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Topic 2.08 · HL only

Transformations of graphs (translation, reflections, stretches): notes and practice questions

Summary
  • Transformations alter position, orientation, or size; use "stretch" with a scale factor.
  • Horizontal translation by vector (a0)\begin{pmatrix} a \\ 0 \end{pmatrix}: y=f(x−a)y = f(x - a), (x,y)→(x+a,y)(x, y) \to (x + a, y), VA: x=k→x=k+ax=k \to x=k+a.
  • Vertical translation by vector (0b)\begin{pmatrix} 0 \\ b \end{pmatrix}: y=f(x)+by = f(x) + b, (x,y)→(x,y+b)(x, y) \to (x, y + b), HA: y=k→y=k+by=k \to y=k+b.
  • Reflection in x-axis: y=−f(x)y = -f(x), (x,y)→(x,−y)(x, y) \to (x, -y), HA: y=k→y=−ky=k \to y=-k.
  • Reflection in y-axis: y=f(−x)y = f(-x), (x,y)→(−x,y)(x, y) \to (-x, y), VA: x=k→x=−kx=k \to x=-k.
  • Vertical stretch (parallel to y-axis) by scale factor pp: y=pf(x)y = p f(x), (x,y)→(x,py)(x, y) \to (x, py), HA: y=k→y=pky=k \to y=pk.
  • Horizontal stretch (parallel to x-axis) by scale factor qq: y=f(xq)y = f\left(\frac{x}{q}\right), (x,y)→(qx,y)(x, y) \to (qx, y), VA: x=k→x=qkx=k \to x=qk.
  • Composite Vertical Transformations (y=af(x)+by = a f(x) + b):
  • First: Vertical stretch by scale factor aa (reflect in x-axis if a<0a<0).
  • Then: Vertical translation by vector (0b)\begin{pmatrix} 0 \\ b \end{pmatrix}.
  • Composite Horizontal Transformations (y=f(ax+b)y = f(ax + b)):
  • First: Horizontal translation by vector (−b0)\begin{pmatrix} -b \\ 0 \end{pmatrix}.
  • Then: Horizontal stretch by scale factor 1a\frac{1}{a} (reflect in y-axis if a<0a<0).

How it is examined

Note the horizontal stretch convention: scale factor 1q\frac{1}{q} for y=f(qx)y = f(qx), so y=f(2x)y = f(2x) is a stretch of scale factor 12\frac{1}{2}, not 2. Getting the reciprocal the wrong way round is the single most common error here. Order matters when a stretch and a translation act in the same direction, and questions do exploit that. Matrix transformations of points are a separate subtopic (AHL 3.9) and use different machinery, so keep the two apart.

Key ideas
  • Translations: y=f(x)+by = f(x) + b and y=f(x−a)y = f(x - a).
  • Reflections: in the xx-axis, y=−f(x)y = -f(x), and in the yy-axis, y=f(−x)y = f(-x).
  • Vertical stretch with scale factor pp: y=p f(x)y = p\,f(x).
  • Horizontal stretch with scale factor 1q\dfrac{1}{q}: y=f(qx)y = f(qx).

Linking questions

  • Other contexts: translating curves to reduce rounding errors for large values.
  • Links to other subjects: shifting supply and demand curves (economics), electromagnetic induction (physics).
  • TOK: is mathematics independent of culture? To what extent are we aware of the impact of culture on what we believe or know?

Practice questions

22 questions · 1 easy · 14 medium · 7 hard
Showing 20 of 20

Question 1

EasyPaper 1 · calculator7 marks

Let f(x)=log⁡2x,f(x) = \log_{2}x, which is translated by (mn)\begin{pmatrix} m \\ n \end{pmatrix} such that it can pass through the points A(4,1) and B(5,2).

Find the value of mm and nn.

Question 2

MediumPaper 1 · calculator7 marks

A digital artist is designing a new fractal pattern. The initial curve for the pattern is defined by the function f(x)=log⁡2xf(x) = \log_2 x. To create a variation, the artist applies a transformation to f(x)f(x), resulting in a new curve g(x)g(x). This transformation involves a horizontal translation of pp units and a vertical translation of qq units.

The new curve g(x)g(x) is observed to pass through the points (6,−2)(6, -2) and (12,0)(12, 0).

Find the value of pp and the value of qq.

