Graph transformations: translations, stretches and reflections

Graph transformations let you sketch a new function from a familiar graph without starting again from a table of values. The central skill is recognising whether a change acts on the output of a function or on its input.

Changes outside $f$ act vertically. Changes inside the brackets of $f$ act horizontally, often in the opposite direction from the sign you first see.

Before you begin

You should be comfortable with:

• function notation;
• coordinates, intercepts and asymptotes;
• the shapes of common graphs from graphs of functions;
• solving simple equations and inequalities.

The coordinate mapping principle

Suppose $(u,v)$ lies on $y=f(x)$. This means

$$v=f(u).$$

For the transformed graph

$$y=af\bigl(b(x-h)\bigr)+k,$$

choose $x$ so that the input to $f$ is still $u$:

$$b(x-h)=u.$$

Therefore

$$x=h+\frac{u}{b}.$$

The new output is

$$y=av+k.$$

So every point is mapped by

$$\boxed{(u,v)\longmapsto\left(h+\frac{u}{b},\,k+av\right)}.$$

This single rule covers translations, stretches, compressions and reflections. It is also the safest way to handle several transformations at once.

The four basic translations and reflections

Starting with $y=f(x)$:

New graph	Transformation	Point mapping
$y=f(x)+a$	Translation up by $a$	$(x,y)\mapsto(x,y+a)$
$y=f(x-a)$	Translation right by $a$	$(x,y)\mapsto(x+a,y)$
$y=-f(x)$	Reflection in the $x$-axis	$(x,y)\mapsto(x,-y)$
$y=f(-x)$	Reflection in the $y$-axis	$(x,y)\mapsto(-x,y)$

Here $a>0$. A negative value reverses the translation. For example, $f(x)-3$ moves down by $3$, while $f(x+3)$ moves left by $3$.

The horizontal translation looks reversed because the whole input must remain unchanged. To find where the old input $u$ appears on $f(x-a)$, solve

$$x-a=u,$$

which gives $x=u+a$.

Vertical translations

For

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

the input $x$ is unchanged and $a$ is added to every output. Every point moves vertically by the same amount.

Worked example 1: translate a quadratic vertically

The graph of

$$y=x^2$$

has vertex $(0,0)$. For

$$y=x^2-5,$$

every output is $5$ smaller, so the graph moves down by $5$ and the vertex becomes $(0,-5)$.

The new roots are not obtained by shifting the old root horizontally. Solve them from the new equation:

$$x^2-5=0$$ $$x=\pm\sqrt5.$$

Vertical translation preserves the graph’s shape, but it need not preserve its intercepts.

Horizontal translations

For

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

the graph moves right by $a$. For $y=f(x+a)$, it moves left by $a$.

Worked example 2: translate a reciprocal graph

Start with

$$y=\frac1x,$$

which has asymptotes $x=0$ and $y=0$. Now consider

$$y=\frac{1}{x-4}+2.$$

The $x-4$ moves the graph right by $4$, and the $+2$ outside the fraction moves it up by $2$. The asymptotes therefore become

$$x=4 \qquad\text{and}\qquad y=2.$$

The centre of the reciprocal graph moves from $(0,0)$ to $(4,2)$.

Vertical stretches and compressions

For

$$y=af(x),$$

every $y$-coordinate is multiplied by $a$:

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

• If $|a|>1$, there is a vertical stretch with scale factor $|a|$.
• If $0<|a|<1$, there is a vertical compression with scale factor $|a|$.
• If $a<0$, there is also a reflection in the $x$-axis.

Points on the $x$-axis stay fixed because multiplying $0$ by $a$ still gives $0$. Therefore $af(x)$ has the same roots as $f(x)$ when $a\ne0$.

Worked example 3: track exact points

Suppose $y=f(x)$ contains the points

$$(-2,3),\qquad(0,-1),\qquad(5,0).$$

On

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

the corresponding points are

$$(-2,-6),\qquad(0,2),\qquad(5,0).$$

The factor $2$ doubles each vertical distance from the $x$-axis. The negative sign reflects the graph in the $x$-axis.

