Mathematics/Lessons/Graphs & Transformations/Translating & Stretching Graphs
Pure Mathematics · Graphs & Transformations · Year 1

Translating & Stretching Graphs

Every curve you already know how to sketch — quadratics, cubics, reciprocals — can be shifted or squeezed into a new one, without any fresh sketching from scratch.

10 min read 4 worked examples

Outside changes vs inside changes

Given the graph of $y=f(x)$, changing the equation in one of four standard ways transforms the curve predictably. The single most useful question to ask is: is the change outside the function, or inside it?

Why inside-the-bracket changes go the "wrong" way

Consider $y=f(x+2)$. To get the same output the original graph had at $x=0$, the new graph only needs $x=-2$ (so that $x+2=0$). Every feature therefore arrives 2 units earlier — the graph shifts left, not right, even though we added. Inside-the-bracket transformations always do the opposite of what the sign suggests, because you're changing the input before the function acts on it.

Translations

$f(x)+a$ (outside): shift up by $a$ (down if $a$ is negative) — every $y$-value increases by $a$, direction matches the sign.
$f(x+a)$ (inside): shift left by $a$ (right if $a$ is negative) — the direction is reversed from what the sign suggests.

Stretches

$a\,f(x)$ (outside): stretch vertically by scale factor $a$ — every $y$-value is multiplied by $a$.
$f(ax)$ (inside): stretch horizontally by scale factor $\dfrac{1}{a}$ — the reciprocal of $a$, and again the direction/size is reversed from the naive reading.

Strategy — outside vs inside is the only question

Outside $\Rightarrow$ vertical and intuitive (the sign or factor does what it looks like it should). Inside $\Rightarrow$ horizontal and reversed (opposite sign for shifts, reciprocal factor for stretches). That one check settles both direction and size every time.

Worked example 1

Describing a translation
The graph of $y=f(x)$ is transformed to $y=f(x-3)+2$. Describe the transformation fully.
Cartesian axes with one curve drawn as a solid line, and a second, shifted version of the same curve drawn as a dashed line beside it, with a short arrow indicating the translation between them.
A curve (solid) and its translated image (dashed), moved by a fixed vector.
MethodSplit into the inside change and the outside change
1
Two changes are present: $-3$ inside the bracket, and $+2$ outside.
Handle input and output separately.
2
Inside: $f(x-3)$ shifts the graph right by 3 (inside works in reverse, so $-3$ means right).
Input change → horizontal, reversed.
3
Outside: $+2$ shifts the graph up by 2.
Output change → vertical, normal.
4
Combined: a translation by the vector $\begin{pmatrix}3\\2\end{pmatrix}$ — right 3, up 2.
Two translations combine into one vector.
▪ described
Examiner's eye

Use the word "translation" and give a column vector $\begin{pmatrix}3\\2\end{pmatrix}$ — the precise language mark schemes want. The reversal — $f(x-3)$ moving right — feels backwards until you remember inside-the-bracket flips the sign.

Worked example 2

Applying a stretch to a specific point
The point $P(4,3)$ lies on $y=f(x)$. Find the image of $P$ under the transformation $y=f(2x)$.
MethodDecide the axis and factor, then transform the coordinate
1
The change $2x$ is inside the bracket, so it's a horizontal stretch, behaving in reverse.
Input change → horizontal.
2
$f(2x)$ is a horizontal stretch with scale factor $\tfrac12$ (the reciprocal of 2).
Inside stretches use the reciprocal factor.
3
A horizontal stretch changes only the $x$-coordinate: $x=4 \to 4\times\tfrac12=2$.
The y-coordinate is untouched.
4
So $P(4,3)$ maps to $(2,3)$.
Halved horizontally, unchanged vertically.
▪ solved
The key insight

The reciprocal is the trap: $f(2x)$ squashes the graph to half its width, not double it — doubling the input means the function reaches each value in half the horizontal distance. A vertical stretch $af(x)$ has no such reciprocal trap — $2f(x)$ really does double every $y$-value directly.

Worked example 3

A vertical stretch applied to a known curve
Sketch $y=3x^2$ as a transformation of $y=x^2$, stating the transformation and the coordinates of the image of the point $(2,4)$.
Graph of y=x^2 (dashed) through (2,4) and its image y=3x^2 (solid) through (2,12), both with vertex at the origin, the solid curve narrower from the vertical stretch.
$y=x^2$ (dashed) stretched vertically by $3$ to $y=3x^2$ (solid) — $(2,4)\to(2,12)$.
MethodIdentify the outside factor and apply it
1
$y=3x^2 = 3f(x)$ where $f(x)=x^2$ — the factor $3$ is outside.
Multiplies the whole function.
2
This is a vertical stretch, scale factor $3$ — every $y$-value on $y=x^2$ is tripled.
Outside factors are intuitive: scale factor matches the number.
3
Applying this to $(2,4)$: the $x$-coordinate is unchanged, the $y$-coordinate triples: $(2, 4\times3) = (2,12)$.
Vertical stretch only affects y.
4
Check directly: $y=3(2)^2=3(4)=12$ ✓ — matches.
Confirms the transformed point lies on the new curve.
▪ solved
Common mistake

Don't confuse $3x^2$ (a vertical stretch of $x^2$) with $(3x)^2=9x^2$ (a horizontal stretch, scale factor $\tfrac13$) — these look superficially similar but are completely different transformations. Always check exactly where the multiplying number sits relative to the function.

Worked example 4

Combining a translation and a stretch · Challenge
The point $A(6,8)$ lies on $y=f(x)$. Find the image of $A$ under the transformation $y=2f(x-1)$, and state the two transformations applied, in the order they act.

Two transformations here — one inside, one outside — and the order in which they're applied to a specific point matters.

MethodApply the inside change first, then the outside change
1
$x-1$ inside the bracket: a translation right by 1 (horizontal, reversed sign).
Read off the inside transformation.
2
$\times2$ outside: a vertical stretch, scale factor 2.
Read off the outside transformation.
3
Apply the inside change to the point's $x$-coordinate first: $x=6 \to 6+1=7$ (shift right 1).
Inside transformations act on x, applied before the outside one.
4
Apply the outside change to the $y$-coordinate: $y=8 \to 8\times2=16$.
Outside transformations act on y, applied last.
5
So $A(6,8)$ maps to $(7,16)$.
Combine both coordinate changes.
▪ solved
Look deeper — inside acts on x, outside acts on y, always

Because a horizontal (inside) change only ever touches the $x$-coordinate and a vertical (outside) change only ever touches the $y$-coordinate, the two updates to a single point can always be done independently and in either order — the coordinates never interact with each other. This is what makes tracking key points through a combined transformation manageable, even with several steps chained together.

Common mistake

When both a horizontal shift and a horizontal stretch act inside the same bracket, such as $f(2x-6)$, applying them in the wrong order changes the result — always factor the inside expression fully first, e.g. $f(2x-6)=f(2(x-3))$, to see the stretch and shift separately before describing them.

Check yourself

The point $B(3,5)$ lies on $y=f(x)$. Find the image of $B$ under $y=f(x+4)-6$, and describe the transformation as a single vector translation.
(Hint: the +4 is inside the bracket, the -6 is outside.)

Method
1
$+4$ inside: shift left by 4, so $x=3 \to 3-4=-1$.
2
$-6$ outside: shift down by 6, so $y=5 \to 5-6=-1$.
3
So $B(3,5)$ maps to $(-1,-1)$, a translation by $\begin{pmatrix}-4\\-6\end{pmatrix}$.
Left 4, down 6.
▪ solved
← Previous
Points of Intersection