Question 3

HardPaper 1 · calculator9 marks
(a)(i)

A landscape architect is designing a decorative water channel. The cross-section of the channel is modelled by the function f(x)=x3−4xf(x) = x^3 - 4x, for −2≤x≤2-2 \leq x \leq 2. The shaded region, RR, represents the cross-sectional area of the channel, bounded by the graph of y=f(x)y = f(x) and the xx-axis.

Graph of y=f(x) = x^3-4x from -2 to 2, with shaded region R between the curve and x-axis. The curve is above the x-axis for x in [-2,0 and below for x in [0,2].]

Write down an integral that represents the area of RR.

[2]
(a)(ii)

Find the area of RR.

[2]
(b)

The architect considers a modified design, where the cross-section is given by g(x)=0.5f(x+1)g(x) = 0.5f(x+1).

On the following set of axes, the graph of y=f(x)y = f(x) has been drawn. On the same set of axes, sketch the graph of y=g(x)y = g(x).

Axes with graph of y=f(x)=x^3-4x drawn from -3 to 3, ready for sketch of y=g(x). The graph of f(x) has roots at -2, 0, 2 and local max/min at approx (-1.15, 3.08) and (1.15, -3.08).
[2]
(c)

The region RR (the original cross-section) is rotated through 2π2\pi radians about the xx-axis to form a three-dimensional decorative element. Find the volume of this element.

[3]

Question 4

MediumPaper 1 · calculator4 marks
(a)

The graph of y=f(x)y = f(x) represents the concentration of a certain chemical in a solution (in mol/L) at time xx (in hours). The graph passes through the points (1,10)(1, 10) and (3,4)(3, 4), and has a horizontal asymptote at y=2y = 2.

Let g(x)=3f(x+1)−5g(x) = 3f(x + 1) - 5 represent the concentration in a different experiment.

Find g(0)g(0).

[2]
(b)

On a new set of axes, sketch the graph of y=g(x)y = g(x), clearly indicating its horizontal asymptote and the yy-intercept. You do not need to show the graph of f(x)f(x).

[2]

Question 5

HardPaper 1 · calculator11 marks
(a)

(a) A designer is creating a prototype for a decorative vase. The cross-section of the vase can be modelled by the function f(x)=x4−x2f(x) = x\sqrt{4-x^2}, for −2≤x≤2-2 \le x \le 2.

Sketch the graph of y=f(x)y = f(x) on the following pair of axes.

graph of y=f(x) on axes from -3 to 3 for x and -3 to 3 for y. The curve passes through the origin, has a maximum in the first quadrant and a minimum in the third quadrant. The curve is symmetric about the origin. The endpoints are at x=-2 and x=2. The maximum is at x=sqrt(2) and y=2, and the minimum is at x=-sqrt(2) and y=-2. The curve is smooth. The x-axis is labelled from -3 to 3 and the y-axis is labelled from -3 to 3.
[2]
(b)(i)

(b) The region enclosed by the graph of y=f(x)y = f(x) and the x-axis is rotated 360∘360^\circ about the x-axis to form the body of the vase.

(i) Write down an integral that represents the volume of this vase.

[2]
(b)(ii)

(ii) Calculate the value of this integral.

[4]
(c)

(c) The designer decides to create a new, larger version of the vase, y=g(x)y = g(x), by applying the following transformations to the original cross-section y=f(x)y = f(x):

  • A horizontal stretch by a scale factor of 3, parallel to the x-axis.
  • A vertical stretch by a scale factor of 0.75, parallel to the y-axis.

Find the volume of this new vase.

[3]

Question 6

MediumPaper 1 · calculator7 marks
(a)

A civil engineer is designing a new road. A preliminary section of the road is modelled by a straight line with equation y=kx+dy = kx + d.

Find the vectors a\mathbf{a} and b\mathbf{b} such that the equation of the line can be expressed in vector form r=a+λb\mathbf{r} = \mathbf{a} + \lambda \mathbf{b} in terms of kk and/or dd.

[2]
(b)

Before construction, a ground transformation is applied to the design. This transformation is described by the matrix T=(4263)T = \begin{pmatrix} 4 & 2 \\ 6 & 3 \end{pmatrix}.

Calculate the value of det⁡T\det T.

[1]
(c)

The preliminary road section y=kx+dy = kx + d (where k≠−2k \neq -2) undergoes the transformation described by matrix TT.

Show that the equation of the resulting transformed path does not depend on kk or dd.

[4]

Question 7

HardPaper 1 · calculator8 marks
(a)(i)

A landscape architect is designing a large decorative fountain for a new public park. The outer structure of the fountain consists of a cylindrical base topped by a conical section. The inner part of the fountain, which holds the water, is a hollow space created by rotating a specific curve around the vertical axis.

