IB Maths AA HLTopic 2 — FunctionsPaper 1 & 2~7 min read
Translations of Graphs
A translation slides a graph around the xy-plane without rotating it, stretching it, or flipping it — the shape stays exactly the same, only the position changes. There are two flavours: horizontal (left/right) and vertical (up/down). The trick is reading off the right vector from the equation, and remembering one counter-intuitive rule: f(x − 3) moves the graph to the right, not the left. Get that one straight and the rest follows.
📘 What you need to know
A translation moves a graph by a vector pq. The shape, size, and orientation are unchanged — only the position shifts.
Horizontal translation: y = f(x − p) shifts the graph by vector p0. Positive p → right, negative p → left.
Vertical translation: y = f(x) + q shifts the graph by vector 0q. Positive q → up, negative q → down.
Combined translation: y = f(x − p) + q shifts by vector pq. Horizontal and vertical translations are independent.
Vertical asymptotes shift horizontally; horizontal asymptotes shift vertically — each follows its matching translation only.
Every point (a, b) on the original graph maps to (a + p, b + q) on the translated graph.
The watch-out: f(x + 3) shifts the graph 3 units left, not right — the sign feels wrong but it’s correct.
What a translation does — slide, don’t bend
Picture the graph cut out of cardboard. A translation is when you slide that cardboard cutout to a new position on the plane — without rotating it, flipping it, or resizing it. Every point shifts by the same amount in the same direction.
That direction-and-distance is captured by a translation vector:
Translation vectorpq means “move p units horizontally, then q units vertically”
Sign of p
p > 0 → right p < 0 → left
horizontal component
Sign of q
q > 0 → up q < 0 → down
vertical component
Effect on points
(a, b) → (a+p, b+q)
every point shifts the same way
Horizontal translations — f(x − p)
Horizontal translationy = f(x − p) shifts y = f(x) by the vector p0
So y = f(x − 5) moves the graph 5 to the right, and y = f(x + 2) moves it 2 to the left. The sign inside the bracket looks backwards from what you’d expect — that’s the most common slip-up in the whole topic.
🤔 Why does the sign flip feel backwards?
The new graph at x = 7 should equal the old graph at x = 2 if we shifted right by 5. So we need f(7 − 5) = f(2). The bracket “subtracts off” the shift to get back to where the original graph had that y-value. That’s why right means minus.
What changes, what stays
Changes
x-coordinates of every point x-coordinates of any vertical asymptote
add p to every x-value
Stays the same
y-coordinates of every point any horizontal asymptote
vertical world is untouched
Vertical translations — f(x) + q
Vertical translationy = f(x) + q shifts y = f(x) by the vector 0q
This one behaves the way you’d expect — y = f(x) + 4 moves the graph 4 up, y = f(x) − 7 moves it 7 down. The constant sits outside the bracket, on the same side as y, so the sign makes sense.
What changes, what stays
Changes
y-coordinates of every point y-coordinates of any horizontal asymptote
add q to every y-value
Stays the same
x-coordinates of every point any vertical asymptote
horizontal world is untouched
Horizontal vs vertical translations of the same graph
When in doubt, pick a single point on the original graph (a vertex, an intercept, an asymptote crossing) and track where that point lands. If it goes where you expect, the rest of the graph will too.
Combined translations — both at once
Combined translationy = f(x − p) + q shifts y = f(x) by the vector pq
Horizontal and vertical translations are independent — they don’t interfere with each other, so you can apply them in either order and get the same result. Just read the two values straight off the equation: subtract inside the bracket gives the x-component, add or subtract outside gives the y-component.
Reading off the vector: from y = f(x − 5) + 2, the vector is 52. Notice the 5 keeps its positive sign because the bracket already does the flip for you. Just match the form f(x − p) + q.
Applying a translation — the recipe
🧭 Recipe — translating a graph
Identify the vectorpq from the question or the equation.
Pick the key features on the original graph: max/min points, intercepts, asymptotes, any labelled points.
Add the vector to each point: (a, b) → (a + p, b + q).
Shift the asymptotes: vertical asymptote x = k becomes x = k + p; horizontal asymptote y = m becomes y = m + q.
Sketch the new graph with the same shape as the original, but in its new position. Label all transformed features.
Worked examples
WE 1
Find the equation after a horizontal translation
The graph of f(x) = x2 − 6x + 5 is translated by the vector 40. Find the equation of the new graph in expanded form.
Step 1: Replace x with (x − 4) in f(x)new equation: y = (x − 4)² − 6(x − 4) + 5Step 2: Expand(x − 4)² = x² − 8x + 16−6(x − 4) = −6x + 24y = x² − 8x + 16 − 6x + 24 + 5y = x² − 14x + 45y = x² − 14x + 45a horizontal shift of +4 means swap every x for (x − 4) — the bracket sign flips because the graph slides right
WE 2
Find the equation after a vertical translation
The graph of g(x) = ex + 1 is translated by the vector 0−5. Find the equation of the new graph and the equation of its horizontal asymptote.
