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

Rational Functions with Quadratics

When a quadratic shows up on top, on bottom, or both, you still hunt for the same four feature types — y-intercept, x-intercepts, vertical asymptotes, end-behaviour asymptote. The differences are: there can now be 0, 1, or 2 of each kind, and the end-behaviour asymptote can be oblique (a slanted line) instead of horizontal. Polynomial division is the new tool you’ll need to find that oblique asymptote.

📘 What you need to know

General form: f(x) = g(x)/h(x) where g and h are polynomials.
Y-intercept: substitute x = 0 → evaluate g(0)/h(0). Defined only if h(0) ≠ 0.
X-intercepts: solve numerator g(x) = 0. Can be 0, 1, or 2 roots depending on the discriminant.
Vertical asymptotes: solve denominator h(x) = 0. Can be 0, 1, or 2 vertical asymptotes.
End-behaviour asymptote (quadratic over linear): oblique asymptote y = px + q — found by polynomial division so that g(x) = h(x)·(px + q) + r.
End-behaviour asymptote (linear over quadratic): horizontal asymptote y = 0 (the bottom grows much faster than the top).
Sketches must label: all asymptotes (with equations), all intercepts (with coordinates), and the general shape of each branch.
Use the GDC on Paper 2 to confirm the shape — but always work the algebra to find exact features.

The general approach — same four features, more cases

Y-intercept

f(0) = g(0)/h(0)

substitute x = 0; defined unless h(0) = 0

X-intercepts

g(x) = 0

solve the numerator; can be 0, 1, or 2 solutions

Vertical asymptotes

h(x) = 0

solve the denominator; can be 0, 1, or 2 solutions

The number of x-intercepts and vertical asymptotes is set by the discriminant of the relevant quadratic. Δ > 0 → two; Δ = 0 → one; Δ < 0 → none. Same rule you used for ordinary quadratics.

Quadratic over linear — the oblique asymptote case

Quadratic over linear f(x) = ax² + bx + cdx + e

For very large |x|, this function behaves like a straight line — but it’s not horizontal anymore. It’s an oblique asymptote (slanted), and it’s found by polynomial division.

Finding the oblique asymptote

Rewrite the numerator in the form ax² + bx + c = (dx + e)(px + q) + r:

Polynomial division identity ax² + bx + cdx + e = (px + q) + rdx + e

As |x| → ∞, the remainder term r/(dx + e) → 0, so the function tends to the line y = px + q. That’s the oblique asymptote.

🧭 Recipe — finding an oblique asymptote

Write the numerator as (denominator)(px + q) + r.
Compare coefficients: x², x, and constant terms must match on both sides.
Solve the system to find p, q, and r.
The oblique asymptote is y = px + q (the remainder r is irrelevant).

Quadratic over linear — vertical asymptote + oblique asymptote

Linear over quadratic — horizontal asymptote at y = 0

Linear over quadratic f(x) = ax + bcx² + dx + e

Now the bottom grows faster than the top — so as |x| gets large, the fraction shrinks toward zero:

Horizontal asymptote y = 0 (the x-axis)

How many vertical asymptotes?

That depends entirely on the denominator’s discriminant. The graph can have 0, 1, or 2 vertical asymptotes:

Δ > 0

two distinct VAs

denominator factors → two values to exclude

Δ = 0

one VA (repeated root)

denominator is a perfect square

Δ < 0

no VAs

denominator never zero on the reals

For linear-over-quadratic graphs, a horizontal line drawn anywhere should hit the curve at most twice. If your sketch shows three crossings on a horizontal line, something’s gone wrong.

The general sketching recipe

🧭 Recipe — sketching a quadratic rational function

Find the y-intercept: substitute x = 0.
Find the x-intercepts: solve numerator = 0.
Find the vertical asymptote(s): solve denominator = 0.
Find the end-behaviour asymptote: horizontal (y = 0) for linear-over-quadratic, oblique (use polynomial division) for quadratic-over-linear.
Draw all asymptotes as dashed lines; mark intercepts as dots with coordinates.
Sketch each branch based on which sides of the asymptotes it sits on. Confirm with your GDC if available.

Worked examples

WE 1

Quadratic over linear — find all features

For f(x) = x² − x − 6x + 1, x ≠ −1, find: (a) y-intercept, (b) x-intercepts, (c) vertical asymptote, (d) oblique asymptote.

(a) y-intercept: f(0) f(0) = (0 − 0 − 6)/(0 + 1) = −6 → (0, −6) (b) x-intercepts: x² − x − 6 = 0 (x − 3)(x + 2) = 0 → x = 3 or x = −2 (3, 0) and (−2, 0) (c) Vertical asymptote: x + 1 = 0 → x = −1 (d) Oblique asymptote — polynomial division write x² − x − 6 = (x + 1)(px + q) + r expand: (x + 1)(px + q) + r = px² + (p + q)x + q + r compare: p = 1, p + q = −1 → q = −2, q + r = −6 → r = −4 so x² − x − 6 = (x + 1)(x − 2) − 4 (a) (0, −6); (b) (−2, 0), (3, 0); (c) VA x = −1; (d) oblique y = x − 2 notice the numerator factors nicely AND has a clean oblique asymptote — typical exam set-up

WE 2

Linear over quadratic — find all features

For g(x) = x − 2x² − 4x + 3, find: (a) y-intercept, (b) x-intercept, (c) vertical asymptote(s), (d) horizontal asymptote.

