Transformation of graphs AI HL

IB Mathematics: Applications and Interpretation HL · Unit 2: Functions

Lesson 2.8 · Transformations of Graphs

Translations, reflections, stretches and combinations of transformations.

Faith connection

The apostle John writes, “let us not love in word, neither in tongue; but in deed and in truth” (1 John 3:18, KJV). Love becomes visible through our actions. In this lesson, a change to a function’s rule becomes visible as a change in its graph. As you learn to connect a rule with its effect, consider ways to make your care for others visible. Come to Jesus, our Savior and Lord, and ask him to help you love others in deed and in truth.

Learning objectives

By the end of this lesson, you should be able to:

translate a graph horizontally and vertically;
reflect a graph in the $x$ -axis or $y$ -axis;
apply vertical and horizontal stretches;
track points, domains and ranges through a transformation;
describe composite transformations in the correct order;
use transformations to interpret mathematical models.

1. Begin with a known graph

Suppose the graph of $y=f(x)$ contains a point $(u,v)$ . A transformation changes its position or shape according to a rule. Tracking known points is often the most reliable way to sketch the new graph.

Worked example · Parent graph.

The graph of $f(x)=x^2$ contains $(-2,4)$ , $(0,0)$ and $(2,4)$ . Its vertex is $(0,0)$ , its axis of symmetry is $x=0$ , its domain is all real numbers, and its range is $y\geq0$ . We can use these features to recognise its transformations.

2. Translations

A translation moves every point by the same amount.

New graph	Movement from $y=f(x)$	Point mapping
$y=f(x)+b$	Up $b$ units when $b>0$	$(u,v)\mapsto(u,v+b)$
$y=f(x-a)$	Right $a$ units when $a>0$	$(u,v)\mapsto(u+a,v)$
$y=f(x-a)+b$	Right $a$ and up $b$	$(u,v)\mapsto(u+a,v+b)$

Negative values reverse the stated direction. A translation by the vector $\begin{pmatrix}3\\-2\end{pmatrix}$ moves the graph $3$ units right and $2$ units down.

Worked example · Translating a parabola.

Translate $y=x^2$ by $\begin{pmatrix}3\\-2\end{pmatrix}$ . The new equation is:

$y=(x-3)^2-2$

The vertex moves from $(0,0)$ to $(3,-2)$ . The point $(2,4)$ moves to $(5,2)$ . The new range is $y\geq-2$ .

Remember: The horizontal sign looks opposite to the direction of movement: $f(x-3)$ moves right by $3$ .

3. Reflections

A negative sign outside the function changes each output’s sign. A negative sign inside changes each input’s sign.

New graph	Reflection	Point mapping
$y=-f(x)$	In the $x$ -axis	$(u,v)\mapsto(u,-v)$
$y=f(-x)$	In the $y$ -axis	$(u,v)\mapsto(-u,v)$

Worked example · Reflecting a cubic graph.

Let $f(x)=(x-1)^3+2$ . The point $(1,2)$ lies on its graph. Under reflection in the $x$ -axis:

$y=-f(x)=-(x-1)^3-2$

The point becomes $(1,-2)$ . Under reflection in the $y$ -axis:

$y=f(-x)=(-x-1)^3+2$

The original point $(1,2)$ becomes $(-1,2)$ .

If a graph is already symmetric about the relevant axis, its reflection may look unchanged. For example, $f(x)=x^2$ satisfies $f(-x)=f(x)$ .

4. Vertical stretches

For $p>0$ , the graph of $y=pf(x)$ multiplies every output by $p$ . The $x$ -coordinates stay the same:

$(u,v)\mapsto(u,pv)$

If $p>1$ , the graph stretches vertically. If $0<p<1$ , it is compressed vertically. Points on the $x$ -axis stay fixed.

Worked example · Vertical stretch.

