y = f(x − a) + b is a translation by the vector (a, b), written as a column vector with a on top and b below.

Figure 1. Translation by the vector (3, 2).

WORKED EXAMPLE 1 Describing a translation (basic)

Describe the transformation from y = x3 to y = (x + 4)3 − 1.

Solution

x replaced by x + 4 = x − (−4): 4 units leftA1

−1 outside: 1 unit down; translation by the vector (−4, −1)A1

Practice exercise 1

1. Describe y = √(x − 2) + 3 as a transformation of y = √x.

2. Write the equation of y = ex after a translation by the vector (1, −2).

3. The point (3, 5) lies on y = f(x). Find its image on y = f(x + 2) − 4.

2 Reflections

Mirror images in the axes.

y = −f(x) is a reflection in the x-axis: every y-coordinate changes sign, (x, y) → (x, −y). Roots are unchanged.

y = f(−x) is a reflection in the y-axis: every x-coordinate changes sign, (x, y) → (−x, y). The y-intercept is unchanged.

Even functions satisfy f(−x) = f(x), so they are unchanged by reflection in the y-axis.

Figure 2. Reflections of y = ex.

Practice exercise 2

1. Find the image of (2, −3) under y = −f(x) and under y = f(−x).

2. Write the equation after y = ln x is reflected in the y-axis. State its domain.

3. Which functions are unchanged by y = f(−x)?

3 Vertical stretches

Scaling the outputs.

y = pf(x) is a vertical stretch with scale factor p (parallel to the y-axis): every y-coordinate is multiplied by p, (x, y) → (x, py). Points on the x-axis do not move, so the roots are unchanged. If 0 < p < 1, the graph is compressed towards the x-axis. If p < 0, combine with a reflection in the x-axis.

Figure 3. A vertical stretch with scale factor 3 changes the amplitude.

4 Horizontal stretches

Scaling the inputs, the opposite way.

y = f(qx) is a horizontal stretch with scale factor 1/q (parallel to the x-axis): every x-coordinate is divided by q, (x, y) → (x/q, y). Points on the y-axis do not move, so the y-intercept is unchanged.

Like translations inside the bracket, this acts the 'opposite way': y = f(2x) compresses the graph towards the y-axis by factor ½, because the new function reaches each output at half the x-value. For example, y = sin 2x has period π, half the period of sin x.

Figure 4. y = sin 2x is a horizontal stretch with scale factor ½.

Equation	Transformation	Effect on point (x, y)
y = f(x) + b	vertical translation b	(x, y + b)
y = f(x − a)	horizontal translation a	(x + a, y)
y = −f(x)	reflection in x-axis	(x, −y)
y = f(−x)	reflection in y-axis	(−x, y)
y = pf(x)	vertical stretch, factor p	(x, py)
y = f(qx)	horizontal stretch, factor 1/q	(x/q, y)

Common mistake

Saying y = f(2x) is a horizontal stretch with factor 2. The factor is ½: the graph is compressed towards the y-axis.

Practice exercise 3

1. Find the image of (6, 4) under y = 2f(x) and under y = f(3x).

2. Describe y = cos(x/2) as a transformation of y = cos x, and state its period.

3. The roots of f are −2 and 6. Find the roots of y = f(2x).

5 Composite transformations and order

Order matters when transformations interact.

When several transformations are combined, the order can change the result. A reliable method is to treat vertical changes (outside the function) and horizontal changes (inside the bracket) separately.

Vertical: for y = pf(x) + b, stretch by p first, then translate by b. (If you translated first, the stretch would also multiply the translation.) A reflection −f(x) counts as a stretch with factor −1.

Horizontal: for y = f(qx + c), first factorize the bracket: f(q(x + c/q)). This is a horizontal stretch with factor 1/q and a translation of c/q to the left. If the stretch is applied first, the translation is c/q; if the translation is applied first, it is c, followed by the stretch.

Horizontal and vertical transformations are independent, so they can be done in either order relative to each other.

WORKED EXAMPLE 2 Describing a sequence (standard)

Describe a sequence of transformations that maps y = f(x) to y = 3f(2x + 4) − 1.

Solution

y = 3f(2(x + 2)) − 1M1

Horizontal stretch scale factor ½, then translation 2 units leftA1

Vertical stretch scale factor 3, then translation 1 unit downA1

WORKED EXAMPLE 3 Order changes the result (exam level)

The graph of y = f(x) is translated 2 units up and then stretched vertically by scale factor 3. Write the new equation, and compare with doing the stretch first.