Horizontal stretches and compressions

For

$$y=f(bx),$$

the $x$-coordinates are divided by $b$:

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

The horizontal scale factor is therefore

$$\frac{1}{|b|},$$

not $|b|$.

• If $|b|>1$, the graph is compressed horizontally.
• If $0<|b|<1$, the graph is stretched horizontally.
• If $b<0$, there is also a reflection in the $y$-axis.

This reciprocal effect is another consequence of preserving the input. If the old point uses input $u$, then on $f(bx)$ we need $bx=u$, so $x=u/b$.

Worked example 4: horizontal compression

The graph of $y=f(x)$ contains a turning point at $(6,-4)$ and a root at $(2,0)$. Find the corresponding points on

$$y=f(3x).$$

Divide every $x$-coordinate by $3$:

$$(6,-4)\mapsto(2,-4),$$ $$(2,0)\mapsto\left(\frac23,0\right).$$

The graph is compressed towards the $y$-axis with scale factor $1/3$.

A useful warning about special functions

Different-looking transformations can sometimes produce the same final equation. For example,

$$f(x)=x^2$$

gives

$$f(2x)=(2x)^2=4x^2=4f(x).$$

For this particular $f$, a horizontal compression by scale factor $1/2$ produces the same graph as a vertical stretch by scale factor $4$. The general transformation rules are still different. Do not infer a general identity from one special function.

Reflections

Reflection in the $x$-axis

For $y=-f(x)$, outputs change sign. Points above the $x$-axis move the same distance below it, and vice versa.

For example,

$$y=-e^x$$

is the reflection of $y=e^x$ in the $x$-axis. Its range is $y<0$, and it still has horizontal asymptote $y=0$.

Reflection in the $y$-axis

For $y=f(-x)$, inputs change sign. Each point $(x,y)$ moves to $(-x,y)$.

For example,

$$y=2^{-x}=\left(\frac12\right)^x$$

is the reflection of $y=2^x$ in the $y$-axis.

If $f$ is even, then $f(-x)=f(x)$, so reflection in the $y$-axis makes no visible change. If $f$ is odd, then $f(-x)=-f(x)$, so reflection in either axis gives the same resulting graph.

Combining transformations

For

$$y=af\bigl(b(x-h)\bigr)+k,$$

use the coordinate map

$$(u,v)\mapsto\left(h+\frac{u}{b},k+av\right).$$

This form is clearer than trying to remember a verbal order. It handles every original point in one calculation.

Worked example 5: a full coordinate mapping

Suppose $(6,-2)$ lies on $y=f(x)$. Find its corresponding point on

$$y=-3f\bigl(2(x-4)\bigr)+5.$$

Here

$$a=-3,\qquad b=2,\qquad h=4,\qquad k=5.$$

The new $x$-coordinate is

$$4+\frac62=7.$$

The new $y$-coordinate is

$$5+(-3)(-2)=11.$$

Therefore

$$(6,-2)\mapsto(7,11).$$

You can check directly. At $x=7$, the input to $f$ is

$$2(7-4)=6,$$

so the original output is $f(6)=-2$, and

$$-3(-2)+5=11.$$

Worked example 6: transform a complete quadratic

Starting with $f(x)=x^2$, sketch

$$y=-2f\left(\frac{x-3}{4}\right)+1.$$

Rewrite the input as

$$\frac{x-3}{4}=\frac14(x-3).$$

Thus $a=-2$, $b=1/4$, $h=3$ and $k=1$. The coordinate map is

$$(u,v)\mapsto(3+4u,1-2v).$$

Track three useful points from $y=x^2$:

$$(0,0)\mapsto(3,1),$$ $$(1,1)\mapsto(7,-1),$$ $$(-1,1)\mapsto(-1,-1).$$

The result is a downward-opening parabola with vertex $(3,1)$ and axis of symmetry $x=3$.

The transformations are a horizontal stretch by scale factor $4$, a translation right by $3$, a vertical stretch by scale factor $2$, a reflection in the $x$-axis, and a translation up by $1$.

Does the order matter?

Some transformations commute, meaning their order does not affect the result. Others do not.

