IB Maths AA HL Topic 2 — Functions Paper 1 & 2 ~9 min read

Reciprocal & Rational Functions

A rational function is just a fraction whose top and bottom are both polynomials. The simplest is y = 1/x — the reciprocal function. The general linear-over-linear form (ax + b)/(cx + d) builds on it by stretching, shifting, and reflecting that basic shape. The four things you’ll always need to find: y-intercept, x-intercept, vertical asymptote, horizontal asymptote. Once you have these four, sketching the graph is automatic.

📘 What you need to know

Reciprocal function: f(x) = 1/x, with x ≠ 0. Domain and range both ℝ \ {0}. Self-inverse: f⁻¹(x) = f(x).
Reciprocal graph: two asymptotes (x = 0 and y = 0); two axes of symmetry (y = x and y = −x); no intercepts; no max/min; two branches in opposite quadrants.
Linear rational function: f(x) = (ax + b)/(cx + d), with x ≠ −d/c.
Y-intercept: substitute x = 0 → b/d. X-intercept: solve numerator = 0 → x = −b/a.
Vertical asymptote: solve denominator = 0 → x = −d/c.
Horizontal asymptote: y = a/c (ratio of leading coefficients — what the function tends to as x → ±∞).
Domain: ℝ except x = −d/c. Range: ℝ except y = a/c.
The inverse is also a rational function — find it by the swap-and-rearrange method (no need to memorise a formula).

The reciprocal function — y = 1/x

Reciprocal function f(x) = 1x, x ≠ 0
domain: ℝ \ {0} · range: ℝ \ {0}

Key features of y = 1/x

Vertical asymptote

x = 0

y-axis

Horizontal asymptote

y = 0

x-axis

Symmetry axes

y = ±x

two diagonals

Self-inverse

f = f⁻¹

flip a number, flip again

y = 1/x — the basic reciprocal graph

🤔 Why is it self-inverse?

Apply the function twice: f(f(x)) = 1/(1/x) = x. Flipping a number, then flipping again, gets you back to the original. So 1/x is its own inverse — it equals its mirror image in y = x.

Linear rational functions — (ax + b)/(cx + d)

Linear rational function f(x) = ax + bcx + d, x ≠ −d/c

This is the general linear-over-linear shape — same two-branch structure as 1/x, but shifted so the asymptotes aren’t on the axes anymore. The four key features come from simple substitutions:

y-intercept

y = b/d

substitute x = 0

x-intercept

x = −b/a

solve numerator = 0

Vertical asymptote

x = −d/c

solve denominator = 0

Horizontal asymptote

y = a/c

ratio of leading coefficients

A general (ax + b)/(cx + d) graph with all four features

Memorise the rule “denominator = 0 → vertical asymptote; ratio of leading coefficients → horizontal asymptote“. This works for every rational function in this section.

How to sketch a linear rational function

🧭 Recipe — sketching y = (ax + b)/(cx + d)

Find the y-intercept: substitute x = 0 → coordinate (0, b/d).
Find the x-intercept: set numerator = 0 → coordinate (−b/a, 0).
Find the vertical asymptote: set denominator = 0 → vertical line x = −d/c.
Find the horizontal asymptote: y = a/c — horizontal line.
Sketch: draw the asymptotes as dashed lines; mark the intercepts; draw two branches that approach the asymptotes without crossing them.
Label everything on your sketch — examiners check for asymptote equations and intercept coordinates.

Domain, range, and the inverse

Domain & range

domain: x ≠ −d/c
range: y ≠ a/c

excludes the vertical asymptote x-value; range excludes the horizontal asymptote y-value

Inverse

also a rational function

find by swap-and-rearrange (no formula to memorise)

Reflection check: the graph of f⁻¹(x) is the reflection of f(x) in the line y = x. So vertical and horizontal asymptotes swap when you take the inverse.

Worked examples

WE 1

Find the four key features of a linear rational function

For f(x) = 3x − 6x + 2, x ≠ −2, find: (a) the y-intercept, (b) the x-intercept, (c) the vertical asymptote, (d) the horizontal asymptote.

(a) y-intercept: substitute x = 0 f(0) = (0 − 6)/(0 + 2) = −6/2 = −3 → (0, −3) (b) x-intercept: numerator = 0 3x − 6 = 0 → x = 2 → (2, 0) (c) Vertical asymptote: denominator = 0 x + 2 = 0 → x = −2 (d) Horizontal asymptote: ratio a/c a = 3, c = 1 → y = 3 y-int (0, −3); x-int (2, 0); VA x = −2; HA y = 3 always do all four — the marks come from labelling, not algebra

WE 2

Sketch a rational function with all features labelled

Sketch the graph of f(x) = 2x + 4x − 1, marking the intercepts and asymptotes.