Solution

Translate: y = f(x) + 2; then stretch: y = 3(f(x) + 2) = 3f(x) + 6A1

Stretch first: y = 3f(x); then translate: y = 3f(x) + 2A1

The results differ, so order mattersR1

Practice exercise 4

1. Describe the transformations from y = f(x) to y = 2f(x − 1) + 5.

2. Describe the transformations from y = f(x) to y = f(3x − 6).

3. Write the equation when y = f(x) is reflected in the x-axis then translated 4 units up.

4. Describe y = −2(x + 1)2 + 3 as a transformation of y = x2.

6 Transforming points, asymptotes and key features

Follow each coordinate.

To find the image of a point under several transformations, apply each change to the coordinates in the correct order. Asymptotes, turning points and intercepts transform in exactly the same way: vertical asymptotes x = k transform like x-coordinates; horizontal asymptotes y = k like y-coordinates.

WORKED EXAMPLE 4 Image of a point (standard)

The point (4, 6) lies on y = f(x). Find its image on (a) y = 2f(x − 1) + 3; (b) y = f(2x); (c) y = −f(½x).

Solution

(a) (4 + 1, 2 × 6 + 3) = (5, 15)A1

(b) (4/2, 6) = (2, 6)A1

WORKED EXAMPLE 5 Asymptotes (standard)

y = f(x) has asymptotes x = 2 and y = 1. Find the asymptotes of y = 3f(x + 3) − 4.

Solution

Vertical: x = 2 − 3 = −1A1

Horizontal: y = 3(1) − 4 = −1A1

Practice exercise 5

1. (−2, 5) lies on y = f(x). Find its image on y = 4 − f(x).

2. y = f(x) has a minimum at (3, −2). Find the minimum of y = f(x − 1) + 5 and the maximum of y = −f(x).

3. y = g(x) has asymptote y = 6. Find the asymptote of y = ½g(x) − 1.

7 Finding equations of transformed graphs

Work forwards from a description, or backwards from a graph.

WORKED EXAMPLE 6 From description to equation (standard)

y = ln x is translated 2 units right, stretched vertically by factor 3, then reflected in the x-axis. Write the new equation.

Solution

Translation: y = ln(x − 2)A1

Stretch: y = 3 ln(x − 2)A1

Reflection: y = −3 ln(x − 2)A1

Figure 5. y = ln x and its transformed image, with asymptote x = 2.

WORKED EXAMPLE 7 From a graph (exam level)

The graph of y = a ex + c passes through (0, 5) and has horizontal asymptote y = 2. Find a and c.

Solution

Asymptote y = c, so c = 2A1

At x = 0: a + 2 = 5M1

a = 3A1

Practice exercise 6

1. Write the equation after y = 1/x is translated by (3, −1).

2. y = x2 is stretched horizontally by factor 2, then translated 1 unit up. Write the equation.

3. A parabola y = a(x − h)2 + k has vertex (−1, 4) and passes through (1, −4). Find a, h and k.

8 Transformations of standard functions

Recognize the parameters.

Function	Form	Parameters
Quadratic	a(x − h)2 + k	vertex (h, k); stretch a
Reciprocal	a/(x − h) + k	asymptotes x = h, y = k
Exponential	aeb(x − h) + k	asymptote y = k
Logarithmic	a ln(b(x − h)) + k	asymptote x = h
Sine	a sin(b(x − c)) + d	amplitude \|a\|, period 2π/b, phase shift c, principal axis y = d

The trigonometric form is studied in detail in 3.7: a sin(b(x − c)) + d is a vertical stretch (amplitude |a|), a horizontal stretch (period 2π/b), a horizontal translation by c and a vertical translation by d of y = sin x.

9 Translations in practice

Vectors (a, b).

y = f(x − a) + b is a translation of y = f(x) by the vector (a, b). Every point (x, y) moves to (x + a, y + b), and asymptotes, turning points and intercepts move with it (intercepts must then be recalculated).

WORKED EXAMPLE 8 Translating a cubic (standard)

y = x3 is translated by (2, −1). Write the new equation and find its y-intercept.

Solution

y = (x − 2)3 − 1A1

y-intercept −8 − 1 = −9A1

WORKED EXAMPLE 9 Finding the vector (standard)

Describe the transformation from y = √x to y = √(x + 4) + 2.

Solution

Translation by (−4, 2)A1

Practice exercise 7

1. Write y = ex after a translation by (−1, 3) and give its asymptote.

2. The point (3, 7) lies on y = f(x). Find its image on y = f(x + 5) − 2.

10 Vertical and horizontal stretches in practice

Scale factors.

y = pf(x): vertical stretch with scale factor p (y-values multiplied by p). y = f(qx): horizontal stretch with scale factor 1/q (x-values divided by q). Points on the axis of the stretch stay fixed.