Let $f(x)=x^2-4$ . The graph contains $(1,-3)$ and has zeros at $x=-2$ and $x=2$ . For $y=3f(x)=3x^2-12$ , the point $(1,-3)$ moves to $(1,-9)$ ; the zeros remain at $-2$ and $2$ . Its vertex moves from $(0,-4)$ to $(0,-12)$ .

If the multiplier is negative, combine a vertical stretch by its absolute value with a reflection in the $x$ -axis.

5. Horizontal stretches

For $q>0$ , the graph of $y=f(qx)$ changes each horizontal coordinate by the factor $1/q$ :

$(u,v)\mapsto\left(\frac{u}{q},v\right)$

If $q>1$ , the graph is compressed horizontally. If $0<q<1$ , it stretches horizontally. Points on the $y$ -axis stay fixed.

Worked example · Horizontal compression.

Let $f(x)=\sqrt{x-2}$ . The graph begins at $(2,0)$ and contains $(6,2)$ . For $y=f(2x)=\sqrt{2x-2}$ , divide the old $x$ -coordinates by $2$ :

$(2,0)$ moves to $(1,0)$ , and $(6,2)$ moves to $(3,2)$ . The new domain is $x\geq1$ .

If the inside multiplier is negative, a reflection in the $y$ -axis is also involved. Check the full expression before describing the result.

6. Track domains, ranges and asymptotes

Translations and stretches move important graph features along with the points. For example, a vertical asymptote $x=u$ follows the horizontal point mapping, while a horizontal asymptote $y=v$ follows the vertical mapping.

Worked example · Reciprocal graph.

Start with $f(x)=1/x$ , whose asymptotes are $x=0$ and $y=0$ . Consider:

$g(x)=\frac{2}{x-3}+4=2f(x-3)+4$

Move the original graph right $3$ units, stretch its outputs by factor $2$ , then move it up $4$ units. Its vertical asymptote is $x=3$ , its horizontal asymptote is $y=4$ , its domain is $x\neq3$ , and its range is $y\neq4$ .

7. Composite transformations and their order

Several transformations can be combined. Some orders give different final graphs, especially when a stretch and a translation act in the same direction. Describe a sequence that actually produces the stated equation.

Worked example · Stretch, then translate.

Starting with $y=x^2$ , obtain $y=3x^2+2$ by stretching vertically by factor $3$ and then translating up $2$ :

$x^2\ \longrightarrow\ 3x^2\ \longrightarrow\ 3x^2+2$

If you translate up $2$ first and then stretch, you obtain $3(x^2+2)=3x^2+6$ , a different graph.

Worked example · Horizontal and vertical changes.

Starting with $y=\sin x$ , obtain $y=4\sin(2x)$ by compressing horizontally by factor $1/2$ and stretching vertically by factor $4$ . In degree mode, the period changes from $360^\circ$ to $180^\circ$ , and the amplitude changes from $1$ to $4$ .

Worked example · Mapping a known point.

Suppose $(2,5)$ lies on $y=f(x)$ . For $g(x)=-2f(3(x-4))+7$ , match the inside input to the old input $2$ :

$3(x-4)=2\quad\Longrightarrow\quad x=\frac{14}{3}$

The new output is $-2(5)+7=-3$ . Therefore $\left(\frac{14}{3},-3\right)$ lies on the graph of $g$ .

8. Transformations in a model

Changes to a graph can describe changes to a real situation. Interpret the units before describing a translation or stretch.

Worked example · Delayed production.

Let $P(t)$ be a factory’s output, in items per day, $t$ days after a production plan begins. A second factory follows the same pattern but begins $4$ days later and produces twice as many items on corresponding days. Its model is:

$Q(t)=2P(t-4)$

The replacement $t-4$ represents a delay of $4$ days. The outer factor $2$ doubles the output. If the original model applies for $0\leq t\leq20$ , this translated model applies for $4\leq t\leq24$ .