Find all four features y-int: f(0) = 4/(−1) = −4 → (0, −4) x-int: 2x + 4 = 0 → x = −2 → (−2, 0) VA: x − 1 = 0 → x = 1 HA: y = 2/1 = 2 Sketch draw asymptotes x = 1 and y = 2 as dashed lines left branch passes through (−2, 0) and (0, −4), approaches both asymptotes right branch sits in upper-right region, also approaching both sketch with VA x = 1, HA y = 2, x-int (−2, 0), y-int (0, −4) the two branches sit on opposite sides of where VA and HA meet — visualise that intersection point as the “centre”

WE 3

Find the inverse of a linear rational function

Find f⁻¹(x) for f(x) = x + 4x − 3 and state its domain.

Step 1: Set y = f(x) and swap x and y y = (x + 4)/(x − 3) swap → x = (y + 4)/(y − 3) Step 2: Multiply by (y − 3) and rearrange x(y − 3) = y + 4 xy − 3x = y + 4 xy − y = 3x + 4 y(x − 1) = 3x + 4 y = (3x + 4)/(x − 1) Step 3: Domain of f⁻¹ = range of f → exclude HA y = 1 f⁻¹(x) = (3x + 4)/(x − 1), x ≠ 1 note: HA of f was y = 1 (a/c = 1/1) — that becomes the excluded x-value for f⁻¹

WE 4

State domain and range

Find the domain and range of f(x) = 5x − 12x + 3.

Domain: exclude denominator = 0 2x + 3 = 0 → x = −3/2 domain: x ∈ ℝ, x ≠ −3/2 Range: exclude horizontal asymptote HA: y = 5/2 range: f(x) ∈ ℝ, f(x) ≠ 5/2 domain: x ≠ −3/2; range: f(x) ≠ 5/2 remember: domain excludes VA x-value; range excludes HA y-value

WE 5

Show 1/x is self-inverse

Show that f(x) = 1/x, x ≠ 0 is self-inverse, and find f(f(7)).

Compute f(f(x)) f(f(x)) = f(1/x) = 1 / (1/x) = x ✓ so f⁻¹ = f → self-inverse Apply at x = 7 f(7) = 1/7, then f(1/7) = 7 f is self-inverse; f(f(7)) = 7 flip, flip, back to start — works for any non-zero input

WE 6

Identify a rational function from its features

A rational function of the form f(x) = (ax + b)/(x + d) has vertical asymptote x = 4, horizontal asymptote y = 3, and y-intercept (0, −2). Find a, b, and d.

Step 1: VA gives d x = 4 means denominator zero at x = 4 → x + d = 0 → d = −4 Step 2: HA gives a y = a/c, with c = 1 (coefficient of x in denom) → a = 3 Step 3: y-intercept gives b f(0) = b/d = b/(−4) = −2 b = 8 a = 3, b = 8, d = −4 → f(x) = (3x + 8)/(x − 4) three features, three unknowns — work them out one at a time using the right-side relationships

💡 Top tips

The four features are the whole game: y-intercept (substitute x = 0); x-intercept (numerator = 0); VA (denominator = 0); HA (ratio of leading coefficients).
Asymptotes are dashed on a sketch. The curve approaches but doesn’t cross them.
Always label asymptote equations (x = …, y = …) and intercept coordinates on your sketch.
For domain: exclude any x that makes the denominator zero.
For range: exclude the horizontal asymptote y-value.
1/x is self-inverse — and so is any function of the form (ax + b)/(cx − a) (where the constant in denom is the negative of a).
Sketch with the GDC first if you’re unsure — then transfer the key features to your hand sketch.

⚠ Common mistakes

Drawing asymptotes as solid lines. They should always be dashed — the curve doesn’t actually touch them.
Forgetting to label intercepts on the sketch — examiners deduct marks for incomplete sketches.
Confusing horizontal asymptote rule: it’s the ratio of leading coefficients (a/c), not the ratio of constants.
Using b/d for the x-intercept. The y-intercept is b/d; the x-intercept is −b/a.
Sketching the curve crossing the asymptote. Linear-over-linear graphs never touch their asymptotes.
Forgetting the range excludes the horizontal asymptote. Domain ≠ range.
Not stating the domain restriction when finding an inverse — the inverse has its own VA-related exclusion.

The next note steps up the complexity to rational functions with quadratics — fractions with quadratics on top, on bottom, or both. The number of asymptotes can change (you may get two, one, or none), and you’ll meet a new type — oblique asymptotes — which require a polynomial division to find.

Need help with Reciprocal & Rational Functions?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →