Key Idea: Rational functions are two-branch hyperbolas that hug a vertical and a horizontal line called asymptotes. The exam asks you to find those asymptotes, the intercepts, and sketch — a Paper 1, by-hand skill where everything comes from simple algebra, not the GDC.
📈 The two shapes you must know
- horizontal shift — sets the vertical asymptote x = h
- vertical shift — sets the horizontal asymptote y = k
- leading coefficients — their ratio sets the horizontal asymptote y = a/c
| Reciprocal y = 1/(x − h) + k | Rational y = (ax + b)/(cx + d) | |
|---|---|---|
| Vertical asymptote | x = h (where x − h = 0) | denominator = 0, i.e. cx + d = 0 |
| Horizontal asymptote | y = k (the shift up) | y = a/c (ratio of leading coefficients) |
| Domain / range | x ≠ h, y ≠ k | x ≠ (the vertical asymptote); y ≠ a/c |
| x-intercept | set y = 0 and solve | numerator = 0, i.e. ax + b = 0 |
| y-intercept | put x = 0 | put x = 0, giving b/d |
To sketch: draw both asymptotes as dashed lines, mark any intercepts, then draw the two branches hugging the asymptotes (never crossing the vertical one). A reciprocal-type graph crosses each axis at most once.
✏️ IB-style worked examples
IB-style question — asymptotes of a shifted reciprocal
State the asymptotes of y = 1/(x − 4) + 3.
Step by step:
Vertical: set the denominator to zero.
Horizontal: the curve levels off at the shift up.
Vertical asymptote x = 4, horizontal asymptote y = 3.
🔒 Animated graph
Watch the graph build step by step in study mode.
IB-style question — asymptotes of a rational function
Find the vertical and horizontal asymptotes of y = (3x + 1)/(x − 5).
Step by step:
Vertical: denominator = 0.
Horizontal: ratio of leading coefficients (a = 3, c = 1).
Vertical asymptote x = 5, horizontal asymptote y = 3 (domain x ≠ 5).
🔒 Animated graph
Watch the graph build step by step in study mode.
IB-style question — intercepts for a full sketch
For y = (3x + 1)/(x − 5), find the x- and y-intercepts.
Step by step:
x-intercept: a fraction is zero only when the numerator is zero.
y-intercept: put x = 0, so y = b/d.
x-intercept (−1/3, 0); y-intercept (0, −1/5). With asymptotes x = 5 and y = 3, the sketch is complete.
🔒 Animated graph
Watch the graph build step by step in study mode.
Important: The vertical asymptote is where the denominator = 0, not the value you read off. 1/(x − 4) gives x = +4; (x + 6) on the bottom gives x = −6. And the horizontal asymptote is the ratio a/c, not just a — for (3x + 1)/(2x − 5) it's y = 3/2, not y = 3.
Tap each card to reveal the answer.
What are the asymptotes of y = 1/x? x = 0 and y = 0 — the two axes; domain x ≠ 0, range y ≠ 0.
Asymptotes of y = 1/(x − 2) + 5? Vertical x = 2, horizontal y = 5 — h shifts the vertical line, k the horizontal.
Vertical asymptote of y = (4x − 1)/(2x + 6)? Set 2x + 6 = 0 → x = −3; that x is also excluded from the domain.
Horizontal asymptote of y = (5x + 2)/(2x − 1)? Ratio of leading coefficients: y = 5/2.
Where does (2x + 1)/(x − 4) cross the x-axis? Numerator = 0: 2x + 1 = 0 → x = −1/2, so (−1/2, 0).
First step when sketching a rational function? Draw both asymptotes as dashed lines, then add intercepts and the two branches.
Exam tips
- Vertical asymptote = where the denominator equals 0 (and that x is barred from the domain).
- Horizontal asymptote: y = k for 1/(x − h) + k; y = a/c (leading coefficients) for (ax + b)/(cx + d).
- x-intercept: set the numerator = 0. y-intercept: put x = 0 (gives b/d).
- Watch the sign: (x − 4) → x = +4; (x + 4) → x = −4.
- Sketch order: dashed asymptotes first, then intercepts, then the two branches hugging them.