Rational functions | Year 12

01 / LEARN

Understand the idea.

Factor before sketching

A rational function is a quotient of polynomials. Record the denominator zeros before cancelling factors; these are excluded inputs of the original function. Algebraic simplification changes the expression, not those original exclusions.

After complete cancellation, an excluded input where the simplified rule has a finite limit is a hole. A remaining denominator zero with nonzero numerator gives a pole. Numerator zeros give horizontal intercepts only when the inputs are allowed.

Use division and derivatives

Polynomial division can expose an asymptote by writing the function as a polynomial plus a remainder term. For a horizontal or oblique asymptote, that remainder term tends to zero as the input grows in magnitude.

Use the simplified rule to differentiate on the original domain. Separate the real line at denominator exclusions and critical inputs before checking derivative signs. To justify a range, establish global bounds and whether the endpoint values are actually attained.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A removable hole

Find the hole in $f (x) = (x^{2} - 1) / (x - 1)$ .

Worked solution

1
Record the original exclusion.
$x - 1 \neq 0 \Rightarrow x \neq 1.$
The original denominator is zero at one.
2
Factor and cancel only on the domain.
$f (x) = \frac{(x - 1) (x + 1)}{x - 1} = x + 1, x \neq 1.$
Cancellation is valid only when its common factor is nonzero.
3
Find the missing point on the line.
$lim_{x \to 1} f (x) = 1 + 1 = 2.$
The finite limiting ordinate is two, but the original function has no value at one; the hole is $(1, 2)$ .

Answer

$(1, 2)$ .

EXAMPLE 02

A rational function with extrema

Find the stationary points and range of $f (x) = x / (x^{2} + 1)$ .

Worked solution

1
Establish the domain and derivative.
$x^{2} + 1 > 0, f^{'} (x) = \frac{(x^{2} + 1) - 2 x^{2}}{{(x^{2} + 1)}^{2}} = \frac{1 - x^{2}}{{(x^{2} + 1)}^{2}} .$
The function and derivative exist for every real input.
2
Locate and classify stationary inputs.
$\begin{aligned} f^{'} (x) = 0 & \Rightarrow x = - 1, 1, \\ f^{'} (x) & < 0 & for ∣ x ∣ > 1, \\ f^{'} (x) & > 0 & for ∣ x ∣ < 1. \end{aligned}$
The sign changes from negative to positive at $- 1$ and positive to negative at $1$ .
3
Calculate the stationary points.
$f (- 1) = - \frac{1}{2}, f (1) = \frac{1}{2} .$
The local minimum is $(- 1, - \frac{1}{2})$ and the local maximum is $(1, \frac{1}{2})$ .
4
Prove these are global bounds.
$\begin{aligned} \frac{1}{2} - f (x) & = \frac{{(x - 1)}^{2}}{2 (x^{2} + 1)} \geq 0, \\ f (x) + \frac{1}{2} & = \frac{{(x + 1)}^{2}}{2 (x^{2} + 1)} \geq 0. \end{aligned}$
The nonnegative squares show that no value lies outside the stationary ordinates.
5
State the full range.
$range (f) = [- \frac{1}{2}, \frac{1}{2}] .$
Both bounds are attained, and continuity on $[- 1, 1]$ supplies every intermediate value.

Answer

Minimum $(- 1, - 1 / 2)$ , maximum $(1, 1 / 2)$ ; range $[- 1 / 2, 1 / 2]$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

For $(x^{2} - 9) / (x - 3)$ , is x=3 allowed?

Show answer & reasoning

No.

The factorisation $x^{2} - 9 = (x - 3) (x + 3)$ gives the rule $x + 3$ only for $x \neq 3$ . The original denominator excludes three, leaving a hole at $(3, 6)$ .

Keep practising

6 worksheet questions · 20 marks

Open worksheet Worked solutions

Read the questions online

Find the hole in $f (x) = \frac{x^{2} - 1}{x - 1}$ .
3 marks
Divide $\frac{x^{2} + 1}{x - 1}$ and state the oblique asymptote.
3 marks
Analyse $f (x) = \frac{2 x + 1}{x - 3}$ : give asymptotes and intercepts.
4 marks
Find the stationary points and range of $f (x) = \frac{x}{x^{2} + 1}$ .
4 marks
For $f (x) = \frac{x - 2}{x^{2} - 4}$ , state the domain, simplified rule and discontinuities.
4 marks
Can a graph cross a horizontal asymptote? Use $f (x) = \frac{x}{x^{2} + 1}$ to justify your answer.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the hole in $f (x) = \frac{x^{2} - 1}{x - 1}$ .
3 marks
For $x$ not equal to 1, simplify to $x + 1$ .
The excluded original input remains excluded.
$(1, 2)$ .
Divide $\frac{x^{2} + 1}{x - 1}$ and state the oblique asymptote.
3 marks
Polynomial division gives quotient $x + 1$ and remainder 2.
The remainder term tends to zero as the magnitude of $x$ grows.
$x + 1 + \frac{2}{x - 1}$ ; asymptote $y = x + 1$ .
Analyse $f (x) = \frac{2 x + 1}{x - 3}$ : give asymptotes and intercepts.
4 marks
Rewrite as $2 + \frac{7}{x - 3}$ .
Set the numerator to zero for the x-intercept; substitute $x = 0$ for the other.
Asymptotes $x = 3$ , $y = 2$ ; intercepts $(- \frac{1}{2}, 0)$ and $(0, - \frac{1}{3})$ .
Find the stationary points and range of $f (x) = \frac{x}{x^{2} + 1}$ .
4 marks
$f^{'} (x) = \frac{1 - x^{2}}{{(x^{2} + 1)}^{2}}$ .
The sign changes at $- 1$ and 1 establish the extrema.
Minimum $(- 1, - \frac{1}{2})$ , maximum $(1, \frac{1}{2})$ ; range $[- \frac{1}{2}, \frac{1}{2}]$ .
For $f (x) = \frac{x - 2}{x^{2} - 4}$ , state the domain, simplified rule and discontinuities.
4 marks
Factor the denominator into $(x - 2) (x + 2)$ .
Cancellation does not restore the excluded input 2.
Domain excludes $\pm 2$ ; $f (x) = \frac{1}{x + 2}$ for allowed $x$ ; pole at $- 2$ and hole at $(2, \frac{1}{4})$ .
Can a graph cross a horizontal asymptote? Use $f (x) = \frac{x}{x^{2} + 1}$ to justify your answer.
2 marks
A horizontal asymptote describes end behaviour, not a forbidden y-value.
Yes; it crosses its horizontal asymptote $y = 0$ at the origin.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist functions and sketching graphs. Calculus-based analysis assumes concurrent Methods differentiation; real-polynomial division detail is taught in the separate Real Polynomials chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

Factor before sketching

Use division and derivatives

See the method in action.

A removable hole

Record the original exclusion.

Factor and cancel only on the domain.

Find the missing point on the line.

A rational function with extrema

Establish the domain and derivative.

Locate and classify stationary inputs.

Calculate the stationary points.

Prove these are global bounds.

State the full range.

Try it yourself.

Check your understanding

Keep practising