Mathematical Methods · Units 3 & 4

Transformations of Graphs

Master transformations of graphs the easy way, with plain English intuition, the correct order to apply translations, dilations and reflections, the image of a point, mapping notation, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Bend a graph, slide it sideways, flip it over, stretch it tall. Every messy looking curve in this course is really just a friendly basic shape in disguise, pushed around the page. Transformations are the rules for that pushing. Learn to read them straight off an equation and the scariest function suddenly tells you exactly where it sits and which way it points.

The four moves, in plain English

Every transformation in Methods is one of three ideas, and one of them comes in two flavours.

A translation is a slide. The shape stays exactly the same size and the same way up, it just moves to a new spot. A dilation is a stretch or a squash, either taller or wider. A reflection is a flip across an axis, turning the picture into its mirror image.

The trick is knowing which part of the equation controls which move. Here is the one rule that unlocks everything.

So in $y = A\,f\big(b(x-c)\big) + d$ , the $A$ and $d$ live outside and act on $y$ in the obvious way, while the $b$ and $c$ live inside next to $x$ and act in reverse.

Reading each letter

Take the general transformed graph $y = A\,f\big(b(x-c)\big) + d$ and read it piece by piece.

The $A$ is a dilation by a factor of $|A|$ from the $x$ axis, a vertical stretch. If $A$ is negative it also reflects the graph in the $x$ axis. The $d$ is a translation of $d$ units up.

The $b$ is a dilation by a factor of $\tfrac{1}{b}$ from the $y$ axis, a horizontal stretch. Notice the reciprocal, because $b$ is an inside change. If $b$ is negative it also reflects the graph in the $y$ axis. The $c$ is a translation of $c$ units in the positive $x$ direction, again the opposite of the minus sign you see.

y = A\,f\big(b(x-c)\big) + d

The single biggest mark loser in real exams is sloppy language. The examiner reports say it plainly. You must name the factor, the axis, and the direction every single time. Say dilation by a factor of $2$ from the $x$ axis, not just “stretch by $2$ ”. For a vertical move say in the positive direction of the $y$ axis or up, never just “across”.

The basic parabola y = x² (grey, dashed) becomes y = 2(x−3)² + 1 (blue): a dilation by factor 2 from the x axis, then a translation 3 right and 1 up, which carries the turning point from (0, 0) to (3, 1).

Why the order can matter

Here is the part that trips up the most students in the exam. If you stretch first and then slide, you can land somewhere different from sliding first and then stretching. The safe rule for the standard form is this.

Apply dilations and reflections first, then translations last. That matches the way the rule is built, because the $A$ and $b$ act on the bare shape before the $c$ and $d$ shuffle it into place.

There is one freedom. A vertical translation can be slotted in at any stage without changing the answer, and so can be described anywhere in the sequence. But a horizontal translation must come after any horizontal dilation or reflection, otherwise the amount of the shift gets stretched too and your final position is wrong. When a question asks you to list a sequence, keep the reflections and dilations ahead of the matching translations and you will be safe.

The image of a point and mapping notation

You do not always need the full rule. Often you just need to know where one point ends up. This is where mapping notation earns its keep. A mapping is a compact instruction that takes any point $(x,y)$ and tells you its image.

For the general transformation above, the mapping is

(x,y) \to \left( \tfrac{1}{b}x + c,\; A\,y + d \right)

To find the image of a specific point, push its coordinates through the mapping. Treat the $x$ part and the $y$ part separately, since horizontal moves only touch $x$ and vertical moves only touch $y$ . A common exam blunder is to transform the wrong point, for example moving the turning point when the question fixed a different one, so always check you are mapping the point the question actually names.

Learn it, then fade the help

1 · I do — fully worked

Map a point through a transformation The graph $y = f(x)$ maps to $y = f(2x) - 4$ . The inside $2$ gives a horizontal dilation by factor $\tfrac12$ ( $x \to \tfrac12 x$ ) and the outside $-4$ gives $y \to y - 4$ , so the mapping is $(x,y) \to \left(\tfrac12 x,\; y - 4\right)$ . The image of $(6, 5)$ is $\left(\tfrac12 \times 6,\; 5 - 4\right) = (3, 1)$ .

2 · We do — your turn, guided

Same steps, your turn. The point $(2, 5)$ is sent by the mapping $(x, y) \to (x - 3,\; 2y)$ . Apply each part separately: image $= (2 - 3,\; 2 \times 5) = (?,\; 10)$ .

Check your step

$(-1, 10)$ — subtract $3$ from the $x$ coordinate and double the $y$ coordinate.

3 · You do — solo

No scaffolding now: $y = f(x)$ has a local minimum at $(1, -2)$ . Find the local minimum of $y = f(x - 3) + 5$ .