Calculator and sketching tips

Graph the original function and its transformed version on the same axes to check direction and scale.
Mark transformed intercepts, vertices, endpoints and asymptotes when applicable.
When finding a transformed point, work from the equation or map its coordinates; avoid estimating by eye.
Use degree mode for sinusoidal examples whose periods are given in degrees.
Restrict the graph to the physically meaningful domain in a modelling question.

IB-style Paper 1 · Skill practice

Question 1 · Describe transformations

Describe a sequence of transformations taking $y=x^2$ to $y=-2(x+3)^2+5$ . State the new vertex and range.

Show solution

Translate left $3$ , stretch vertically by factor $2$ , reflect in the $x$ -axis, then translate up $5$ . The vertex is $(-3,5)$ and the range is $y\leq5$ . The vertical stretch and reflection can be performed in either order here.

Question 2 · Map a point

The point $(6,-4)$ lies on $y=f(x)$ . Find the corresponding point on $y=3f(2(x-1))-5$ .

Show solution

For the original input $6$ , solve $2(x-1)=6$ to obtain $x=4$ . The new output is $3(-4)-5=-17$ . The point is $(4,-17)$ .

Question 3 · Domain and range

Let $f(x)=\sqrt{x}$ , with domain $x\geq0$ . For $g(x)=2\sqrt{3-x}-4$ , state the domain and range, and identify the transformed endpoint.

Show solution

Require $3-x\geq0$ , so the domain is $x\leq3$ . Since $\sqrt{3-x}\geq0$ , the range is $y\geq-4$ . The endpoint is $(3,-4)$ .

Question 4 · Asymptotes

Starting from $f(x)=1/x$ , describe the transformations leading to $g(x)=-3/(x+2)-1$ . State the asymptotes and the domain and range.

Show solution

Translate left $2$ , stretch vertically by factor $3$ and reflect in the $x$ -axis, then translate down $1$ . The vertical asymptote is $x=-2$ and the horizontal asymptote is $y=-1$ . The domain is $x\neq-2$ and the range is $y\neq-1$ .

Question 5 · Order matters

Starting with $y=f(x)$ , write the equation obtained by:

translating up $4$ and then stretching vertically by factor $3$ ;
stretching vertically by factor $3$ and then translating up $4$ .

Show solution

Part (a) gives $y=3(f(x)+4)=3f(x)+12$ . Part (b) gives $y=3f(x)+4$ . The vertical positions differ by $8$ units.

Question 6 · Transforming a sine graph

In degree mode, consider $g(x)=5\sin(3x)-2$ . State the amplitude, period, equation of the midline, maximum and minimum values. Describe how $y=\sin x$ was transformed.

Show solution

The amplitude is $5$ , the period is $360^\circ/3=120^\circ$ , and the midline is $y=-2$ . The maximum is $3$ and the minimum is $-7$ . Compress horizontally by factor $1/3$ , stretch vertically by factor $5$ , and translate down $2$ .

Applications and interpretation

Application 1 · A delayed delivery schedule

A company models daily deliveries by $D(t)=20+4t$ for $0\leq t\leq10$ , where $t$ is days after a schedule starts. A second route begins $3$ days later and handles $1.5$ times the deliveries on corresponding days.

Write a model $R(t)$ for the second route.
State its domain.
Find $R(7)$ and interpret the result.

Show solution

$R(t)=1.5D(t-3)=1.5[20+4(t-3)]$ for $3\leq t\leq13$ . At $t=7$ , $R(7)=1.5D(4)=1.5(36)=54$ . The model predicts $54$ deliveries on day $7$ after the first schedule began.

Application 2 · A new Ferris wheel

A wheel’s passenger height is modelled in metres by $h(t)=18-16\cos(45t)$ for $0\leq t\leq8$ minutes, with the calculator in degree mode. A replacement wheel has twice the radius, the same midline height and takes twice as long for one revolution. Both rides begin at their lowest point.

Write a model $H(t)$ for the replacement wheel.
State its period and range.
Explain the transformations from $h$ to $H$ .