(a) y-intercept: g(0) g(0) = (−2)/3 = −2/3 → (0, −2/3) (b) x-intercept: x − 2 = 0 → x = 2 → (2, 0) (c) Vertical asymptotes: x² − 4x + 3 = 0 (x − 1)(x − 3) = 0 → x = 1 or x = 3 two VAs: x = 1 and x = 3 (d) Horizontal asymptote: linear over quadratic → y = 0 (a) (0, −2/3); (b) (2, 0); (c) VAs x = 1 and x = 3; (d) HA y = 0 three vertical lines (the two VAs and the y-axis) cut the plane into regions — work out the sign of g in each

WE 3

Find an oblique asymptote by comparing coefficients

Show that f(x) = 3x² + 7x − 1x + 2 can be written as f(x) = px + q + rx + 2, and hence find the oblique asymptote.

Step 1: Set up identity 3x² + 7x − 1 = (x + 2)(px + q) + r Step 2: Expand RHS (x + 2)(px + q) = px² + (q + 2p)x + 2q + r → 3x² + 7x − 1 = px² + (q + 2p)x + (2q + r) Step 3: Compare coefficients x²: p = 3 x: q + 2p = 7 → q + 6 = 7 → q = 1 const: 2q + r = −1 → 2 + r = −1 → r = −3 f(x) = 3x + 1 − 3/(x + 2); oblique asymptote: y = 3x + 1 the remainder term r/(x+2) → 0 as x → ±∞, so the function settles onto the line y = 3x + 1

WE 4

Determine the number of vertical asymptotes

State the number of vertical asymptotes of each function:
(a) f(x) = x + 5x² − 6x + 8 (b) g(x) = 2x − 1x² + 4x + 4 (c) h(x) = x + 3x² + 1

(a) discriminant of x² − 6x + 8 Δ = 36 − 32 = 4 > 0 → two distinct roots (a) 2 vertical asymptotes (b) discriminant of x² + 4x + 4 Δ = 16 − 16 = 0 → one repeated root (b) 1 vertical asymptote (c) discriminant of x² + 1 Δ = 0 − 4 = −4 < 0 → no real roots (c) 0 vertical asymptotes the discriminant of the denominator alone tells you everything about VAs — no need to plot

WE 5

Sketch a quadratic-over-linear graph

Sketch f(x) = x² − 4x − 3, marking all intercepts and asymptotes.

Find features y-int: f(0) = (−4)/(−3) = 4/3 → (0, 4/3) x-ints: x² − 4 = 0 → x = ±2 → (−2, 0) and (2, 0) VA: x − 3 = 0 → x = 3 Oblique asymptote — polynomial division x² − 4 = (x − 3)(x + 3) + 5 f(x) = x + 3 + 5/(x − 3) oblique: y = x + 3 Sketch elements dashed VA x = 3, dashed oblique y = x + 3 left branch passes through (−2, 0), (0, 4/3), (2, 0); follows oblique downward to lower-left, dives down toward VA right branch comes from top of VA, follows oblique up to top-right sketch with VA x = 3, oblique y = x + 3, x-ints (−2, 0) and (2, 0), y-int (0, 4/3) when both intercepts lie on the same branch (left of VA here), the curve is bowed — usually with a local max/min between them

WE 6

Find a function from its features

A function of the form f(x) = k(x − 1)(x + 4) has y-intercept (0, 1/2). Find k and state the equations of the vertical asymptotes.

Step 1: Use y-intercept f(0) = k / [(−1)(4)] = k/(−4) = 1/2 k = −2 Step 2: Vertical asymptotes — denominator zero (x − 1)(x + 4) = 0 → x = 1 or x = −4 k = −2; VAs: x = 1 and x = −4 denominator already factored — read off the VAs immediately, then use the y-intercept to nail down k

💡 Top tips

Always check both top and bottom for the discriminant: it tells you the number of x-intercepts (numerator) and vertical asymptotes (denominator).
Quadratic over linear → oblique asymptote. Find it by polynomial division or by comparing coefficients.
Linear over quadratic → horizontal asymptote y = 0. The bottom grows faster than the top.
Compare-coefficients method beats long division for most exam questions — it’s faster and less error-prone.
The remainder doesn’t matter for the asymptote — the asymptote is just the polynomial part of the division.
Plot on the GDC first when uncertain — see the shape, then check it matches your algebraic features.
Label everything on the sketch: each asymptote with its equation, each intercept with its coordinates. Marks come from labels.

⚠ Common mistakes

Forgetting the oblique asymptote when the numerator’s degree exceeds the denominator’s. It’s there, just slanted.
Using the wrong horizontal asymptote rule: y = 0 only works when the bottom degree exceeds the top degree.
Ignoring the discriminant of the denominator. It controls how many VAs the graph has — sometimes zero.
Including the remainder in the asymptote equation. Only the polynomial part (px + q) is the asymptote.
Confusing “no real roots” with “no graph”. A quadratic with negative discriminant just has no real x-intercepts — the curve still exists.
Sketching the curve crossing a vertical asymptote. It can’t — that’s where the function is undefined.
Forgetting to label asymptotes as dashed lines. Solid lines on a sketch suggest the curve passes through them.

And that closes Section 2.8 — Reciprocal & Rational Functions. The next section, Transformations of Graphs, takes any function you’ve already met and shifts, stretches, or flips it. The patterns learned there apply across every later topic — trig graphs, polynomial graphs, modulus graphs, all of them.

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