Reveal solution

The rule gives the mapping $(x, y) \to (x + 3,\; y + 5)$ , so the minimum moves to $(1 + 3,\; -2 + 5) = (4, 3)$ .

How to actually do it

When a question hands you words and wants a rule, or hands you a rule and wants words, follow the same short recipe.

Spot the dilations and reflections, the $A$ and $b$ . Apply them to the basic shape first.
Spot the translations, the $c$ and $d$ . Apply them last, horizontal before you forget the sign flip.
To move a point, write the mapping $(x,y) \to \left(\tfrac{1}{b}x + c,\; Ay + d\right)$ and substitute.
State every dilation with its factor and axis, and every translation with its amount and direction.

See this recipe in action in the Worked Examples tab, then test yourself in Try It.

Lock it in with active recall

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

Do changes outside the function act on

x

y

, and how do they behave?

y = 3f(x)

, what is the transformation?

y = f(4x)

, what is the transformation?

What translation turns

y = x^2

into

y = (x+4)^2 - 7

What order do you apply transformations in?

Write the mapping for

y = A\,f(b(x-c)) + d

Recall · Circular Functions

Why does

y = \cos\!\left(x - \tfrac{\pi}{4}\right)

shift the cosine right?

Recall · Inverse Functions

Reflecting a graph in the line

y = x

produces what?

Worked Example 1A sequence of transformations, from a real exam

Let $f:R\to R$ , $f(x)=e^x+e^{-x}$ and $g:R\to R$ , $g(x)=\tfrac{1}{2}f(2-x)$ . A possible sequence of transformations that maps $f$ to $g$ begins with a dilation of factor $\tfrac{1}{2}$ from the $x$ axis. State the remaining transformations.

1
Rewrite $g$ so the inside change is in the standard form. Since $g(x)=\tfrac12 f(2-x)=\tfrac12 f(-(x-2))$ , after the $\tfrac12$ vertical dilation we need $x\to-(x-2)$ .
$g(x)=\tfrac12 f\big(-(x-2)\big)$
2
The factor $-1$ inside is a reflection in the $y$ axis, and the $(x-2)$ is a translation $2$ units in the positive $x$ direction. Order the reflection before the horizontal translation.
$x\mapsto -x\ \text{then}\ x\mapsto x+2\ (\text{2 units right})$
3
Because $f$ is even, translating $2$ units right alone also works, but the examiner report stresses precise wording such as "reflect in the $y$ axis".
$g(x)=\tfrac12 f(2-x)$

Answer

\text{Reflect in the } y \text{ axis, then translate } 2 \text{ units in the positive } x \text{ direction.}

VCAA 2023 Mathematical Methods Exam 2, Section B Q5a

Worked Example 2Dilation through two points, from a real exam

The graph of $y=F(x)$ , where $F(x)=\tfrac{x^2}{4}+7x+c$ , can be dilated by a factor of $m$ from the $x$ axis so that its image passes through both $(-12,1)$ and $(2,8)$ . Find the values of $m$ and $c$ .

1
A dilation by factor $m$ from the $x$ axis multiplies every $y$ value by $m$ , so the image rule is $y=mF(x)$ .
$y=m\left(\tfrac{x^2}{4}+7x+c\right)$
2
Substitute each point to form two equations. At $(-12,1)$ , $\tfrac{(-12)^2}{4}+7(-12)=36-84=-48$ , and at $(2,8)$ , $\tfrac{2^2}{4}+7(2)=1+14=15$ .
$m(-48+c)=1, \qquad m(15+c)=8$
3
Subtract the first from the second to eliminate $c$ , since $m(15+c)-m(-48+c)=63m$ .
$63m=7 \ \Rightarrow\ m=\tfrac{1}{9}$
4
Substitute $m=\tfrac19$ back into $m(15+c)=8$ to find $c$ .
$\tfrac19(15+c)=8 \ \Rightarrow\ 15+c=72 \ \Rightarrow\ c=57$

Answer

m=\dfrac{1}{9}, \quad c=57

VCAA 2025 Mathematical Methods Exam 2, Section B Q2fii

Worked Example 3Build the rule from words

The graph of $y = x^2$ is dilated by a factor of $2$ from the $x$ axis, then translated $3$ units in the positive $x$ direction and $1$ unit up. Find the rule of the image.

1
Dilation by factor $2$ from the $x$ axis multiplies every $y$ value by $2$ , so it multiplies the whole rule by $2$ .
$y = 2x^2$
2
Translation $3$ units in the positive $x$ direction replaces $x$ with $x-3$ (you subtract to move right).
$y = 2(x-3)^2$
3
Translation $1$ unit up adds $1$ to the whole rule.
$y = 2(x-3)^2 + 1$
4
Quick check with the point $(1,1)$ , which is on $y=x^2$ . Its image should satisfy the new rule at $x=4$ .
$2(4-3)^2 + 1 = 2 + 1 = 3$

Answer

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

Worked Example 4Order matters, reflection then dilation then translation

The graph of $f(x) = \sqrt{x}$ is reflected in the $x$ axis, then dilated by a factor of $3$ from the $x$ axis, then translated $2$ units right and $1$ unit up. Find the rule of the image, then find the image of the point $(4,2)$ .