The shape of the inner hollow is based on a transformation of the graph y=−x3y = -x^3. The curve that defines the profile of the inner hollow is given by y=40−0.064x3y = 40 - 0.064x^3. This transformation involves a vertical translation of aa units and a stretch parallel to the x-axis with a scale factor of bb.

(a.i) Write down the value of aa.

[1]
(a)(ii)

(a.ii) Find the value of bb.

[2]
(b)

The cylindrical base of the fountain has a radius of 1515 m and a height of 55 m. The conical section on top has the same base radius of 1515 m and a height of 88 m. The inner hollow, described by the curve y=40−0.064x3y = 40 - 0.064x^3, extends from the base of the fountain (y=0y=0) up to the total height of the outer structure.

Find the volume of the solid material that makes up the fountain (i.e., the volume of the outer structure minus the volume of the inner hollow).

[5]

Question 8

MediumPaper 1 · calculator5 marks
(a)

A civil engineer is designing a section of a new roller coaster track. She collects data points from a preliminary design sketch to model a specific curved section.

Here are the coordinates of five points on the track's profile:

xxyy
29.70
410.77
611.33
88.70
10-0.05

The engineer thinks a cubic curve will be a good model for this section of the track.

Find the equation of the cubic regression curve for this data.

[2]
(b)

For another section of the roller coaster, the engineer first creates a small-scale model. The equation for the profile of this model section is given by y=0.02x3−0.3x2+1.5x+5y = 0.02x^3 - 0.3x^2 + 1.5x + 5.

The full-size roller coaster track will be an enlargement of this model, with a scale factor of 5, centered at the origin (0,0).

Determine the equation of the cubic curve that models the full-size roller coaster track.

[3]

Question 9

HardPaper 2 · calculator21 marks
(a)(i)

The "SkyGazer" is a new observation wheel in a city park. The wheel has a diameter of 7070 m. To begin the ride, a passenger enters a capsule at the lowest point on the wheel, which is 33 m above the ground. A ride consists of multiple revolutions, and the wheel makes 22 revolutions per minute.

The height of a capsule above the ground, hh, measured in metres, during a ride on the SkyGazer can be modelled by the function h(t)=−acos⁡(bt)+dh(t) = -a \cos (bt) + d, where tt is the time, in seconds, since a passenger began their ride.

(a) Calculate the value of

(i) aa;

[2]
(a)(ii)

(a)(ii) bb;

[3]
(a)(iii)

(a)(iii) dd.

[2]
(b)

(b) A ride on the SkyGazer lasts for 1010 minutes in total.

Calculate the number of revolutions of the wheel per ride.

[2]
(c)(i)

(c) For exactly one ride on the SkyGazer, suggest

(i) an appropriate domain for h(t)h(t);

[2]
(c)(ii)

(c)(ii) an appropriate range for h(t)h(t).

[2]
(d)

(d) A 2020 metre-tall building stands on the horizontal ground next to the SkyGazer.

By considering the graph of h(t)h(t), determine the length of time during one revolution of the wheel for which the capsule is higher than the building.

[5]
(e)(i)

(e) There is a plan to relocate the SkyGazer onto a taller platform which will increase the maximum height of the wheel to 7575 m. This will change the value of one parameter, aa, bb or dd, found in part (a).

(i) Identify which parameter will change.

[1]
(e)(ii)

(e)(ii) Find the new value of the parameter identified in part (e)(i).

[2]

Question 10

MediumPaper 1 · calculator10 marks
(a)

A drone is launched vertically upwards from a platform. Its height, hh metres above the ground, tt seconds after launch, is given by the formula h=c+ut−5t2h = c + ut - 5t^2, where cc is the initial height of the platform and uu is the initial upward velocity.

The drone is launched from a platform 2.52.5 m above the ground with an initial upward velocity of 4040 ms−1^{-1}.

Calculate the time taken for the drone to return to the height of the launch platform.

[2]
(b)

Find the time it takes for the drone to reach its maximum height.

[2]
(c)

Calculate the maximum height of the drone.

[2]
(d)

Another drone operator, Liam, launches an identical drone with the same initial upward velocity, but from a platform 3.03.0 m above the ground.

By considering the difference in the two height-time graphs for the drones, write down the answers to parts (a), (b), and (c) for Liam's drone.