WORKED EXAMPLE 10 Stretches (standard)

(4, −6) lies on y = f(x). Find its image under y = 3f(x) and y = f(2x).

Solution

(4, −18)A1

(2, −6)A1

Practice exercise 8

1. Describe y = sin(3x) as a transformation of y = sin x.

2. The roots of f are −6 and 9. Find the roots of y = f(3x).

11 Reflections in practice

In the axes.

y = −f(x) reflects in the x-axis ((x, y) → (x, −y)); y = f(−x) reflects in the y-axis ((x, y) → (−x, y)). Even functions satisfy f(−x) = f(x); odd functions satisfy f(−x) = −f(x).

WORKED EXAMPLE 11 Combined reflection (standard)

Find the image of (2, 5) under y = −f(−x).

Solution

(−2, −5)A1

This is a rotation of 180° about the originR1

Practice exercise 9

1. Show that f(x) = x3 − x is odd.

2. Write y = ln x reflected in the y-axis and state the domain.

12 Order of composite transformations

Inside (horizontal) and outside (vertical).

Vertical transformations act on the output: for y = pf(x) + b, stretch first then translate. Horizontal transformations act on the input: for y = f(q(x − a)), stretch then translate; if written f(qx − c), factorize first. Horizontal and vertical changes are independent of each other.

WORKED EXAMPLE 12 Sequence (exam level)

Describe the transformations mapping y = f(x) to y = 2f(3x + 6) − 1.

Solution

f(3(x + 2)): horizontal stretch factor 1/3, then translation 2 leftM1 A1

Vertical stretch factor 2, then translation 1 downA1

Common mistake

Reading y = f(3x + 6) as a translation 6 units left. Factorize: f(3(x + 2)) is a translation 2 units left (after the stretch).

Practice exercise 10

1. Write the equation of y = f(x) after: horizontal stretch factor 2, then translation 3 right.

2. Describe y = −f(x/2) + 4.

13 Images of points and key features

Track the coordinates.

To find the image of a point, apply the horizontal transformations to x (in order) and the vertical ones to y (in order). Maxima remain maxima unless there is a reflection in the x-axis, which turns them into minima.

WORKED EXAMPLE 13 Maximum point (standard)

y = f(x) has a maximum at (3, 8). Find the corresponding point on y = 5 − 2f(x + 1).

Solution

x: 3 − 1 = 2; y: 5 − 16 = −11M1 A1

(2, −11), now a minimumA1 R1

Practice exercise 11

1. (−2, 4) lies on y = f(x). Find its image on y = ½f(2x) + 3.

2. y = f(x) has a minimum at (1, −2). Find the minimum of y = 3f(x − 4) + 1.

14 Transforming asymptotes

Lines move too.

Vertical asymptotes x = c transform like x-coordinates, horizontal asymptotes y = d like y-coordinates. For example, if f has asymptotes x = 1 and y = 2, then y = 3f(x + 2) − 1 has asymptotes x = −1 and y = 5.

WORKED EXAMPLE 14 Asymptotes (standard)

y = f(x) has asymptotes x = −2 and y = 3. Find the asymptotes of y = −f(2x) + 1.

Solution

x = −1A1

y = −3 + 1 = −2A1

Practice exercise 12

1. Find the asymptotes of y = 4/(x − 3) + 2 by transforming y = 1/x.

2. Find the asymptote of y = 3 − 2ex−1.

15 Sketching transformed graphs

Key points and asymptotes.

To sketch a transformed graph: identify the parent function, apply the transformations to its key points and asymptotes, and draw a smooth curve of the correct shape. Label intercepts and turning points of the new graph.

Figure 6. y = x2 transformed: stretch, reflection, translation.

WORKED EXAMPLE 15 Sketch (standard)

Sketch y = −2(x − 1)2 + 3, labelling the vertex and intercepts.

Solution

Vertex (1, 3), maximumA1

y-intercept (0, 1); x-intercepts 1 ± √1.5: −0.225 and 2.22A1 A1

Practice exercise 13

1. Sketch y = 2x−1 − 4, labelling the asymptote and intercepts.

16 Transformations of trigonometric graphs

Amplitude, period, phase, axis.

For y = a sin(b(x − c)) + d: amplitude |a| (vertical stretch), period 2π/b (horizontal stretch factor 1/b), phase shift c (translation right), principal axis y = d (translation up).

WORKED EXAMPLE 16 Trig transformation (standard)

Describe y = 3 sin(2x − π/3) + 1 as transformations of y = sin x.

