Intercepts, stationary behaviour and asymptotes | Year 12

01 / LEARN

Understand the idea.

Collect exact features

Collect the domain, intercepts, symmetry and asymptotes before using calculus. A derivative zero is a stationary input only if the original function is defined there. Denominator exclusions must never be labelled stationary points.

At an allowed derivative zero, test the signs on its two sides. Positive to negative gives a local maximum; negative to positive gives a local minimum. No sign change means it is not an extremum; further concavity evidence may establish an inflection.

Reason branch by branch

Separate the real line at every pole and stationary input. Record the derivative sign on each resulting interval and draw each continuous domain branch separately. A pole interrupts a branch even if derivative signs match across it.

Asymptotes describe limits rather than ordinary points of the graph. An oblique asymptote comes from a remainder tending to zero after division. Use exact extrema, limits or a discriminant argument when the range also needs justification.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Extrema around a pole

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

Worked solution

1
Divide to expose the end behaviour.
$f (x) = \frac{x^{2}}{x - 1} = x + 1 + \frac{1}{x - 1}, x \neq 1.$
The denominator excludes one and the remainder term tends to zero at either infinite end.
2
State the asymptotes.
$x = 1, y = x + 1.$
The surviving pole gives a vertical asymptote; the vanishing difference from $x + 1$ gives an oblique asymptote.
3
Differentiate and solve on the domain.
$\begin{aligned} f^{'} (x) & = \frac{2 x (x - 1) - x^{2}}{{(x - 1)}^{2}} \\ = \frac{x (x - 2)}{{(x - 1)}^{2}}, \\ f^{'} (x) = 0 & \Rightarrow x = 0, 2. \end{aligned}$
The denominator is positive wherever the function is defined.
4
Check signs on intervals separated by the pole.
$\begin{aligned} f^{'} & > 0 & on (- \infty, 0) and (2, \infty), \\ f^{'} & < 0 & on (0, 1) and (1, 2) . \end{aligned}$
There is a positive-to-negative change at zero and a negative-to-positive change at two.
5
Calculate and classify the points.
$f (0) = 0, f (2) = 4; local maximum (0, 0), local minimum (2, 4) .$
The two extrema belong to separate domain branches, so their ordinates do not conflict.

Answer

Asymptotes $x = 1$ , $y = x + 1$ ; local maximum $(0, 0)$ , local minimum $(2, 4)$ .

EXAMPLE 02

A stationary point without an extremum

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

Worked solution

1
Verify the origin is a stationary point.
$\begin{aligned} f (0) & = 0, \\ f^{'} (x) & = \frac{3 x^{2} (x^{2} - 1) - 2 x^{4}}{{(x^{2} - 1)}^{2}} \\ = \frac{x^{2} (x^{2} - 3)}{{(x^{2} - 1)}^{2}}, f^{'} (0) = 0. \end{aligned}$
The function is defined at zero and its tangent there is horizontal.
2
Check the derivative immediately on both sides.
$0 < ∣ x ∣ < 1 : x^{2} > 0, x^{2} - 3 < 0, {(x^{2} - 1)}^{2} > 0.$
Thus $f^{'} (x) < 0$ on both sides of zero, so the curve decreases through the stationary point.
3
Rule out a local extremum.
$f^{'} : - → - across x = 0.$
There is no change from increasing to decreasing or vice versa, so the origin is neither a local maximum nor a local minimum.
4
Confirm the inflection classification.
$f^{''} (x) = \frac{2 x (x^{2} + 3)}{{(x^{2} - 1)}^{3}} .$
Near zero this is positive on the left and negative on the right, so concavity changes and the origin is a stationary inflection.

Answer

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

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

No.

For $f (x) = \frac{x^{2}}{x - 1}$ , the input $x = 1$ makes the denominator zero. Since there is no graph point at that input, it cannot be stationary.

Keep practising

6 worksheet questions · 26 marks

Open worksheet Worked solutions

Read the questions online

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

Read the worked solutions

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

For $f (x) = \frac{x^{2}}{x - 1}$ , find asymptotes and stationary points.
5 marks
Division gives $x + 1 + \frac{1}{x - 1}$ .
$f^{'} (x) = \frac{x (x - 2)}{{(x - 1)}^{2}}$ .
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 $f (x) = \frac{x^{2} - 4}{x^{2} - 1}$ , give intercepts, asymptotes and its stationary point.
5 marks
Rewrite as $1 - \frac{3}{x^{2} - 1}$ .
$f^{'} (x) = \frac{6 x}{{(x^{2} - 1)}^{2}}$ , 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 $f (x) = x + \frac{1}{x}$ and classify them.
4 marks
$f^{'} (x) = 1 - \frac{1}{x^{2}}$ .
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 $f (x) = \frac{x^{3}}{x^{2} - 1}$ , explain why the origin is not a local extremum.
4 marks
$f^{'} (x) = \frac{x^{2} (x^{2} - 3)}{{(x^{2} - 1)}^{2}}$ .
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 $f (x) = \frac{x + 1}{x^{2} - 1}$ 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 $f (x) = \frac{x^{2}}{x - 1}$ 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 ↗

In this lesson

Understand the idea.

Collect exact features

Reason branch by branch

See the method in action.

Extrema around a pole

Divide to expose the end behaviour.

State the asymptotes.

Differentiate and solve on the domain.

Check signs on intervals separated by the pole.

Calculate and classify the points.

A stationary point without an extremum

Verify the origin is a stationary point.

Check the derivative immediately on both sides.

Rule out a local extremum.

Confirm the inflection classification.

Try it yourself.

Check your understanding

Keep practising