Show solution

The original amplitude is $16$ m and period is $360/45=8$ minutes. The new amplitude is $32$ m and period is $16$ minutes. Keeping the midline at $18$ m gives:

$H(t)=18-32\cos(22.5t),\qquad 0\leq t\leq16$

The mathematical range is $-14\leq H(t)\leq50$ m. A negative height is physically impossible, so a wheel with twice the radius cannot keep this midline and remain above ground. The proposed replacement requires a higher centre. This check exposes an unrealistic design assumption.

Application 3 · Cost after a policy change

A transport provider models a fare by $F(d)=3+0.8d$ euros for journeys of $0\leq d\leq40$ km. A new policy adds €2 to every fare and doubles the per-kilometre part, while retaining the original €3 starting charge.

Show that the new fare is $G(d)=5+1.6d$ .
Express $G$ using $F$ .
Explain why simply writing $G(d)=2F(d)+2$ would give the wrong policy.

Show solution

The new fixed charge is $3+2=5$ euros, and the distance rate is $2(0.8)=1.6$ euros per kilometre, so $G(d)=5+1.6d$ . Since $F(2d)=3+1.6d$ , one expression is $G(d)=F(2d)+2$ , for $0\leq d\leq20$ if the original model’s stated domain must be respected. Alternatively, $G(d)=2F(d)-1$ for $0\leq d\leq40$ . Writing $2F(d)+2$ also doubles the original €3 starting charge, resulting in an incorrect fixed amount of €8.

Application 4 · Comparing periodic signals

A sensor records $S(t)=3\sin(30t)+5$ , where $t$ is measured in seconds and angles are in degrees. An upgraded sensor records the same underlying pattern with twice the amplitude, half the period and a baseline that is $4$ units higher.

Construct an equation for the upgraded signal.
State its maximum and minimum readings and its period.
Find its reading at $t=1$ second.

Show solution

The original amplitude is $3$ , baseline is $5$ , and period is $360/30=12$ seconds. The upgraded amplitude is $6$ , baseline is $9$ , and period is $6$ seconds. Thus $U(t)=6\sin(60t)+9$ . Its maximum is $15$ , minimum is $3$ , and $U(1)=6\sin(60^\circ)+9=9+3\sqrt3\approx14.2$ .

Paper 1 checklist

For an outside change, track the output; for an inside change, track the input. Check the direction of a horizontal translation carefully. Map a known point to verify the result. When several changes are involved, state an order that produces the exact equation. In applications, check that the transformed graph remains meaningful.

Lesson 2.8 · Transformations of Graphs

Faith connection

Learning objectives

1. Begin with a known graph

2. Translations

3. Reflections

4. Vertical stretches

5. Horizontal stretches

6. Track domains, ranges and asymptotes

7. Composite transformations and their order

8. Transformations in a model

Calculator and sketching tips

IB-style Paper 1 · Skill practice

Question 1 · Describe transformations

Question 2 · Map a point

Question 3 · Domain and range

Question 4 · Asymptotes

Question 5 · Order matters

Question 6 · Transforming a sine graph

Applications and interpretation

Application 1 · A delayed delivery schedule

Application 2 · A new Ferris wheel

Application 3 · Cost after a policy change

Application 4 · Comparing periodic signals

Paper 1 checklist

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IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Transformation of graphs AI HL

Faith connection

Learning objectives

1. Begin with a known graph

2. Translations

3. Reflections

4. Vertical stretches

5. Horizontal stretches

6. Track domains, ranges and asymptotes

7. Composite transformations and their order

8. Transformations in a model

Calculator and sketching tips

IB-style Paper 1 · Skill practice

Question 1 · Describe transformations

Question 2 · Map a point

Question 3 · Domain and range

Question 4 · Asymptotes

Question 5 · Order matters

Question 6 · Transforming a sine graph

Applications and interpretation

Application 1 · A delayed delivery schedule

Application 2 · A new Ferris wheel

Application 3 · Cost after a policy change

Application 4 · Comparing periodic signals

Paper 1 checklist

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