Perpendicular transformations

A purely horizontal transformation and a purely vertical transformation can be handled independently. For example, translating right and then translating up gives the same result as translating up and then right.

Transformations in the same direction

Order can matter when both transformations act on the same coordinate.

Start with a $y$-coordinate $v$.

• Stretch vertically by factor $2$, then move up by $3$: $v\mapsto2v+3$.
• Move up by $3$, then stretch vertically by factor $2$: $v\mapsto2(v+3)=2v+6$.

These are not the same.

The expression

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

means multiply the original output by $2$, then add $3$. Brackets are decisive:

$$y=2\bigl(f(x)+3\bigr)=2f(x)+6$$

describes a different graph.

For combined transformations, use algebra or the coordinate mapping rather than relying on a memorised list of operations.

Transforming intercepts and asymptotes

Important features transform like other points and lines, but intercepts require care.

Intercepts

A vertical stretch $y=af(x)$ preserves the roots because $af(x)=0$ exactly when $f(x)=0$, provided $a\ne0$. A vertical translation generally changes the roots.

A horizontal transformation moves roots according to the input rule. If $f(r)=0$, then

$$f\bigl(b(x-h)\bigr)=0$$

when

$$b(x-h)=r,$$

so the new root is

$$x=h+\frac rb.$$

The old $y$-intercept does not usually map to the new $y$-intercept after a horizontal translation. Find the new $y$-intercept by substituting $x=0$ into the transformed equation.

Asymptotes

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

$$y=af\bigl(b(x-h)\bigr)+k$$

has corresponding horizontal asymptote

$$y=aL+k.$$

If $y=f(x)$ has vertical asymptote $x=c$, the transformed input reaches $c$ when

$$b(x-h)=c.$$

Therefore the new vertical asymptote is

$$x=h+\frac cb.$$

Worked example 7: transform asymptotes

The graph of $y=f(x)$ has asymptotes $x=2$ and $y=-1$. Find the corresponding asymptotes of

$$y=3f\bigl(-2(x+1)\bigr)+4.$$

For the vertical asymptote, solve

$$-2(x+1)=2.$$

This gives

$$x=-2.$$

For the horizontal asymptote, transform the old output $-1$:

$$y=3(-1)+4=1.$$

The new asymptotes are $x=-2$ and $y=1$.

Domain and range under transformations

Domain concerns allowable inputs. Range concerns possible outputs.

If the domain of $f$ is $u\in D$, then the domain of

$$af\bigl(b(x-h)\bigr)+k$$

contains precisely those $x$ for which

$$b(x-h)\in D.$$

If the range of $f$ is $v\in R$, the transformed range consists of

$$av+k$$

for $v\in R$. If $a<0$, remember that multiplying an inequality by $a$ reverses its direction.

Worked example 8: transform a restricted domain and range

Suppose $f$ has domain $0\le x\le6$ and range $-2\le f(x)\le5$. Define

$$g(x)=-2f\bigl(3(x-1)\bigr)+4.$$

For the domain, require

$$0\le3(x-1)\le6.$$

Divide by $3$:

$$0\le x-1\le2.$$

Add $1$:

$$1\le x\le3.$$

For the range, start with

$$-2\le f(x)\le5.$$

Multiplying by $-2$ reverses both inequality signs:

$$-10\le-2f(x)\le4.$$

Then add $4$:

$$-6\le g(x)\le8.$$

Modulus transformations

Modulus transformations are best understood by asking which coordinates must become non-negative.

The graph of $\boldsymbol{y=|f(x)|}$

The modulus acts on the output. Parts of $y=f(x)$ below the $x$-axis are reflected above it. Parts on or above the $x$-axis stay fixed:

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

The resulting range contains no negative values.

The graph of $\boldsymbol{y=f(|x|)}$

The modulus acts on the input. Keep the part of $y=f(x)$ for $x\ge0$, reflect it in the $y$-axis, and discard the original part for $x<0$:

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

so the result is always an even function.

Worked example 9: two different modulus graphs

Let

$$f(x)=x^2-4.$$

For

$$y=|f(x)|=|x^2-4|,$$

the part of the parabola between $x=-2$ and $x=2$ lies below the $x$-axis, so that section reflects upwards. The roots remain $x=\pm2$, and the point $(0,-4)$ becomes $(0,4)$.