Step 1: Vertical translation — add −5 outsidenew equation: y = e^x + 1 − 5 = e^x − 4Step 2: Horizontal asymptoteoriginal HA: y = 1 (since e^x → 0 from above)shift down by 5: y = 1 − 5 = −4y = e^x − 4; HA: y = −4vertical translations behave the way the sign reads — minus 5 outside the bracket means down 5
WE 3
Sketch the image of a graph under a translation
The graph of y = f(x) has a maximum at P(−2, 7), a minimum at Q(4, −1), and crosses the x-axis at the origin.
(a) Find the image of P and Q under the translation y = f(x − 3).
(b) Find the image of P and Q under the translation y = f(x) + 4.
(a) y = f(x − 3): translation by vector (3, 0) — right by 3P(−2, 7) → (−2 + 3, 7) = (1, 7)Q(4, −1) → (4 + 3, −1) = (7, −1)(a) max at (1, 7); min at (7, −1)(b) y = f(x) + 4: translation by vector (0, 4) — up by 4P(−2, 7) → (−2, 7 + 4) = (−2, 11)Q(4, −1) → (4, −1 + 4) = (4, 3)(b) max at (−2, 11); min at (4, 3)the shape never changes — a max stays a max, a min stays a min — only the coordinates shift
WE 4
Identify the translation vector
The graph of y = f(x) is translated to give the graph of y = f(x + 6) − 2. State the translation vector.
Match to the form y = f(x − p) + qf(x + 6) = f(x − (−6)), so p = −6−2 outside the bracket gives q = −2Read off the vectortranslation vector = (p, q) = (−6, −2)vector = (−6, −2) — left 6, down 2the bracket sign flips when reading off p — that’s why f(x + 6) corresponds to a shift of −6 in the x-direction
WE 5
Translation involving asymptotes
The graph of y = 1x is translated by the vector 2−3.
(a) Find the equation of the new graph.
(b) State the equations of the two asymptotes.
(a) Apply the translationreplace x with (x − 2): y = 1/(x − 2)then add −3 outside: y = 1/(x − 2) − 3(a) y = 1/(x − 2) − 3(b) Shift the asymptotesoriginal VA x = 0 → shifts right by 2 → x = 2original HA y = 0 → shifts down by 3 → y = −3(b) VA: x = 2; HA: y = −3VAs follow the horizontal shift, HAs follow the vertical shift — each asymptote moves only with its matching component
WE 6
Find unknowns from a translation
The graph of y = f(x) passes through the point (1, 8) and has a vertical asymptote at x = −4. The graph of y = g(x) is the image of y = f(x) after a translation by the vector −53.
(a) Find a point on the graph of g. (b) State the equation of its vertical asymptote.
(a) Image of (1, 8) under translation (−5, 3)x: 1 + (−5) = −4y: 8 + 3 = 11(a) (−4, 11) lies on g(x)(b) Vertical asymptote shifts only with the horizontal componentx = −4 → x = −4 + (−5) = −9(b) VA: x = −9add the vector components to every x-y pair — including any feature with an x-value (like a vertical asymptote)
💡 Top tips
Memorise the bracket flip: f(x − p) shifts right by p; f(x + p) shifts left by p.
Vertical translations behave normally: f(x) + q shifts up by q when q is positive.
Horizontal and vertical are independent — handle each component separately, then put them together.
Track a single point: pick a labelled feature like a vertex or intercept, apply the vector, and the rest of the graph follows.
Vertical asymptotes shift horizontally only; horizontal asymptotes shift vertically only. Each follows its own component.
Use proper vector notation in your answers: write the translation as a column vector pq, not as words.
Check with a specific value: substitute one x into both the old and new equations and confirm the y-values match the shift you intended.
⚠ Common mistakes
Mixing up the bracket sign: thinking f(x + 3) shifts right when it actually shifts left.
Applying the vertical shift inside the bracket: putting f(x + 4) when you meant f(x) + 4.
Shifting horizontal asymptotes horizontally — they only respond to vertical translations.
Forgetting to translate intercepts when sketching: the new x– and y-intercepts are usually different from the originals.
Writing the translation as a row vector or just words. Examiners want a column vector — clear and unambiguous.
Translating a max into a min (or vice versa). The shape doesn’t change — a max stays a max.
Forgetting that “vector” answers need both components: stating just “right by 3” loses marks. Give 30.
Translations are the gentlest of the four transformations — they only slide the graph around. The next note covers reflections, where the graph gets flipped across the x– or y-axis. Same idea of tracking how each point moves, but now the orientation changes too.
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