[4]

Question 11

HardPaper 2 · calculator28 marks
(a)(i)

(a) (i) Consider the function f(x)=x3f(x) = x^3. Find f′(x)f'(x).

[1]
(a)(ii)

(ii) The first section of the stone wall's profile is given by y=f(x)y = f(x) for 0≤x≤0.50 \le x \le 0.5. A straight glass panel is to be installed tangent to this section of the wall at the point where x=0.5x=0.5. Find the equation of this tangent line.

[3]
(b)

The full profile of the stone wall, F(x)F(x), is defined by:

F(x)={x30≤x≤0.50.75x−0.250.5<x≤1.0F(x) = \begin{cases} x^3 & 0 \le x \le 0.5 \\ 0.75x - 0.25 & 0.5 < x \le 1.0 \end{cases}

A smaller, decorative stone insert is designed using a transformation of F(x)F(x). The graph of G(x)G(x) is obtained from the graph of F(x)F(x) by:

  • a stretch scale factor of 12\frac{1}{2} in the xx direction,
  • followed by a stretch scale factor of 12\frac{1}{2} in the yy direction,
  • followed by a translation of 0.50.5 units to the right.

Point P lies on the graph of F(x)F(x) and has coordinates (1.0,0.5)(1.0, 0.5). Point Q is the image of P under the given transformations and has coordinates (qx,qy)(q_x, q_y).

Find the value of qxq_x and the value of qyq_y.

[3]
(c)(i)

The piecewise function G(x)G(x) is given by

G(x)={k(x)c≤x≤dmx+nd<x≤qxG(x) = \begin{cases} k(x) & c \le x \le d \\ mx + n & d < x \le q_x \end{cases}

(c) Find

(i) an expression for k(x)k(x).

[4]
(c)(ii)

(ii) the value of dd.

[2]
(c)(iii)

(iii) the value of nn.

[3]
(d)(i)

(d) (i) Calculate the total area of the profile of the stone wall, enclosed by y=F(x)y = F(x), the xx-axis, and the line x=1.0x = 1.0.

[7]
(d)(ii)

The decorative insert G(x)G(x) is placed within the main wall profile F(x)F(x). The region of the main wall profile that is not covered by the insert is to be painted a contrasting colour. This region is bounded by y=F(x)y=F(x), the xx-axis, and the lines x=0x=0 and x=1x=1, excluding the area under G(x)G(x) from x=0.5x=0.5 to x=1.0x=1.0. Find the area of this region.

[5]

Question 12

MediumPaper 2 · calculator12 marks
(a)

Consider the function f(x)=∣x∣f(x) = |x|.

(a) The graph of the function is translated vertically to define a new function g(x)=f(x)+kg(x) = f(x) + k. Find the value of kk such that the graph of g(x)g(x) passes through the point (2,7)(2, 7).

[2]
(b)

(b) The graph of the function is translated horizontally to define a new function h(x)=f(x−l)h(x) = f(x - l). Find the values of ll such that the graph of h(x)h(x) passes through the point (5,3)(5, 3).

[3]
(c)

(c) The graph of the function is translated to define a new function m(x)=f(x−r)+sm(x) = f(x - r) + s. Find the values of rr and ss such that the minimum point of m(x)m(x) is (−1,6)(-1, 6).

[3]
(d)

(d) The graph of the function is transformed to define a new function n(x)=af(x−b)+cn(x) = af(x - b) + c. Find the values of a,ba, b and cc such that the maximum point of n(x)n(x) is (4,10)(4, 10) and the graph passes through the point (6,6)(6, 6).

[4]

Question 13

HardPaper 1 · calculator13 marks
(a)

A design studio is manufacturing a decorative wall sconce. The cross-sectional profile of the sconce is defined by two mathematical curves. The sconce is formed by rotating the region between these curves through π\pi radians about the xx-axis, creating a flat rear surface that mounts flush against a wall. All linear dimensions are in centimetres.

The curve of the outer profile is given by f(x)=6cos⁡(π12x)f(x) = 6 \cos\left(\frac{\pi}{12}x\right), for 0≤x≤60 \le x \le 6.

The curve of the inner profile, g(x)g(x), is formed by translating the graph of ff by 0.60.6 units to the left and 1.21.2 units down.

(a) Write down an expression for g(x)g(x).

[2]
(b)

(b) The curve g(x)g(x) intersects the xx-axis at (a,0)(a, 0), where 0≤x≤a0 \le x \le a defines the inner boundary of the sconce.

Find the value of aa. Give your answer to three significant figures.

