Lesson 4 ~40 min Variation A · Path +85 XP

Graphs of Inverse Variation

The graph of an inverse relationship is a curve, not a line, and it never meets either axis. Recognising it by eye is unreliable, so this lesson also gives the reliable test: straighten the curve and see whether what is left is a line through the origin.

Today's hook: Plenty of curves fall steeply and then flatten out. Only some of them are inverse variation, and no amount of squinting at a sketch will tell you which. There is a trick that turns the question into one you already know how to answer, and it takes one extra column in a table.

0/5QUESTS

Think First

warm-up

Work out $\dfrac{12}{x}$ for $x = 0.5$, $x = 1$, $x = 12$ and $x = 1000$. Now say what happens to the answer as $x$ gets very large, and what happens as $x$ gets very close to zero. In each case, does the answer ever actually reach a particular value?

Record your answer in your workbook.

The Big Idea

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The graph of $y = \dfrac{k}{x}$ is a curve in two separate branches, called a hyperbola. It approaches both axes without ever touching either, and the two branches never join.

$$y = \frac{k}{x} \quad \text{a curve, not a line}$$

Eyeballing a curve is not a test. The reliable check is to plot $y$ against $\dfrac{1}{x}$: if the original relationship is inverse variation, that new graph is a straight line through the origin, because $y = k \times \dfrac{1}{x}$ is direct variation in disguise.

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

Two branches

One for positive $x$, one for negative. The curve cannot cross between them.

Never touches an axis

The axes are asymptotes. The curve gets arbitrarily close and never arrives.

Straighten to check

Plot $y$ against $\tfrac{1}{x}$. A line through the origin confirms inverse variation.

What You'll Master

objectives

Know

That the graph of $y = \tfrac{k}{x}$ is a curve in two branches
That the curve approaches both axes without meeting them
That plotting $y$ against $\tfrac{1}{x}$ straightens an inverse relationship

Understand

Why the curve cannot touch either axis
Why the straightening test works, and why it is more reliable than looking at the curve

Can Do

Sketch the graph of an inverse relationship from its equation
Recognise inverse variation from a graph or a table
Use the reciprocal plot to test whether data show inverse variation

Words You Need

vocabulary

HyperbolaThe curve produced by an inverse relationship, in two branches.

BranchOne connected piece of a graph that comes in separate parts.

AsymptoteA line the graph approaches without ever reaching.

Reciprocal plotA graph of $y$ against $\tfrac{1}{x}$, used to straighten an inverse relationship.

LineariseTo transform data so that a curved relationship becomes a straight line.

The Shape, and Why

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The graph of $y = \dfrac{k}{x}$ with $k > 0$ has three features worth naming.

It is a curve, not a straight line. Between $x = 1$ and $x = 2$ the value drops from $k$ to $\tfrac{k}{2}$, a fall of $\tfrac{k}{2}$. Between $x = 5$ and $x = 6$ it falls only from $\tfrac{k}{5}$ to $\tfrac{k}{6}$, a fall of about $\tfrac{k}{30}$. Equal steps in $x$ produce unequal steps in $y$, and that is exactly what makes a graph curved rather than straight.

It never touches either axis. It cannot touch the vertical axis, because $x = 0$ makes the expression undefined. It cannot touch the horizontal axis either, because $y = 0$ would require $\tfrac{k}{x} = 0$, and a fraction is zero only when its numerator is, which here it never is. The two axes are asymptotes: lines the curve approaches arbitrarily closely without arriving.

It comes in two branches. One where $x$ is positive and $y$ is positive, and one where both are negative, since $k$ is positive and a positive divided by a negative is negative. The curve cannot cross from one branch to the other, because getting between them would mean passing through $x = 0$, where nothing exists.

In most contexts only the first branch is meaningful, since speeds, worker counts and volumes are positive.

Sketching One

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A sketch of $y = \dfrac{k}{x}$ needs only a few points and the two asymptotes.

1. Draw the axes lightly and remember they are the asymptotes.
2. Plot three or four points on the positive branch, choosing $x$ values that divide into $k$ neatly.
3. Join them with a smooth curve that flattens towards the horizontal axis on the right and rises steeply towards the vertical axis on the left.
4. Add the second branch by reflecting through the origin, if negative values are meaningful.

For $y = \dfrac{12}{x}$, convenient points are $(1,12)$, $(2,6)$, $(3,4)$, $(4,3)$, $(6,2)$ and $(12,1)$. Notice they pair up: $(2,6)$ and $(6,2)$, $(3,4)$ and $(4,3)$. That symmetry is a consequence of the product being fixed, and it is a fast way to double your points.

Two drawing errors to avoid. Do not let the curve flatten into a horizontal line at the right, because it is still falling, just very slowly. And do not let it meet the vertical axis at the top, however steep it becomes: leave a visible gap so the sketch says what the mathematics says.