Solution

sin(2(x − π/6)): horizontal stretch 1/2, translation π/6 rightM1 A1

Vertical stretch 3, translation 1 upA1

Practice exercise 14

1. State the period and amplitude of y = −4 cos(x/3).

2. Write y = sin x after a horizontal stretch factor 2 and translation by (0, −1).

17 Transformations and the equation of a model

Fitting a model by transforming.

Models are often built by transforming a standard graph to fit data: shift a parabola so its vertex matches the maximum height of a projectile, or stretch and shift a sine curve to fit tide data. Identify the transformation from key features of the data.

WORKED EXAMPLE 17 Projectile (exam level)

A ball reaches a maximum height of 12 m after 1.5 s and lands at t = 3.06 s. Model h by transforming y = −x2.

Solution

h = −a(t − 1.5)2 + 12M1

h(3.06) = 0: a = 12/1.562 = 4.93A1

h = −4.93(t − 1.5)2 + 12A1

Practice exercise 15

1. A curve y = ae−x + c passes through (0, 10) and approaches 2. Find a and c.

18 Modulus transformations (HL)

y = |f(x)| and y = f(|x|).

y = |f(x)| reflects the parts of the graph below the x-axis upwards. y = f(|x|) keeps the right-hand part (x ≥ 0) and reflects it in the y-axis, replacing the left-hand part. These are covered fully in booklet 2.15.

WORKED EXAMPLE 18 Two moduli (exam level)

For f(x) = x2 − 4x, describe y = |f(x)| and y = f(|x|).

Solution

|f(x)|: the part between 0 and 4 is reflected above the axisA1

f(|x|): even function with minima at x = ±2A1

Practice exercise 16

1. Sketch y = |ln x|.

2. Find the number of solutions of |x2 − 4| = 3.

19 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem (exam level)

The graph of y = f(x) passes through (0, 4), (2, 0) and has asymptote y = 1. Find the corresponding features of y = 2f(x − 1) − 3.

Solution

(0, 4) → (1, 5); (2, 0) → (3, −3)A1 A1

Asymptote y = 2 − 3 = −1A1

Practice exercise 17

1. Find the transformation that maps y = x2 to y = x2 − 6x + 4.

2. y = f(x) is reflected in the y-axis then translated 3 right. Write the equation.

20 Additional exam-style practice

Paper 1 questions.

WORKED EXAMPLE 20 Paper 1 question (exam level)

The graph of y = g(x) is obtained from y = ln x by a vertical stretch factor 2 followed by a translation by (3, −1). (a) Write g(x). (b) Find the x-intercept of g. (c) State the domain of g.

Solution

(a) g(x) = 2 ln(x − 3) − 1A2

(b) ln(x − 3) = 0.5: x = 3 + e0.5 = 4.65A1

Practice exercise 18

1. Describe the transformation from y = 2x to y = 2x+3 as a vertical stretch.

21 Transformations with technology

Sliders and checking.

Dynamic graphing software lets you vary parameters with sliders and watch the effect. Use it to check a written description: graph the original and the transformed function together and check that key points have moved as predicted.

WORKED EXAMPLE 21 Checking (standard)

Check that (1, 0) on y = ln x maps to (4, −1) on y = 2 ln(x − 3) − 1.

Solution

2 ln(1) − 1 = −1 ✓A1

GDC tip

Graph y = f(x) and the transformed function together; use the trace or table to compare a key point and its image.

22 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 12 minutes)

The graph of y = f(x) passes through A(−2, 4) and has a horizontal asymptote y = 2. g(x) = −f(x − 1) + 5.

(a) Describe the transformations mapping f to g. [3]

(b) Find the image of A on g. [2]

(d) h(x) = f(3x). Find the image of A on h. [1]

Markscheme

Part	Answer	Marks
(a)	translation 1 unit right	A1
	reflection in the x-axis	A1
	translation 5 units up (after the reflection)	A1
(b)	x = −1; y = −4 + 5 = 1	M1
	(−1, 1)	A1
(c)	y = 3	A1
(d)	(−2/3, 4)	A1

Examiner's comments. In (a), the reflection must precede the vertical translation.

Question 2 (Paper 1 style) (about 12 minutes)

Let f(x) = x2 − 6x + 13.

(a) Write f(x) in the form (x − h)2 + k. [2]

(b) Hence describe f as a transformation of y = x2. [2]

(c) The graph of f is stretched vertically by factor 2 and then reflected in the x-axis to give g. Find g(x) and the coordinates of its vertex. [3]

Markscheme

Part	Answer	Marks
(a)	(x − 3)2 + 4	A2
(b)	translation by the vector (3, 4)	A2
(c)	g(x) = −2((x − 3)2 + 4)	M1
	= −2(x − 3)2 − 8	A1
	vertex (3, −8)	A1

Examiner's comments. Applying the stretch to the whole of f(x), including the constant, is essential in (c).