[3]
(c)(i)

(c.i) Write down an expression for the volume of the solid formed.

[5]
(c)(ii)

(c.ii) Hence find the volume of material used in the sconce. Give your answer to three significant figures.

[3]

Question 14

MediumPaper 1 · calculator9 marks
(a)

The depth of water, DD metres, at the entrance to a harbour can be modelled by the function D(t)=A+Bcos⁡(Ct)D(t) = A + B\cos(Ct), where tt is the time in hours after midnight and CC is a constant in radians per hour. The graph below shows the depth of water over a 24-hour period.

Graph of water depth over 24 hours. The x-axis represents time (t) in hours from 0 to 24. The y-axis represents depth (D) in metres. The graph is a cosine wave. It starts at a maximum depth of 12m at t=0, reaches a minimum depth of 4m at t=6, a maximum of 12m at t=12, a minimum of 4m at t=18, and a maximum of 12m at t=24. The midline is at D=8.

(a) On the same axes, sketch the graph of the function D2(t)=(A−1)+B2cos⁡(Ct)D_2(t) = (A-1) + \frac{B}{2}\cos(Ct).

[2]
(b)

(b) Determine the values of the constants AA, BB, and CC.

[7]

Question 15

HardPaper 1 · calculator13 marks
(a)

A design studio is developing a wall-mounted decorative bracket. The bracket is designed to mount flush against a flat vertical wall. The cross-sectional profile of the bracket in the xyx y-plane is defined by two mathematical curves, where all linear dimensions are measured in centimetres.

The curve of the outer profile is given by f(x)=8cos⁡(π16x)f(x) = 8 \cos\left(\frac{\pi}{16}x\right), for 0≤x≤80 \le x \le 8.

The curve of the inner profile, g(x)g(x), is formed by translating the graph of ff by 0.80.8 units to the left and 1.21.2 units downwards.

(a) Write down an expression for g(x)g(x).

[2]
(b)

The inner profile curve g(x)g(x) intersects the xx-axis at the point (a,0)(a, 0). The relevant portion of the inner profile is restricted to 0≤x≤a0 \le x \le a.

(b) Find the value of aa. Give your answer to three significant figures.

[3]
(c)(i)

The bracket is modelled by the solid formed when the region between the profiles is rotated through π\pi radians about the xx-axis. This region is bounded by f(x)f(x) from x=0x = 0 to x=8x = 8, the line x=0x = 0, the xx-axis, and g(x)g(x) from x=0x = 0 to x=ax = a.

(c.i) Write down an expression for the volume of the solid formed.

[5]
(c)(ii)

(c.ii) Hence find the volume of material used in the bracket. Give your answer to three significant figures.

[3]

Question 16

MediumPaper 2 · calculator5 marks
(a)

(a) The population PP of a rare species of orchid in a botanical garden, tt years after its introduction, is modelled by the logistic function P(t)=1201+e−0.5t+10P(t) = \frac{120}{1+e^{-0.5t+10}}.

Determine the time tt (in years) at which the rate of population growth of the orchid species is at its maximum.

[2]
(b)

(b) Hence, find the coordinates of the point of inflexion for a modified model P(t−3)P(t-3), which represents the population if the initial conditions were set 3 years earlier.

[3]

Question 17

MediumPaper 1 · calculator7 marks
(a)

(a) A Ferris wheel has a radius of 2020 metres, and its centre is 2222 metres above the ground. A passenger boards the wheel at its highest point at t=0t = 0 minutes. The height, HH metres, of the passenger above the ground at time tt minutes is modelled by the function H(t)=22+20cos⁡(π5t)H(t) = 22 + 20 \cos\left(\frac{\pi}{5}t\right).

Write down the amplitude of the Ferris wheel's motion.

[1]
(b)

(b) Calculate the minimum height above the ground a passenger reaches. Determine the first positive time, in minutes, this height occurs.

[4]
(c)

(c) Find the period of the Ferris wheel's rotation.

[2]

Question 18

MediumPaper 1 · calculator10 marks
(a)

A company monitors the daily temperature fluctuations in a specialized industrial oven. The temperature TT, in degrees Celsius, at time tt hours after midnight, is modelled by the function T(t)=A−Bcos⁡(Ct)T(t) = A - B \cos(Ct), where AA, BB, CC are positive constants. The graph below shows the temperature fluctuations over a 24-hour period.

