IB Maths AI HL Transformations of Graphs Paper 1 & 2 Translation vectors, asymptotes ~8 min read

Translations of Graphs

A translation slides the entire graph — shape, size and orientation all preserved. y = f(x − a) shifts right by a; y = f(x) + b shifts up by b. Watch the sign: f(x − a) goes right, not left.

📘 What you need to know

Translation vector (a; b): a is horizontal (positive = right, negative = left); b is vertical (positive = up, negative = down).
Horizontal: y = f(x − a) translates by (a; 0). Note the sign flip — minus inside gives a positive horizontal shift.
Vertical: y = f(x) + b translates by (0; b). The sign here is natural.
Combined: y = f(x − a) + b translates by (a; b).
Points: a point (p, q) on f moves to (p + a, q + b) on the translated graph.
Asymptotes shift too: vertical asymptote x = k → x = k + a; horizontal asymptote y = k → y = k + b.

Horizontal vs vertical translation

A horizontal translation acts on the input: replacing x with (x − a) shifts the entire curve to the right by a. The sign flip catches everyone out — y = f(x − 3) is 3 to the right, not the left. A vertical translation acts on the output: adding b to the whole formula lifts the curve up by b, and the sign behaves naturally. x-coordinates change for horizontal shifts only; y-coordinates change for vertical shifts only.

Translating points and asymptotes

Every point on y = f(x) gets the translation vector added to it: (p, q) becomes (p + a, q + b). The vertex of a parabola, the maximum or minimum of any curve, intercepts — they all move by the same vector. Asymptotes move too: a vertical asymptote at x = k becomes x = k + a (horizontal shifts move vertical asymptotes), and a horizontal asymptote at y = k becomes y = k + b.

The vertex (0, −2) moves to (3, −2) under f(x − 3) and to (0, 2) under f(x) + 4. Every point on the curve translates by the same vector.

Translation at a glance y = f(x − a) + b ↔ vector (a; b) point (p, q) → (p + a, q + b) · asymptotes shift too

Quick practice with the sign rule

The horizontal sign flip is worth drilling. f(x − 5) shifts right 5; f(x + 5) shifts left 5. One sanity check: if (4, 7) is on f, then it lands on f(x − 5) at (4 + 5, 7) = (9, 7) — substitute back and check: at x = 9, the new graph evaluates f(9 − 5) = f(4) = 7 ✓. Always do this check when you’re unsure.

The “−” trick: in y = f(x − a), the inner bracket has to equal the same input that gave you the original y. So whatever x-input you needed before, you now need a more — hence shift right.

🧭 Recipe — translating a graph

Read the translation vector: (a; b) where a = horizontal shift, b = vertical shift.
Adjust the equation: replace x with (x − a) inside f, and add b outside.
Translate every key point: vertex, intercepts, marked points — add (a, b) to each.
Translate the asymptotes: vertical x = k moves to x = k + a; horizontal y = k moves to y = k + b.
Sketch the new curve — same shape, new position.

Worked examples

WE 1

Describe each translation

For the graph of y = f(x), state the translation vector that maps it to each new graph:
(a) y = f(x − 4)
(b) y = f(x) + 5
(c) y = f(x + 2)
(d) y = f(x) − 7
(e) y = f(x − 1) + 6

match each form to a vector (a) (4; 0) — 4 right (b) (0; 5) — 5 up (c) (−2; 0) — 2 left (d) (0; −7) — 7 down (e) (1; 6) — 1 right, 6 up five translation vectors above remember: inside (x − a) gives shift +a, not −a.

