Resources For Teachers For Tutors For Students & Parents Pricing
Year 12 Maths - Methods (Unit 3 & Unit 4) Transformations

Translations

20 practice questions 0 video lessons Theory + worked examples
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

A translation slides a graph horizontally and/or vertically without changing its shape or size. Writing \(y=f(x-h)\) shifts the graph \(h\) units horizontally (right when \(h>0\)), \(y=f(x)+k\) shifts it \(k\) units vertically (up when \(k>0\)), and the combined form \(y=f(x-h)+k\) maps every point \((x,y)\mapsto(x+h,\,y+k)\).

A translation is a transformation that slides a graph to a new position without changing its shape, size or orientation. The image is congruent to the original — every point moves by the same horizontal and vertical amount, so intercepts, turning points and asymptotes all move together.

A vertical translation is produced by adding a constant: \(y=f(x)+k\) moves the graph up by \(k\) when \(k>0\) and down when \(k<0\). A horizontal translation comes from replacing \(x\) with \(x-h\): \(y=f(x-h)\) moves the graph right by \(h\) when \(h>0\) and left when \(h<0\). The horizontal shift looks “backwards” because the bracket must equal the original input, so \(x\) has to be larger.

Combining both gives \(y=f(x-h)+k\), under which each point transforms by the mapping \((x,y)\mapsto(x+h,\,y+k)\). Applying this rule to the key points of a standard curve — and shifting any asymptotes, the domain and the range by the same amounts — is the fastest way to sketch the image.

Key idea. \(y=f(x-h)+k\) translates the graph of \(y=f(x)\) by \(h\) horizontally and \(k\) vertically, so \((x,y)\mapsto(x+h,\,y+k)\). Remember the sign flip: \(x-h\) moves the graph right by \(h\), while \(+k\) moves it up.
Parabola translated 2 right and 1 upThe graph of y=f(x)=x squared and its image y=f(x-2)+1=(x-2) squared+1. The turning point moves from (0,0) to (2,1); every point (x,y) maps to (x+2,y+1). x y y=f(x) y=f(x-2)+1 (0,0) (2,1)
\(y=f(x-2)+1\): the turning point moves \((0,0)\mapsto(2,1)\) — right \(2\), up \(1\)
Exponential curve translated 2 up, asymptote shiftsThe curve y=f(x)=2 to the x with horizontal asymptote y=0, and its image y=f(x)+2 with asymptote y=2. Shifting up 2 units raises the asymptote by 2. x y y=f(x) y=f(x)+2 y=0 y=2
A vertical shift up \(2\) raises the horizontal asymptote from \(y=0\) to \(y=2\)

The general translated graph — \(h\) horizontal, \(k\) vertical:

\[y=f(x-h)+k\]
y=f(xh)+k

The point mapping — every point moves by \((h,k)\):

\[(x,\,y)\;\mapsto\;(x+h,\;y+k)\]
(x,y)(x+h,y+k)

Asymptotes shift by the same amounts — a vertical asymptote by \(h\), a horizontal asymptote by \(k\):

\[x=a\;\to\;x=a+h \qquad y=b\;\to\;y=b+k\]
x=ax=a+h
Sign convention. Inside the bracket the sign is reversed: \(x-h\) shifts right by \(h\) and \(x+h\) shifts left by \(h\). Outside, \(+k\) shifts up and \(-k\) shifts down.