Graph of oven temperature over 24 hours, showing a cosine wave starting at a minimum of 130 degrees Celsius at t=0, rising to a maximum of 170 degrees Celsius at t=12, and returning to 130 degrees Celsius at t=24. The midline is at 150 degrees Celsius.

On the same axes, sketch the graph of Tnew(t)=A+B2cos⁡(Ct)T_{\text{new}}(t) = A + \frac{B}{2} \cos(Ct).

[3]
(b)

Determine the values of the constants AA, BB and CC.

[7]

Question 19

MediumPaper 2 · calculator10 marks
(a)

The temperature, TT in degrees Celsius, in a controlled environment can be modelled by a sinusoidal function of time, hh hours after monitoring began. At h=100h=100 hours, the temperature reaches its maximum of 15∘C15^{\circ}C. At h=250h=250 hours, the temperature drops to its minimum of 5∘C5^{\circ}C. The monitoring takes place over 600600 hours.

(a) Find the equation of the function in the form T(h)=Acos⁡(B(h−C))+KT(h) = A \cos(B(h-C) ) + K, for h∈[0,600]h \in [0, 600].

[6]
(b)

During the monitoring period, the temperature T(h)=βT(h) = \beta is recorded exactly five times. Find the value of β\beta.

[4]

Question 20

MediumPaper 1 · calculator13 marks
(a)

A company is designing a decorative stand for a new line of luxury smart speakers. The profile of the stand's base is defined by two mathematical curves. The stand itself is formed by rotating the region between these curves through π\pi radians about the xx-axis. All dimensions are in centimetres.

The curve of the outer profile is given by f(x)=5cos⁡(π10x)f(x) = 5 \cos\left(\frac{\pi}{10}x\right), for 0≤x≤50 \le x \le 5.

The curve of the inner profile, g(x)g(x), is formed by translating the graph of ff by 0.40.4 units to the left and 0.60.6 units down.

(a) Write down an expression for g(x)g(x).

[2]
(b)

The inner profile curve g(x)g(x) intersects the xx-axis at (a,0)(a, 0). The relevant portion of g(x)g(x) for the stand is restricted to 0≤x≤a0 \le x \le a.

(b) Find the value of aa. Give your answer to three significant figures.

[3]
(c)(i)

The decorative stand is modelled by the solid formed when the region RR is rotated through π\pi radians about the xx-axis. The region RR is defined by the area under f(x)f(x) from x=0x=0 to x=5x=5, excluding the area under g(x)g(x) from x=0x=0 to x=ax=a.

(c.i) Write down an expression for the volume of the solid formed.

[5]
(c)(ii)

(c.ii) Hence find the volume of material used in the stand. Give your answer to three significant figures.

[3]

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What does Transformations of graphs (translation, reflections, stretches) cover in IB Maths AI?

Transformations alter position, orientation, or size; use "stretch" with a scale factor. Horizontal translation by vector beginpmatrix a \ 0 endpmatrix: y = f(x - a), (x, y) → (x + a, y), VA: x=k → x=k+a. Vertical translation by vector beginpmatrix 0 \ b endpmatrix: y = f(x) + b, (x, y) → (x, y + b), HA: y=k → y=k+b.

Is Transformations of graphs (translation, reflections, stretches) SL or HL?

Transformations of graphs (translation, reflections, stretches) is HL only. SL students are not examined on it.

How do I revise Transformations of graphs (translation, reflections, stretches) for IB Maths AI?

Start from the core idea: transformations alter position, orientation, or size; use "stretch" with a scale factor. In the exam: note the horizontal stretch convention: scale factor (1)/(q) for y = f(qx), so y = f(2x) is a stretch of scale factor (1)/(2), not 2. Getting the reciprocal the wrong way round is the single most common error here. Then practise exam-style questions, easiest first, writing out every step of your working before you check it.

How does FourtyFive help me practise Transformations of graphs (translation, reflections, stretches)?

FourtyFive has 22 Transformations of graphs (translation, reflections, stretches) questions. Every answer you write is marked mark by mark, IB-style, and you see where each mark was won or lost. Every part has a hint, the AI tutor helps you through the step you are stuck on, and your Study Profile picks what to practise next.

Is FourtyFive free for Transformations of graphs (translation, reflections, stretches) practice?

Yes. A free account gives you 50 marked answers a month, and you do not need a card to sign up.

Can I handwrite Transformations of graphs (translation, reflections, stretches) answers on an iPad?

Yes. In the FourtyFive iPad app you write your working by hand with Apple Pencil, the way you would on paper, and it is marked the same way.

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