Graph Transformations

Scope Label

Core 9758 graphing support. Transformations of graphs are part of the functions-and-graphs strand. This note supports Graphing Techniques by giving a focused method for moving from a known graph $y=f(x)$ to a transformed graph.

This note focuses mainly on transformations of graphs that can be written in the form

$$y=f(x)$$

and related forms such as

$$y=|f(x)|,y=f(|x|).$$

Standard conics such as parabolas, ellipses, and hyperbolas are treated in the main graphing and conics notes, where their centres, axes, vertices, asymptotes, intersections, and restrictions are handled explicitly. The transformation ideas here are still useful when interpreting shifted or scaled standard forms, but this note does not replace the conics treatment.

The main aim is not to memorise isolated rules. The main aim is to know whether a change acts on:

• the output of the function, which moves or rescales $y$-values
• the input of the function, which moves or rescales $x$-values

This distinction explains most graph-transformation rules.

The Central Idea

Start with a known graph

$$y=f(x).$$

A transformation changes how points on this graph are read.

If a point $(x,y)$ lies on $y=f(x)$, then

$$y=f(x).$$

After a transformation, the corresponding point may become $(X,Y)$ on the new graph.

The cleanest way to think is:

• output changes affect $Y$
• input changes affect $X$

For example, if the new graph is

$$Y=f(X)+3,$$

then every output is $3$ larger. The graph moves up.

If the new graph is

$$Y=f(X-2),$$

then the function is fed the input $X-2$. To get the same old input $x$, we need

$$X-2=x,$$

so

$$X=x+2.$$

The graph moves right by $2$.

This is why horizontal transformations can feel reversed: the algebra is changing the input before the function acts.

Caption: Output transformations change $y$-values directly, while input transformations change which $x$-value is fed into the original function.

Transformation as point mapping

A transformation can often be understood by asking:

If $(x,y)$ is an old point, where does it go on the new graph?

For example, suppose $(x,y)$ lies on $y=f(x)$.

For the transformed graph

$$Y=2f(X-3)+1,$$

we first match the old input:

$$X-3=x.$$

Hence

$$X=x+3.$$

Then transform the old output:

$$Y=2y+1.$$

So the point mapping is

$$(x,y)\mapsto (x+3,2y+1).$$

This is more reliable than trying to describe the graph using only words such as “shift” and “stretch”.

Output Transformations

Output transformations change the value of $f(x)$ after the input has already been used.

They are usually easier to read because the visible change agrees with the sign or factor in the formula.

Vertical translation

The graph of

$$y=f(x)+a$$

is obtained by translating the graph of $y=f(x)$ vertically by $a$ units.

• if $a>0$, move up by $a$
• if $a<0$, move down by $|a|$

Point mapping:

$$(x,y)\mapsto (x,y+a).$$

The domain is unchanged. The range is shifted by $a$.

For example, if $y=f(x)$ has range $[-2,5]$, then $y=f(x)+3$ has range

$$[1,8].$$

Vertical stretch

The graph of

$$y=af(x)$$

is obtained by multiplying every $y$-value by $a$.

Point mapping:

$$(x,y)\mapsto (x,ay).$$

If $a>1$, the graph is stretched vertically. If $0<a<1$, the graph is compressed vertically.

If $a<0$, there is also a reflection in the $x$-axis.

If $a=0$, the expression $y=af(x)$ collapses to the constant graph $y=0$, so it is no longer a stretch of the graph in the usual sense.

For example,

$$y=-2f(x)$$

means:

1. multiply all $y$-values by $2$
2. reflect in the $x$-axis

Equivalently, each point $(x,y)$ becomes

$$(x,-2y).$$

Reflection in the x-axis

The graph of

$$y=-f(x)$$

is the reflection of $y=f(x)$ in the $x$-axis.

Point mapping:

$$(x,y)\mapsto (x,-y).$$

The domain is unchanged. The range changes sign.

If the original range is

$$[m,M],$$

then the reflected range is

$$[-M,-m].$$

Output transformations and intercepts

Output transformations can change $x$-intercepts because the new graph may cross the $x$-axis at different places.

