Inverse-proportion graphs and asymptotes | Year 11

01 / LEARN

Understand the idea.

A constant product

For inverse proportion, the product $x y = k$ is a fixed nonzero constant, so $y = \frac{k}{x}$ . Multiplying the input by a nonzero factor divides the output by that factor. Determine $k$ by multiplying a known input-output pair.

Equal additions to the input do not give equal output changes. To distinguish inverse proportion from a decreasing line, calculate products for the supplied pairs. Constant products support an inverse-proportion model for the supplied pairs; a downward trend alone does not. A finite set of pairs does not prove the rule at every other input.

x y = k, y = \frac{k}{x}

The products are constant for y=24/x
x	y	xy
-6	-4	24
-3	-8	24
2	12	24
4	6	24
8	3	24

Branches and asymptotes

For $y = \frac{k}{x}$ with $k \neq 0$ , neither $x$ nor $y$ can be zero. The axes are asymptotes: the curve approaches them without meeting them. Positive $k$ gives matching signs, while negative $k$ gives opposite signs.

Context can restrict the mathematical graph to one branch. Travel time and speed for a fixed positive distance are both positive. A shifted reciprocal such as $y = \frac{3}{x - 2} + 1$ has shifted asymptotes and does not make the product $x y$ constant.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find an inverse rule

y is inversely proportional to x, and y=8 when x=3. Find y when x=12.

Worked solution

1
Find the constant product from the known pair.
$k = x y = 3 (8) = 24$
The input-output product is the same for every pair in an inverse-proportion model.
2
Write the rule with its algebraic restriction.
$y = \frac{24}{x}, x \neq 0$
Dividing the constant product by the input gives the output.
3
Evaluate the new input and check the scale change.
$y = \frac{24}{12} = 2$
The input rose from three to twelve, a factor of four, so the output is one quarter of eight.

Answer

$y = 2$

EXAMPLE 02

Travel time

A 180 km trip is travelled at constant speed v km/h. Write the travel time t, then compare 60 km/h and 90 km/h.

Worked solution

1
Use the distance-speed-time relationship.
$180 = v t$
Constant speed multiplied by travel time gives the trip distance in kilometres.
2
Isolate the time and state its domain.
$t = \frac{180}{v}, v > 0$
A positive travel speed is required for this fixed positive-distance trip; the output is in hours.
3
Evaluate the time at each speed.
$\begin{aligned} t (60) & = \frac{180}{60} = 3 h \\ t (90) & = \frac{180}{90} = 2 h \end{aligned}$
Dividing kilometres by kilometres per hour leaves hours.
4
Compare the two journeys.
$\frac{90}{60} = \frac{3}{2}, \frac{2}{3} = \frac{1}{3 / 2}$
Multiplying speed by three halves reduces travel time by the reciprocal factor two thirds; the negative branch has no meaning here.

Answer

$t = 180 / v$ ; times are 3 h and 2 h.

03 / YOUR TURN

Try it yourself.

Check your understanding

If $y = 30 / x$ , what is the output at x=-5 and which inputs are excluded?

Show answer & reasoning

$y = - 6$ ; only x=0 is excluded from the algebraic domain.

Substituting gives $y = \frac{30}{- 5} = - 6$ . The algebraic rule excludes only $x = 0$ , since that is the only input that makes its denominator zero.

Keep practising

6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

$y$ is inversely proportional to $x$ , with $y = 6$ at $x = 4$ . Find its rule.
2 marks
For $y = \frac{- 18}{x}$ , find $y$ at $x = 3$ and $x = - 6$ .
2 marks
Determine whether $(2, 12)$ , $(3, 8)$ , $(6, 4)$ fit one inverse-proportion rule.
2 marks
For $y = \frac{5}{x + 2} - 3$ , state its vertical and horizontal asymptotes.
2 marks
A rectangular garden has fixed area 48 m². Express width $w$ in terms of length $l$ , and find $w$ when $l = 12$ m.
2 marks
Explain why $y = 12 - 2 x$ is not inverse proportion, even while it is positive and decreasing.
2 marks

Read the worked solutions

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

$y$ is inversely proportional to $x$ , with $y = 6$ at $x = 4$ . Find its rule.
2 marks
The constant product is $4 \times 6 = 24$ .
$y = \frac{24}{x}$
For $y = \frac{- 18}{x}$ , find $y$ at $x = 3$ and $x = - 6$ .
2 marks
Substitute each non-zero input separately.
-6 and 3.
Determine whether $(2, 12)$ , $(3, 8)$ , $(6, 4)$ fit one inverse-proportion rule.
2 marks
Every pair has product 24.
Yes: $y = \frac{24}{x}$ .
For $y = \frac{5}{x + 2} - 3$ , state its vertical and horizontal asymptotes.
2 marks
The denominator vanishes at -2. As the reciprocal term approaches zero, $y$ approaches -3.
$x = - 2$ and $y = - 3$ .
A rectangular garden has fixed area 48 m². Express width $w$ in terms of length $l$ , and find $w$ when $l = 12$ m.
2 marks
Use $l w = 48$ , then divide by positive $l$ .
$\frac{48}{12} = 4$ .
$w = \frac{48}{l}$ , with $l > 0$ ; $w = 4$ m.
Explain why $y = 12 - 2 x$ is not inverse proportion, even while it is positive and decreasing.
2 marks
At $x = 1$ , $y = 10$ and $x y = 10$ ; at $x = 2$ , $y = 8$ and $x y = 16$ .
Its product $x y$ is not constant.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics, Topic 1: Functions and graphs. Prerequisite algebra and coordinate geometry are embedded as supporting lessons.

Official curriculum source ↗

In this lesson

Understand the idea.

A constant product

Branches and asymptotes

See the method in action.

Find an inverse rule

Find the constant product from the known pair.

Write the rule with its algebraic restriction.

Evaluate the new input and check the scale change.

Travel time

Use the distance-speed-time relationship.

Isolate the time and state its domain.

Evaluate the time at each speed.

Compare the two journeys.

Try it yourself.

Check your understanding

Keep practising