2.13 Further rational functions (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Which rational functions are on the syllabus?

At HL you meet two new forms:

$f(x) = \dfrac{ax + b}{cx^{2} + dx + e}$ — linear over quadratic;
$f(x) = \dfrac{ax^{2} + bx + c}{dx + e}$ — quadratic over linear.

For each you must find all asymptotes (vertical, horizontal or oblique) and any intercepts with the axes, and describe the graph from them.

Key termsrational function

Section 2

Vertical asymptotes and intercepts

Vertical asymptotes occur where the denominator is zero and the numerator is not. For $\frac{2x + 3}{x^{2} - x - 6} = \frac{2x + 3}{(x - 3)(x + 2)}$ they are $x = 3$ and $x = -2$ .

If the quadratic denominator has no real roots (negative discriminant), there is no vertical asymptote.

$x$ -intercepts: numerator $= 0$ .
$y$ -intercept: substitute $x = 0$ (if $0$ is in the domain).

Key termsvertical asymptote

Common mistake

Taking the sign straight from the factor: $(x + 2)$ gives the asymptote $x = -2$ , not $x = 2$ .

Exam tip

Check the sign of $f$ just either side of a vertical asymptote to decide whether the graph goes up or down.

Section 3

Linear over quadratic: horizontal asymptote y = 0

When the numerator has lower degree than the denominator, $f(x) \to 0$ as $x \to \pm\infty$ , so the horizontal asymptote is $y = 0$ .

Unlike a vertical asymptote, the graph can cross a horizontal asymptote: $\frac{2x + 3}{x^{2} - x - 6}$ crosses $y = 0$ at $x = -\frac{3}{2}$ .

To find the range, set $f(x) = k$ , form a quadratic in $x$ and ask when its discriminant is $\ge 0$ (treat $k = 0$ separately). For $\frac{2x + 4}{x^{2} + 2x - 8}$ the discriminant $36k^{2} + 8k + 4$ is always positive, so the range is $\mathbb{R}$ .

Key termshorizontal asymptote

Common mistake

Thinking a graph can never cross a horizontal asymptote — it can, for finite $x$ .

Section 4

Quadratic over linear: oblique asymptotes

When the numerator has degree one more than the denominator, divide to write $f(x) = mx + c + \frac{r}{dx + e}.$ As $x \to \pm\infty$ the last term $\to 0$ , so the graph approaches the oblique asymptote $y = mx + c$ . There is also a vertical asymptote at $x = -\frac{e}{d}$ (provided the remainder $r \ne 0$ ).

Example: $\frac{x^{2} + 3x - 1}{x - 2} = x + 5 + \frac{9}{x - 2}$ has asymptotes $y = x + 5$ and $x = 2$ . There is no horizontal asymptote.

Key termsoblique asymptote

Exam tip

The sign of the remainder term tells you which side of the oblique asymptote the graph is on: $\frac{9}{x - 2} > 0$ for $x > 2$ , so the curve is above $y = x + 5$ there.

Common mistake

Writing the oblique asymptote from the leading terms only, e.g. y = x + 3 from (x² + 3x)/x, without dividing properly.

Section 5

Rational functions as models

Average-cost models often take the form $C(x) = \frac{\text{quadratic}}{\text{linear}}$ . Rewriting as $C(x) = x + 2 + \frac{64}{x + 2}$ shows the long-run behaviour (oblique asymptote) and makes calculus and inequalities simpler: the minimum is at $(x + 2)^{2} = 64$ .

When solving $C(x) \le k$ , multiply through by the denominator only when you know its sign (here $x + 2 > 0$ because $x > 0$ ).

Exam tip

Interpret asymptotes in context: the oblique asymptote gives the approximate cost for large x; the vertical asymptote is usually outside the model's domain.

Must know

Vertical asymptotes: denominator zero (numerator non-zero).
Linear over quadratic: horizontal asymptote $y = 0$ ; the graph may cross it.
Quadratic over linear: divide to find the oblique asymptote $y = mx + c$ ; no horizontal asymptote.
Always give the $x$ - and $y$ -intercepts.
Range: solve $f(x) = k$ as a quadratic and use the discriminant.

That's the notes covered.

Carry on to the next subtopic.