For example, if

$$g(x)=f(x)+3,$$

then the $x$-intercepts of $g$ are found by solving

$$f(x)+3=0,$$

not by simply moving the old $x$-intercepts upward.

However, a vertical stretch by a non-zero factor preserves the old $x$-intercepts. If

$$g(x)=af(x),a\ne 0,$$

then

$$g(x)=0⟺f(x)=0.$$

So the $x$-intercepts remain at the same $x$-values.

Input Transformations

Input transformations change the value of $x$ before the function rule is applied.

They require more care because the visible horizontal movement is determined by solving for the old input.

Horizontal translation

The graph of

$$y=f(x-a)$$

is obtained by translating $y=f(x)$ right by $a$ units.

Point mapping:

$$(x,y)\mapsto (x+a,y).$$

The graph of

$$y=f(x+a)$$

is obtained by translating $y=f(x)$ left by $a$ units.

Point mapping:

$$(x,y)\mapsto (x-a,y).$$

The reason is structural. To reproduce the old output $f(x)$ using the new input $X$, solve:

$$X-a=x.$$

So

$$X=x+a.$$

This is the safest way to avoid the common sign error.

Horizontal stretch

For $a\ne 0$, the graph of

$$y=f(ax)$$

is obtained by multiplying all $x$-coordinates by $\frac{1}{a}$.

Point mapping:

$$(x,y)\mapsto \left(\frac{x}{a},y\right).$$

If $a>1$, the graph is compressed horizontally by factor $a$.

If $0<a<1$, the graph is stretched horizontally by factor $\frac{1}{a}$.

If $a<0$, there is also a reflection in the $y$-axis.

This can be understood by solving for the old input:

$$aX=x$$

so

$$X=\frac{x}{a}.$$

Again, the visible horizontal change is the reciprocal of the coefficient inside the function.

Reflection in the y-axis

The graph of

$$y=f(-x)$$

is the reflection of $y=f(x)$ in the $y$-axis.

Point mapping:

$$(x,y)\mapsto (-x,y).$$

This is an input transformation because the negative sign is applied to $x$ before the function acts.

Input transformations and the domain

Input transformations affect which $x$-values are allowed.

Suppose $f$ has domain

$$x\ge 0.$$

Then $f(x-2)$ is defined only when

$$x-2\ge 0,$$

so the transformed graph has domain

$$x\ge 2.$$

For

$$f(2x+1),$$

we require

$$2x+1\ge 0,$$

so

$$x\ge -\frac{1}{2}.$$

The safest method is not to guess the new domain from the picture. Instead, impose the original domain condition on the new input.

Summary Table

New graph	Point mapping from $y=f(x)$	Description
$y=f(x)+a$	$(x,y)\mapsto (x,y+a)$	vertical translation by $a$
$y=f(x-a)$	$(x,y)\mapsto (x+a,y)$	horizontal translation right by $a$
$y=af(x)$	$(x,y)\mapsto (x,ay)$	vertical scale factor $a$
$y=f(ax)$, $a\ne 0$	$(x,y)\mapsto \left(\frac{x}{a},y\right)$	horizontal scale factor $\frac{1}{a}$, with reflection if $a<0$
$y=-f(x)$	$(x,y)\mapsto (x,-y)$	reflection in the $x$-axis
$y=f(-x)$	$(x,y)\mapsto (-x,y)$	reflection in the $y$-axis

Caption: Basic transformations are most reliable when read as point mappings from the original graph.

Modulus-Related Graphs

Modulus graphs require special care because the modulus symbol changes signs or inputs.

The two common forms

$$y=|f(x)|$$

and

$$y=f(|x|)$$

do different things.

The graph of y = |f(x)|

For

$$y=|f(x)|,$$

the output is made non-negative.

So:

• parts of $y=f(x)$ above or on the $x$-axis stay where they are
• parts below the $x$-axis are reflected in the $x$-axis

This transformation acts on outputs.

Point meaning:

$$(x,y)\mapsto (x,|y|).$$

The $x$-intercepts remain in the same positions, but the graph may form sharp corners there.

The range of $y=|f(x)|$ is always non-negative:

$$y\ge 0.$$

However, the precise range still depends on the original graph. For example, if $f(x)$ never reaches $0$, then $|f(x)|$ may have range such as $y\ge 2$ rather than $y\ge 0$.

The graph of y = f(|x|)

For

$$y=f(|x|),$$

the input is made non-negative before being used.

So:

• for $x\ge 0$, the graph is unchanged
• for $x<0$, the graph mirrors the right-hand part across the $y$-axis

This transformation acts on inputs.

The left-hand part of the original graph is ignored unless it is already the mirror of the right-hand part.

This is a common source of mistakes. The graph of $y=f(|x|)$ is not obtained by reflecting the negative part below the $x$-axis. That is the behaviour of $y=|f(x)|$, not $y=f(|x|)$.

Caption: $y=|f(x)|$ reflects negative outputs upwards, while $y=f(|x|)$ mirrors the right-hand part of the original graph.

A concrete comparison

Let

$$f(x)=x+1.$$

Then

$$y=|f(x)|=|x+1|.$$

This graph is obtained from the line $y=x+1$ by reflecting the part below the $x$-axis upward. The corner occurs at the original $x$-intercept:

$$x=-1.$$

But

$$y=f(|x|)=|x|+1.$$

This graph keeps the right-hand part of $y=x+1$ and mirrors it across the $y$-axis. The corner occurs at

$$x=0.$$

So even for the simple base graph $f(x)=x+1$, the two modulus graphs are different:

$$|f(x)|=|x+1|,$$

whereas

$$f(|x|)=|x|+1.$$

Caption: A concrete comparison using $f(x)=x+1$: $y=|f(x)|$ reflects negative outputs, while $y=f(|x|)$ mirrors the right-hand part of the original graph.

Combining Transformations

When several transformations appear together, do not rely on vague phrases such as “shift then stretch” without checking which expression is being transformed.

For example, compare

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

with

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

The first graph is easier:

• $x-3$ means shift right by $3$
• multiplying by $2$ stretches vertically by factor $2$
• adding $1$ shifts upward by $1$

The second graph needs extra care because

$$2x-3=2\left(x-\frac{3}{2}\right).$$

The inside transformation is not simply “right by $3$”. It combines a horizontal scale with a horizontal translation.

A reliable method is to use point mapping.

Suppose $(x,y)$ lies on the original graph. For

$$Y=2f(2X-3)+1,$$

we want

$$2X-3=x.$$

Solving gives

$$X=\frac{x+3}{2}.$$

The output is

$$Y=2y+1.$$

So the point mapping is

$$(x,y)\mapsto \left(\frac{x+3}{2},2y+1\right).$$

This method avoids arguing about which horizontal operation happened first.

More generally, for

$$Y=af(bX+c)+d,$$

with $b\ne 0$, solve

$$bX+c=x$$

to get

$$X=\frac{x-c}{b}.$$

The output becomes

$$Y=ay+d.$$

So the point mapping is

$$(x,y)\mapsto \left(\frac{x-c}{b},ay+d\right).$$

This formula is useful, but it should not be used mechanically. The values of $a,b,c,d$ must be read from the exact transformed expression after any needed factorisation inside the function.

Caption: For combined transformations, solving for the old input gives the safest point mapping.

Order versus point mapping

Students often ask whether the graph should be shifted first or stretched first.

For many exam questions, the safest answer is:

Use the point mapping. It automatically gives the final position.

For example, for

$$Y=f(2X-3),$$

matching old input gives

$$2X-3=x,$$

so

$$X=\frac{x+3}{2}.$$

This says every old $x$-coordinate becomes

$$\frac{x+3}{2}.$$

There is no need to argue verbally about whether the graph is first shifted or first compressed. The equation itself determines the final coordinate.

What Happens to Key Graph Features

Transformations should not be applied only to the visible curve. They also move the structural features that define the curve.

A complete transformed sketch should consider:

• points
• intercepts
• asymptotes
• endpoints
• holes or excluded points
• domain and range
• symmetry

This is why graph transformations belong naturally inside curve sketching. A sketch is incomplete if the curve is moved but its defining features are left behind.

Intercepts

Intercepts may move, disappear, or appear after transformation.

For example, a vertical translation may change where the graph crosses the $x$-axis. A horizontal translation moves existing $x$-intercepts horizontally, but a vertical stretch by a non-zero factor keeps the same $x$-intercepts.

So intercepts should be checked from the transformed equation if exact values are required.

The safest rule is:

• transformed old intercepts are useful visual clues
• exact intercepts of the new graph should be checked using $y=0$ or $x=0$

For example, if

$$g(x)=f(x)+2,$$

then the $y$-intercept is

$$g(0)=f(0)+2,$$

provided $0$ is in the domain of $f$.

But the $x$-intercepts must satisfy

$$f(x)+2=0.$$

They are not generally obtained by simply shifting the old $x$-intercepts upward, because points on the $x$-axis stop being $x$-intercepts after a vertical shift.

Asymptotes

Asymptotes transform with the graph.

If $y=f(x)$ has vertical asymptote $x=c$, then $y=f(x-a)$ has vertical asymptote

$$x=c+a.$$

If $y=f(x)$ has horizontal asymptote $y=L$, then $y=bf(x)+d$ has horizontal asymptote

$$y=bL+d.$$

The asymptote is not decoration. It is part of the graph’s limiting behaviour, so it must be transformed consistently.

For combined transformations, use the same point-mapping idea.

If an old vertical asymptote is

$$x=c,$$

then solve the input equation

$$bX+c_0=c$$

for $X$, where $bX+c_0$ is the new input fed into $f$.

If an old horizontal asymptote is

$$y=L,$$

then transform the output:

$$Y=aL+d.$$

Domain and range

Input transformations affect the domain. Output transformations affect the range.

For example, if $f$ has domain

$$[1,5],$$

then $f(x-2)$ has domain

$$[3,7].$$

If $f$ has range

$$[-1,4],$$

then $2f(x)-3$ has range

$$[-5,5].$$

The safest rule is:

• transform the domain using the input mapping
• transform the range using the output mapping

Modulus transformations should be checked separately because they may reflect, discard, or duplicate parts of the original graph.

Endpoints and holes

If the original graph has an endpoint, the endpoint is transformed like any other point.

For example, the graph of

$$y=\sqrt{x}$$

has endpoint

$$(0,0).$$

For

$$y=-2\sqrt{x-1}+3,$$

use the old input condition

$$X-1=x.$$

So

$$X=x+1.$$

The output transformation is

$$Y=-2y+3.$$

Thus the endpoint $(0,0)$ becomes

$$(1,3).$$

The transformed domain becomes

$$X\ge 1,$$

and the transformed range becomes

$$Y\le 3.$$

Holes are treated in the same way. If the original graph has a hole at $(p,q)$, then the transformed graph has a hole at the transformed point. The blank circle must be moved with the graph.

Symmetry

Transformations can preserve, move, or destroy symmetry.

For example, if $y=f(x)$ is even, then it is symmetric about the $y$-axis. The graph of

$$y=f(x-2)$$

is symmetric about

$$x=2,$$

not about the $y$-axis.

If $y=f(x)$ is odd, then it is symmetric about the origin. After a translation, the new graph may have rotational symmetry about a different point, or the simple odd-function test may no longer apply.

So symmetry should be checked from the transformed graph or transformed equation, not assumed from the original graph without adjustment.

Worked Examples

Example 1: Transforming a known point

Suppose $(2,5)$ lies on the graph of $y=f(x)$.

Find the corresponding point on

$$y=-3f(x-4)+2.$$

Let the new point be $(X,Y)$.