1
Reflection in the $x$ axis flips the sign of $y$ , so multiply the rule by $-1$ .
$y = -\sqrt{x}$
2
Dilation by factor $3$ from the $x$ axis multiplies $y$ by $3$ .
$y = -3\sqrt{x}$
3
Translate $2$ right and $1$ up, so replace $x$ with $x-2$ and add $1$ .
$y = -3\sqrt{x-2} + 1$
4
For the image of $(4,2)$ , push the point through the same steps. Reflect $y$ , then triple $y$ , then shift.
$(4,2) \to (4,-2) \to (4,-6) \to (4+2,\, -6+1) = (6,-5)$
5
Confirm the image lies on the new rule by substituting $x=6$ .
$-3\sqrt{6-2} + 1 = -3(2) + 1 = -5$

Answer

y = -3\sqrt{x-2}+1, \quad (4,2) \to (6,-5)

Worked Example 5Mapping notation and the image of a point

A transformation maps the graph of $y = f(x)$ to the graph of $y = f(2x) - 4$ . Describe the transformations and write the mapping that sends a point $(x,y)$ to its image. Hence find the image of the point $(6,5)$ .

1
The $2$ inside, next to $x$ , is a horizontal dilation by factor $\tfrac{1}{2}$ from the $y$ axis. Inside changes do the reciprocal.
$x \to \tfrac{1}{2}x$
2
The $-4$ outside is a translation $4$ units down. Outside changes act directly on $y$ .
$y \to y - 4$
3
Write both parts together as a single mapping.
$(x,y) \to \left(\tfrac{1}{2}x,\; y-4\right)$
4
Apply the mapping to $(6,5)$ .
$(6,5) \to \left(\tfrac{1}{2}\times 6,\; 5-4\right) = (3,1)$

Answer

(x,y) \to \left(\tfrac{1}{2}x,\, y-4\right), \quad (6,5) \to (3,1)

Practice test

Try it yourself

6 questions, 8 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.The graph of $y = f(x)$ is transformed to the graph of $y = 3f(x)$ . This transformation is a:

1mark

Need a hint?

The

3

sits outside, multiplying the whole rule, so it acts on

y

. Ask which axis a vertical stretch is measured from.

Q2.The graph of $y = f(x)$ is transformed to the graph of $y = f(4x)$ . The correct dilation is:

1mark

Need a hint?

The

4

is inside, next to

x

, so it acts horizontally. Inside changes do the reciprocal of what they look like.

Q3.The point $(2,5)$ lies on the graph of $y = f(x)$ . Under the mapping $(x,y) \to (x-3,\, 2y)$ , the image of this point is:

1mark

Need a hint?

Substitute the coordinates straight into the mapping. The

x

part is subtraction, the

y

part is multiplication, keep them separate.

Q4.The graph of $y = x^2$ is translated so that the rule becomes $y = (x+4)^2 - 7$ . The translation is:

1mark

Need a hint?

The

+4

is inside next to

x

, so the horizontal direction is the opposite of the sign. The

-7

is outside and acts directly on

y

Q5.The graph of $y = f(x)$ has a local minimum at $(1, -2)$ . The graph of $y = f(x-3) + 5$ has a local minimum at:

1mark

Need a hint?

The

x-3

inside shifts horizontally, sign reversed; the

+5

outside shifts

y

up. Apply both to the minimum coordinates.

Q6.The graph of $y = x^2$ is transformed to the graph of $y = -\tfrac{1}{2}(x-4)^2 + 3$ . Describe a sequence of transformations using correct language, then state the image of the turning point $(0,0)$ .

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

What order do I apply transformations in?

Apply dilations and reflections first, then translations last. The one thing you must not do is apply a horizontal translation before a horizontal dilation, because the shift would get stretched too and your final position would be wrong. A vertical translation can be slotted in anywhere.

Why do inside changes do the opposite of what they look like?

Changes inside the function act on x, and to land a point in a new spot you have to undo the change first. So a plus sign inside moves the graph left, a minus moves it right, and a factor next to x dilates by the reciprocal. Outside changes act directly on y and behave exactly as written.

How do I find the image of a point under a transformation?

Write the mapping that the transformation gives, then push the point's coordinates through it, treating x and y separately. Horizontal moves only touch the x coordinate and vertical moves only touch the y coordinate, so you never mix them up.

Transformations of Graphs

Learn

The four moves, in plain English

Reading each letter

Why the order can matter

The image of a point and mapping notation

How to actually do it

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions