Intercepts, stationary behaviour and asymptotes | Year 12

01 / LEARN

The idea, step by step

Collect exact features

Begin with the domain, intercepts and symmetry. Find poles and end behaviour using the algebraic form. Then solve f prime equals zero only where the function is defined.

An excluded input is never a stationary point. At a valid critical point, a derivative sign change distinguishes a local maximum or minimum from a stationary inflection.

Reason branch by branch

Split the real line at every pole and critical point. Draw the direction of change on each resulting interval, and connect only within the same domain branch.

Use extrema or a discriminant to justify a range. A sketch suggests a range but an exact argument rules out missing branches or falsely included endpoints.

A common trap

Calling every zero derivative a turning point.

Instead: Check whether the derivative changes sign; it may instead be a stationary inflection.

02 / WORKED EXAMPLES

See the thinking

Example 1

Extrema around a pole

For $f (x) = x^{2} / (x - 1)$ , find asymptotes and stationary points.

Division gives $x + 1 + 1 / (x - 1)$ .
$f^{'} (x) = x (x - 2) / {(x - 1)}^{2}$ .
The derivative changes from positive to negative at 0 and negative to positive at 2.

Asymptotes

x = 1

y = x + 1

; local maximum

(0, 0)

, local minimum

(2, 4)

Example 2

A stationary point without an extremum

For $f (x) = x^{3} / (x^{2} - 1)$ , explain why the origin is not a local extremum.

$f^{'} (x) = x^{2} (x^{2} - 3) / {(x^{2} - 1)}^{2}$ .
Its sign does not change at 0; the odd graph has a stationary inflection there.

The derivative vanishes at 0 but is negative immediately on both sides.

Check your understanding

Can x=1 be a stationary point of $x^{2} / (x - 1)$ ?

Show answer

No.

The function is not defined there.

03 / WORKSHEET

Your turn

6 questions · 26 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $f (x) = x^{2} / (x - 1)$ , find asymptotes and stationary points.
5 marks
For $f (x) = (x^{2} - 4) / (x^{2} - 1)$ , give intercepts, asymptotes and its stationary point.
5 marks
Find the stationary points of $f (x) = x + 1 / x$ and classify them.
4 marks
For $f (x) = x^{3} / (x^{2} - 1)$ , explain why the origin is not a local extremum.
4 marks
Find all x-intercepts of $f (x) = (x + 1) / (x^{2} - 1)$ and explain any excluded apparent intercept.
3 marks
Find the range of $f (x) = x^{2} / (x - 1)$ using its extrema or a discriminant.
5 marks

Read the worked solutions

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

For $f (x) = x^{2} / (x - 1)$ , find asymptotes and stationary points.
5 marks
Division gives $x + 1 + 1 / (x - 1)$ .
$f^{'} (x) = x (x - 2) / {(x - 1)}^{2}$ .
The derivative changes from positive to negative at 0 and negative to positive at 2.
Asymptotes $x = 1$ , $y = x + 1$ ; local maximum $(0, 0)$ , local minimum $(2, 4)$ .
For $f (x) = (x^{2} - 4) / (x^{2} - 1)$ , give intercepts, asymptotes and its stationary point.
5 marks
Rewrite as $1 - 3 / (x^{2} - 1)$ .
$f^{'} (x) = 6 x / {(x^{2} - 1)}^{2}$ , changing from negative to positive at 0.
Roots $\pm 2$ ; y-intercept 4; vertical asymptotes $\pm 1$ ; horizontal asymptote 1; local minimum $(0, 4)$ .
Find the stationary points of $f (x) = x + 1 / x$ and classify them.
4 marks
$f^{'} (x) = 1 - 1 / x^{2}$ .
The critical inputs are -1 and 1, with x=0 outside the domain.
Use derivative signs on the separate branches.
Local maximum $(- 1, - 2)$ and local minimum $(1, 2)$ .
For $f (x) = x^{3} / (x^{2} - 1)$ , explain why the origin is not a local extremum.
4 marks
$f^{'} (x) = x^{2} (x^{2} - 3) / {(x^{2} - 1)}^{2}$ .
Its sign does not change at 0; the odd graph has a stationary inflection there.
The derivative vanishes at 0 but is negative immediately on both sides.
Find all x-intercepts of $f (x) = (x + 1) / (x^{2} - 1)$ and explain any excluded apparent intercept.
3 marks
The numerator vanishes only at x=-1, which is excluded by the denominator.
The simplified reciprocal rule has no zero.
There are no x-intercepts.
Find the range of $f (x) = x^{2} / (x - 1)$ using its extrema or a discriminant.
5 marks
$y (x - 1) = x^{2}$ gives $x^{2} - y x + y = 0$ .
Real x requires $y^{2} - 4 y \geq 0$ , so y is at most 0 or at least 4.
The excluded input x=1 is never a root of this equation.
$(- \infty, 0] \cup [4, \infty)$ .

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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 ↗

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