To use the old input $2$, solve

$$X-4=2.$$

So

$$X=6.$$

The old output is $5$, so

$$Y=-3(5)+2=-13.$$

Thus the corresponding point is

$$(6,-13).$$

Example 2: Transforming an asymptote

Suppose $y=f(x)$ has horizontal asymptote

$$y=2.$$

Find the horizontal asymptote of

$$y=4f(x+1)-7.$$

The input change $x+1$ affects horizontal position, not horizontal asymptote height.

The output transformation is

$$Y=4y-7.$$

So

$$Y=4(2)-7=1.$$

Hence the transformed graph has horizontal asymptote

$$y=1.$$

Example 3: Distinguishing two modulus graphs

Suppose $y=f(x)$ has part of its graph below the $x$-axis for $x<0$.

For

$$y=|f(x)|,$$

that negative part is reflected upward.

For

$$y=f(|x|),$$

the graph for $x<0$ is not obtained from the old left-hand part. Instead, it is the mirror image of the original right-hand part.

So the two graphs can be completely different even though both contain modulus signs.

Example 4: Transforming a complete set of graph features

Suppose the graph of $y=f(x)$ has:

• $x$-intercepts $(-1,0)$ and $(1,0)$
• $y$-intercept $(0,2)$
• vertical asymptote $x=3$
• horizontal asymptote $y=-1$
• domain $x\ne 3$
• range $y\ne -1$

Let

$$g(X)=2f(X-1)-3.$$

We want to transform from the old graph to the new graph

$$Y=2f(X-1)-3.$$

Match the old input:

$$X-1=x.$$

So

$$X=x+1.$$

Transform the output:

$$Y=2y-3.$$

Thus the point mapping is

$$(x,y)\mapsto (x+1,2y-3).$$

The old $x$-intercepts become

$$(-1,0)\mapsto (0,-3),(1,0)\mapsto (2,-3).$$

These are not $x$-intercepts of the new graph; they are transformed points from the old $x$-axis.

The old $y$-intercept becomes

$$(0,2)\mapsto (1,1).$$

Again, this is not necessarily the new $y$-intercept. It is the image of the old $y$-intercept.

The vertical asymptote transforms by the input mapping:

$$x=3\mapsto X=4.$$

So the new vertical asymptote is

$$X=4.$$

The horizontal asymptote transforms by the output mapping:

$$y=-1\mapsto Y=2(-1)-3=-5.$$

So the new horizontal asymptote is

$$Y=-5.$$

The domain becomes

$$X\ne 4,$$

and the range becomes

$$Y\ne -5.$$

This example shows the key lesson: transforming the graph means transforming its structure, not only its visible curve.

Caption: A transformed sketch should move points, asymptotes, domain restrictions, range restrictions, and excluded features consistently.

Example 5: Transforming a restricted-domain graph

Consider the base graph

$$y=\sqrt{x}.$$

It has endpoint $(0,0)$, domain $x\ge 0$, and range $y\ge 0$.

Now sketch

$$y=-2\sqrt{x-1}+3.$$

Let the old graph be $y=f(x)=\sqrt{x}$. The new graph is

$$Y=-2f(X-1)+3.$$

Match the old input:

$$X-1=x,$$

so

$$X=x+1.$$

Transform the output:

$$Y=-2y+3.$$

The endpoint becomes

$$(0,0)\mapsto (1,3).$$

The domain comes from

$$X-1\ge 0,$$

so

$$X\ge 1.$$

The old range $y\ge 0$ transforms under

$$Y=-2y+3.$$

Since multiplying by a negative number reverses the inequality,

$$Y\le 3.$$

So the transformed graph starts at $(1,3)$ and extends to the right and downward.

A Practical Sketching Workflow

When transforming a graph, use this order.