23 Common misconceptions and links

Students often think…	Actually…
y = f(x − 3) moves left	it moves right
y = f(2x) stretches by factor 2	factor ½
order never matters	stretch before translate within each direction
y = f(x) + 2 shifts right	it shifts up

How 2.11 connects to the rest of the course

Topic	Connection
2.6 Quadratics	vertex form
3.7 Trigonometric functions	amplitude, period, phase shift
2.15 Modulus functions (HL)	y = \|f(x)\| and y = f(\|x\|)
2.8 Exponential and log	asymptotes

24 Check your understanding

Section A: multiple choice

1. y = f(x) − 4 is a translation

A. 4 units down B. 4 units up C. 4 units left D. 4 units right

2. y = f(x/2) is a horizontal stretch by factor

A. ½ B. −2 C. 2 D. 4

3. The image of (3, 5) under y = f(−x) is

A. (3, −5) B. (−3, 5) C. (−3, −5) D. (5, 3)

4. y = −f(x) is a reflection in the

A. y-axis B. line y = x C. origin D. x-axis

5. y = sin 3x has period

A. 6π B. 2π/3 C. 3 D. π/3

Section B: structured

6. The point (2, −1) lies on y = f(x). Find its image on y = 4 − 2f(x + 3). [2]

7. Describe y = (2x − 6)2 as a transformation of y = x2. [2]

8. Write the equation after y = ex is reflected in the y-axis and translated 3 units up. [2]

Answers

1. A 2. C 3. B 4. D 5. B

6. (−1, 6).

7. Horizontal stretch factor ½, then translation 3 units right (or translation 6 right, then stretch).

8. y = e−x + 3.

25 Summary and topic map

Idea	Remember
Translations	f(x) + b; f(x − a): vector (a, b)
Reflections	−f(x) in x-axis; f(−x) in y-axis
Vertical stretch	pf(x): factor p
Horizontal stretch	f(qx): factor 1/q
Order	stretch/reflect before translate in each direction
Points and asymptotes	transform each coordinate

Before the exam, make sure you…

•use precise transformation language
•factorize inside brackets before describing
•apply transformations to asymptotes
•check with a point

Transformations of Graphs on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 7. Topic map for Transformations of Graphs.

26 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. translation by (2, 3)

2. y = ex−1 − 2

3. (1, 1)

Practice exercise 2

1. (2, 3); (−2, −3)

2. y = ln(−x), x < 0

3. even functions

Practice exercise 3

1. (6, 8); (2, 4)

2. horizontal stretch scale factor 2; period 4π

3. −1 and 3

Practice exercise 4

1. translation 1 right; vertical stretch factor 2; translation 5 up

2. y = f(3(x − 2)): horizontal stretch factor 1/3, then 2 right

3. y = 4 − f(x)

4. translation 1 left; vertical stretch factor 2 with reflection in x-axis; translation 3 up

Practice exercise 5

1. (−2, −1)

2. (4, 3); (3, 2)

3. y = 2

Practice exercise 6

1. y = 1/(x − 3) − 1

2. y = (x/2)2 + 1 = x2/4 + 1

3. h = −1, k = 4, a = −2

Practice exercise 7

1. y = ex+1 + 3; y = 3

2. (−2, 5)

Practice exercise 8

1. horizontal stretch scale factor 1/3

2. −2 and 3

Practice exercise 9

1. f(−x) = −x3 + x = −f(x)

2. y = ln(−x), x < 0

Practice exercise 10

1. y = f((x − 3)/2)

2. horizontal stretch factor 2; reflection in the x-axis; translation 4 up

Practice exercise 11

1. (−1, 5)

2. (5, −5)

Practice exercise 12

1. x = 3, y = 2

2. y = 3

Practice exercise 13

1. asymptote y = −4; y-intercept −3.5; x-intercept x = 3

Practice exercise 14

1. 6π; 4

2. y = sin(x/2) − 1

Practice exercise 15

1. c = 2, a = 8

Practice exercise 16

1. reflect the part for 0 < x < 1 above the axis

2. 4

Practice exercise 17

1. translation by (3, −5)

2. y = f(−(x − 3)) = f(3 − x)

Practice exercise 18

1. 2x+3 = 8(2x): vertical stretch factor 8

revisionprep.comIB-style notes · question bank · mock papers