WE 2

Image of a point

The point P(3, −4) lies on the graph of y = f(x). Find the corresponding point on the graph of:
(a) y = f(x − 2)
(b) y = f(x + 5)
(c) y = f(x) − 6
(d) y = f(x − 1) + 4

add the vector to P (a) (3, −4) + (2, 0) = (5, −4) (b) (3, −4) + (−5, 0) = (−2, −4) (c) (3, −4) + (0, −6) = (3, −10) (d) (3, −4) + (1, 4) = (4, 0) (5, −4) · (−2, −4) · (3, −10) · (4, 0)

WE 3

Translate a quadratic in vertex form

Given f(x) = x² − 4x + 1. (a) Write f(x) in vertex form. (b) The graph is translated by the vector (3; −5). Find the equation of the new graph in vertex form.

(a) complete the square x² − 4x + 1 = (x − 2)² − 4 + 1 f(x) = (x − 2)² − 3, vertex (2, −3) (b) new function g(x) = f(x − 3) − 5 = ((x − 3) − 2)² − 3 − 5 = (x − 5)² − 8 g(x) = (x − 5)² − 8 new vertex (5, −8) = old vertex + (3, −5) ✓

WE 4

Find the translation vector

The graph of y = x² + 1 is translated by the vector (a; b) to give the graph of y = x² − 6x + 11. Find a and b.

put the image in vertex form x² − 6x + 11 = (x − 3)² − 9 + 11 = (x − 3)² + 2 compare vertices f: y = x² + 1, vertex (0, 1) image: vertex (3, 2) subtract to find vector (3 − 0, 2 − 1) = (3, 1) a = 3, b = 1 check: f(x − 3) + 1 = (x − 3)² + 1 + 1 = (x − 3)² + 2 ✓

WE 5

Asymptotes under translation

The graph of y = 1/x has vertical asymptote x = 0 and horizontal asymptote y = 0. State the asymptotes of:
(a) y = 1/(x − 5)
(b) y = 1/x + 4
(c) y = 1/(x + 2) − 3

apply the translation to each asymptote (a) vector (5; 0): vertical x = 5; horizontal y = 0 (b) vector (0; 4): vertical x = 0; horizontal y = 4 (c) vector (−2; −3): vertical x = −2; horizontal y = −3 three pairs of asymptotes above vertical asymptotes move horizontally; horizontal asymptotes move vertically.

WE 6

Applied: delayed launch from a platform

A projectile launched from ground level at time t = 0 has height (m) h(t) = −5t² + 20t. A second projectile is launched 3 s later from a platform 8 m above the ground but with the same trajectory shape. Let g(t) be its height. (a) State the translation vector relating g to h. (b) Find the new maximum height and when it occurs. (c) Find g(4).

(a) 3 s later, 8 m higher translation vector (3; 8) g(t) = h(t − 3) + 8 (b) original max h: vertex at t = 2, h(2) = −20 + 40 = 20 apply (3; 8) new max at t = 2 + 3 = 5, height = 20 + 8 = 28 max height 28 m at t = 5 s (c) g(4) = h(1) + 8 h(1) = −5 + 20 = 15 g(4) = 15 + 8 g(4) = 23 m

💡 Top tips

Use the translation vector notation (a; b) — examiners want correct terminology, not just “moves right”.
Vertex form is your friend for quadratics: read off the original vertex, add the vector, write the new vertex form directly.
Sketch key points, then connect smoothly — vertex, intercepts, marked points like A and B are usually all you need.
For asymptotes: vertical asymptotes follow horizontal shifts; horizontal asymptotes follow vertical shifts.
Sanity check by substituting a known point of f into the new equation.

⚠ Common mistakes

Reading f(x − 3) as “shift left 3” — it’s actually right 3. The minus flips inside the bracket.
Shifting the y-coordinate for a horizontal translation (or vice versa) — only the matching coordinate changes.
Ignoring asymptotes — they translate with the rest of the graph.
Using “moves up” instead of a vector in formal answers — lose marks for sloppy notation.
Adding the vector to the wrong direction: vertical translations don’t change x-coordinates.

Next up: Reflections of Graphs — flipping in the x-axis (y = −f(x)) and the y-axis (y = f(−x)).

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