1. Identify the base graph. Decide what original $y=f(x)$ graph you are transforming.
2. Find the point mapping. Use old coordinates $(x,y)$ and new coordinates $(X,Y)$.
3. Transform key points. Move labelled points, endpoints, holes, and useful intercepts.
4. Transform asymptotes. Move vertical asymptotes using the input mapping and horizontal asymptotes using the output mapping.
5. Update domain and range. Use input restrictions for domain and output mapping for range.
6. Check new intercepts if required. Use $Y=0$ and $X=0$ on the transformed equation.
7. Check symmetry. Decide whether symmetry is preserved, shifted, or lost.
8. Draw the final sketch. The curve and all labels must agree with the transformed features.

This workflow prevents a common error: drawing a plausible shifted curve but leaving the asymptotes, holes, or endpoints in the wrong places.

Common Pitfalls

1. Reversing horizontal shifts

A common error is to say that $f(x-3)$ moves left by $3$ because of the minus sign.

This is wrong. Solve

$$X-3=x.$$

Then

$$X=x+3,$$

so the graph moves right by $3$.

2. Treating f(ax) like af(x)

The expressions

$$f(ax)$$

and

$$af(x)$$

act in different directions.

• $f(ax)$ changes the input, so it is a horizontal transformation.
• $af(x)$ changes the output, so it is a vertical transformation.

3. Forgetting reciprocal horizontal scale

The graph of

$$y=f(2x)$$

is horizontally compressed by factor $2$, not stretched by factor $2$.

Point mapping:

$$(x,y)\mapsto \left(\frac{x}{2},y\right).$$

If the coefficient of $x$ is negative, include the reflection as well. For example, $y=f(-2x)$ maps

$$(x,y)\mapsto \left(-\frac{x}{2},y\right).$$

4. Mixing up |f(x)| and f(|x|)

The expression

$$|f(x)|$$

makes outputs non-negative.

The expression

$$f(|x|)$$

makes inputs non-negative.

Do not treat them as the same transformation.

5. Transforming only the curve, not its features

When the curve moves, its asymptotes, intercepts, endpoints, holes, and domain restrictions move too.

A sketch is incomplete if the curve is transformed but the structural labels remain in their old positions.

6. Forgetting that transformed old intercepts may not be new intercepts

If an old $x$-intercept is moved vertically, it will usually no longer lie on the new $x$-axis.

So transformed old intercepts are useful points, but the new intercepts should be checked from the transformed equation when exact intercepts are required.

7. Treating modulus transformations as ordinary stretches or reflections

The transformations $y=|f(x)|$ and $y=f(|x|)$ are not ordinary one-to-one transformations of the whole graph.

For $y=|f(x)|$, negative outputs are folded upward.

For $y=f(|x|)$, the right-hand part is copied to the left.

This can change sharp corners, range, symmetry, and the number of intersections with a line.

Revision Checklist

Core transformation meaning

• Can I decide whether a transformation acts on input or output?
• Can I use point mapping to transform a point from $y=f(x)$?
• Can I explain why $f(x-a)$ shifts right by $a$?
• Can I explain why $f(ax)$ has horizontal scale factor $\frac{1}{a}$ when $a\ne 0$, with reflection if $a<0$?

Modulus graphs

• Can I sketch $y=|f(x)|$ by reflecting negative outputs upward?
• Can I sketch $y=f(|x|)$ by mirroring the right-hand part of the graph?
• Can I sketch $y=|f(x)|$ and $y=f(|x|)$ without confusing them?
• Can I explain why $y=|f(x)|$ and $y=f(|x|)$ are usually different?

Combined transformations

• Can I use $(x,y)\mapsto \left(\frac{x-c}{b},ay+d\right)$ for $Y=af(bX+c)+d$ after reading the expression correctly?
• Can I handle combined transformations by solving for the old input?
• Can I avoid relying only on ambiguous phrases such as “shift then stretch”?

Transformed graph features

• Can I transform domains, ranges, intercepts, endpoints, holes, and asymptotes consistently?
• Can I distinguish transformed old intercepts from actual intercepts of the new graph?
• Can I update a vertical asymptote using the input transformation?
• Can I update a horizontal asymptote using the output transformation?
• Can I decide whether symmetry is preserved, shifted, or lost?

Related

• Functions
• Graphing Techniques