For

$$y=f(|x|)=|x|^2-4=x^2-4,$$

the graph is unchanged because $f$ was already even.

Now compare this with $f(x)=x-2$. Then

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

has vertex $(2,0)$, while

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

has vertex $(0,-2)$ and roots $x=\pm2$.

Common misconceptions

”Inside changes work in the direction of the sign”

They do not. The graph of $f(x+4)$ moves left by $4$ because the old input $u$ occurs when $x+4=u$, or $x=u-4$.

”$f(2x)$ is a horizontal stretch by factor $2$”

Its horizontal scale factor is $1/2$. Input changes act reciprocally on coordinates.

”$2f(x)$ and $f(2x)$ mean the same thing”

$2f(x)$ doubles outputs, while $f(2x)$ doubles inputs and halves horizontal coordinates. They coincide for a few special functions only by algebraic accident.

”Translate every intercept in the stated direction”

Transform points on the original graph, but do not assume that a transformed old intercept remains an intercept. After a vertical translation, an old root moves away from the $x$-axis. Calculate new intercepts from the new equation.

”For $f(|x|)$, reflect the whole graph”

Only the original right-hand part is retained and reflected. Keeping the old left-hand part as well may give two different $y$-values for the same $x$, which would not define a function.

Check your understanding

Question 1

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

$$y=2f\bigl(3(x-1)\bigr)-5.$$

Question 2

Describe fully the transformation from $y=f(x)$ to

$$y=-f\left(\frac{x+6}{2}\right)+3.$$

Question 3

The graph of $y=f(x)$ has vertical asymptote $x=-4$ and horizontal asymptote $y=2$. Find the corresponding asymptotes of

$$y=-2f(4x-8)+1.$$

Question 4

The domain of $f$ is $-1\le x\le5$. Find the domain of

$$g(x)=f\bigl(2(x+3)\bigr).$$

Question 5

Explain how to construct $y=|f(x)|$ and $y=f(|x|)$ from a sketch of $y=f(x)$.

Answers

1. Here $a=2$, $b=3$, $h=1$ and $k=-5$. Therefore

$$(-3,4)\mapsto\left(1+\frac{-3}{3},-5+2(4)\right)=(0,3).$$

2. Since

$$\frac{x+6}{2}=\frac12(x-(-6)),$$

the graph is stretched horizontally by scale factor $2$, translated left by $6$, reflected in the $x$-axis, then translated up by $3$. The coordinate map is

$$(u,v)\mapsto(2u-6,3-v).$$

3. Rewrite the input as $4(x-2)$. The old vertical asymptote input $-4$ satisfies

$$4(x-2)=-4,$$

so $x=1$. The horizontal asymptote output becomes

$$-2(2)+1=-3.$$

The new asymptotes are $x=1$ and $y=-3$.

4. Require

$$-1\le2(x+3)\le5.$$

Dividing by $2$ and subtracting $3$ gives

$$-\frac72\le x\le-\frac12.$$

5. For $|f(x)|$, keep the part on or above the $x$-axis and reflect every part below it in the $x$-axis. For $f(|x|)$, keep the part for $x\ge0$, reflect it in the $y$-axis, and discard the original part for $x<0$.

Final checklist

Before completing a transformation question, check that you have:

• identified whether each change acts on inputs or outputs;
• reversed horizontal translations correctly;
• used the reciprocal horizontal scale factor $1/|b|$;
• included reflections when $a<0$ or $b<0$;
• transformed points with $(u,v)\mapsto(h+u/b,k+av)$;
• recalculated intercepts rather than moving them blindly;
• transformed asymptotes, domain and range where relevant;
• distinguished $|f(x)|$ from $f(|x|)$.

What to learn next

• Review roots, asymptotes and standard shapes in graphs of functions.
• Apply reflections in $y=x$ in composite and inverse functions.
• Use transformations with exponential and logarithmic graphs in exponential functions.
• Transform sine, cosine and tangent curves in trigonometric graphs.