How to translate a graph and write its equation

  1. Read off \(h\) and \(k\). A horizontal shift of \(h\) (right positive) replaces \(x\) with \(x-h\); a vertical shift of \(k\) (up positive) adds \(k\).
  2. Write the rule. Substitute into \(y=f(x-h)+k\), e.g. “\(y=x^2\), right \(2\), up \(1\)” gives \(y=(x-2)^2+1\).
  3. Map the key points. Apply \((x,y)\mapsto(x+h,\,y+k)\) to intercepts and turning points to locate the image.
  4. Shift asymptotes, domain and range. Move a vertical asymptote and the domain by \(h\); move a horizontal asymptote and the range by \(k\).
  5. Sketch and check. Plot the mapped points and translated asymptotes, then confirm the shape matches the original curve.
Quick check. Substitute the new key point back into your equation — e.g. for \(y=(x-2)^2+1\) the point \((2,1)\) should satisfy it, confirming the turning point is correct.
Example 1 — vertical translation
The graph of \(y=x^{2}\) is translated \(3\) units up. Write the equation of the image and state its turning point.
Solution
Up \(3\) — add \(3\) to the rule (\(k=3\)):
\(y\)\(=\)\(f(x)+3\)
\(=\)\(x^{2}+3\)
Map the turning point — \((x,y)\mapsto(x,\,y+3)\):
\((0,0)\)\(\mapsto\)\((0,\,3)\)
\(\therefore\) \(y=x^{2}+3\); turning point \((0,3)\)
y=x2+3
Example 2 — horizontal translation
The graph of \(y=x^{3}\) is translated \(4\) units to the right. Find the equation of the image and the image of the point \((1,1)\).
Solution
Right \(4\) — replace \(x\) with \(x-4\) (\(h=4\)):
\(y\)\(=\)\(f(x-4)\)
\(=\)\((x-4)^{3}\)
Map the point — \((x,y)\mapsto(x+4,\,y)\):
\((1,1)\)\(\mapsto\)\((5,\,1)\)
Check — substitute \(x=5\):
\((5-4)^{3}\)\(=\)\(1^{3}=1\ \checkmark\)
\(\therefore\) \(y=(x-4)^{3}\); \((1,1)\mapsto(5,1)\)
y=(x4)3
Example 3 — combined translation, domain and range
The graph of \(y=x^{2}\) is translated \(3\) units left and \(2\) units down. Find the equation of the image, its turning point, domain and range.
Solution
Left \(3\) means \(h=-3\), down \(2\) means \(k=-2\) — substitute into \(y=f(x-h)+k\):
\(y\)\(=\)\((x-(-3))^{2}+(-2)\)
\(=\)\((x+3)^{2}-2\)
Map the turning point — \((x,y)\mapsto(x-3,\,y-2)\):
\((0,0)\)\(\mapsto\)\((-3,\,-2)\)
Domain is unchanged; range shifts down \(2\):
domain\(:\)\(x\in\mathbb{R}\)
range\(:\)\(y\ge -2\)
\(\therefore\) \(y=(x+3)^{2}-2\); T.P. \((-3,-2)\); domain \(\mathbb{R}\), range \(y\ge -2\)
y=x^2 translated 3 left and 2 downThe parabola y=x squared and its image y=(x+3) squared-2. The vertex moves from (0,0) to (-3,-2). x y y=x^2 y=(x+3)^2-2 (-3,-2)
y=(x+3)22
Example 4 — asymptotes of a translated reciprocal
The graph of \(y=\dfrac{1}{x}\) is translated \(2\) units right and \(1\) unit up. Find the equation of the image, its asymptotes, domain and range.
Solution
Right \(2\) (\(h=2\)): replace \(x\) with \(x-2\). Up \(1\) (\(k=1\)): add \(1\):
\(y\)\(=\)\(\dfrac{1}{x-2}+1\)
Shift the asymptotes — vertical by \(h\), horizontal by \(k\):
\(x=0\)\(\to\)\(x=2\)
\(y=0\)\(\to\)\(y=1\)
State the domain and range from the asymptotes:
domain\(:\)\(x\ne 2\)
range\(:\)\(y\ne 1\)
\(\therefore\) \(y=\dfrac{1}{x-2}+1\); asymptotes \(x=2,\ y=1\); domain \(x\ne 2\), range \(y\ne 1\)
y=1x2+1

Common pitfalls

The horizontal shift runs opposite to the sign. \(y=f(x-3)\) moves the graph \(3\) units right, not left, because the bracket \(x-3\) must equal the original input.
Do not apply \(k\) inside the function. \(y=f(x)+k\) is a vertical shift; \(y=f(x+k)\) is a horizontal one. \(x^{2}+2\) and \((x+2)^{2}\) are different graphs.
Move the asymptotes and domain/range too. A translation shifts asymptotes with the curve: for \(y=\dfrac{1}{x-2}+1\) the asymptotes are \(x=2\) and \(y=1\), not \(x=0\) and \(y=0\).

Frequently asked questions

What is a translation of a graph?

A translation slides a graph horizontally and/or vertically without changing its shape, size or orientation. Every point moves by the same amount, so the image is congruent to the original.

What does y = f(x - h) do to a graph?

Replacing \(x\) with \(x-h\) translates the graph horizontally by \(h\): right when \(h>0\), left when \(h<0\). So \(y=f(x-3)\) moves it \(3\) units right.

What does y = f(x) + k do to a graph?

Adding \(k\) translates the graph vertically by \(k\): up when \(k>0\), down when \(k<0\). So \(y=f(x)-2\) moves it \(2\) units down.

How do points move under a translation?

For \(y=f(x-h)+k\), every point maps \((x,y)\mapsto(x+h,\,y+k)\). Applying this to intercepts and turning points is the quickest way to draw the image.

How does a translation affect asymptotes, domain and range?

A horizontal shift of \(h\) moves any vertical asymptote and the domain by \(h\); a vertical shift of \(k\) moves any horizontal asymptote and the range by \(k\). So \(x=a\) becomes \(x=a+h\) and \(y=b\) becomes \(y=b+k\).

Why does x - h move the graph to the right?

The curve reaches a given height when \(x-h\) equals the original input, so \(x\) must be \(h\) larger. Because every point needs a larger \(x\), the whole graph shifts \(h\) units right.