The Straightening Test

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Deciding whether a curve is an inverse relationship by looking at it is unreliable. Many decreasing curves flatten out, and a sketch cannot distinguish $\tfrac{k}{x}$ from other shapes.

The reliable method converts the question into one already answered in Lesson 3.

If $y$ varies inversely as $x$, then $y = k \times \dfrac{1}{x}$, which says $y$ varies directly as $\dfrac{1}{x}$. So plotting $y$ against $\dfrac{1}{x}$ should give a straight line through the origin, with gradient $k$.

For the data $x = 2, 3, 4, 6$ with $y = 18, 12, 9, 6$: compute $\tfrac{1}{x} = 0.5, 0.333, 0.25, 0.167$ and plot $y$ against those. The points lie on a straight line through the origin with gradient $36$, so the relationship is inverse variation with $k = 36$.

Equivalently, and with no plotting at all, just check whether the products $xy$ are constant: $36$, $36$, $36$, $36$. The product test and the reciprocal plot are the same test, one done numerically and the other graphically.

Why bother with the plot

With exact values the product test is quicker. With measured data the products will not be exactly equal, and a graph shows at a glance whether the deviations are small random scatter about a line or a systematic bend, which a column of slightly different products does not.

Comparing Direct and Inverse Graphs

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Set the two side by side, since questions frequently ask you to tell them apart.

Direct variation, $y = kx$. A straight line. Passes through the origin. Defined for every $x$, including zero. One connected piece. Equal steps in $x$ give equal steps in $y$. As $x$ grows, $y$ grows without limit.

Inverse variation, $y = \tfrac{k}{x}$. A curve. Passes through neither axis. Undefined at $x = 0$. Two separate branches. Equal steps in $x$ give shrinking steps in $y$. As $x$ grows, $y$ shrinks towards zero without reaching it.

The single question that separates them fastest, given a table, is whether to divide or to multiply:

$$\text{constant } \frac{y}{x} \ \Rightarrow \ \text{direct} \qquad \text{constant } xy \ \Rightarrow \ \text{inverse}$$

Given a graph, the fastest question is whether it is straight. If it is straight, check the origin; if it is curved and approaches both axes, suspect inverse and confirm with the product test.

Common Pitfalls

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Drawing the curve touching or crossing an axis.

Fix: it does neither. Leave a visible gap at both ends of each branch so the sketch matches the mathematics.

Joining the two branches through the origin.

Fix: nothing exists at $x = 0$, so there is nothing to join through. Draw two separate curves with a gap between them.

Concluding a relationship is inverse variation because its graph is a decreasing curve.

Fix: check the products, or plot against $\tfrac{1}{x}$. Plenty of decreasing curves are not inverse relationships at all.

Plotting $\tfrac{1}{y}$ against $x$ instead of $y$ against $\tfrac{1}{x}$.

Fix: either works here, since $xy = k$ is symmetric, but be consistent and label the axis. The gradient differs between the two versions.

Watch Me Solve It · 3 examples

Watch Me Solve It · Sketching from an equation

+15 XP per step

PROBLEM

Sketch $y = \dfrac{18}{x}$ for positive $x$, describing the key features.

1

Choose points that divide into 18 neatly

$(1,18), \ (2,9), \ (3,6), \ (6,3), \ (9,2), \ (18,1)$

Note the pairing: $(2,9)$ and $(9,2)$, $(3,6)$ and $(6,3)$. Each pair costs only one calculation.
2

Describe the left end

As $x$ approaches zero from the right, $y$ grows without bound: at $x = 0.1$ it is $180$, at $x = 0.01$ it is $1800$. The curve rises steeply and hugs the vertical axis without meeting it.
3

Describe the right end

As $x$ grows, $y$ shrinks towards zero: at $x = 100$ it is $0.18$, at $x = 1000$ it is $0.018$. The curve flattens towards the horizontal axis without meeting it.
4

Draw and check

A smooth curve through the plotted points, steep on the left and flat on the right, with a visible gap from both axes. Checking one product: $3 \times 6 = 18$, as required.

AnswerA single branch through $(1,18)$, $(2,9)$, $(3,6)$, $(6,3)$, $(9,2)$, $(18,1)$, approaching both axes without touching either

Watch Me Solve It · Testing data with the reciprocal plot

+15 XP per step

PROBLEM

A set of measurements gives $x = 2, 4, 5, 10$ with $y = 30, 15, 12, 6$. (a) Test whether the relationship is inverse variation. (b) Describe what a plot of $y$ against $\tfrac{1}{x}$ would look like, and state its gradient.

1

(a) Compute the products

$2(30) = 60, \quad 4(15) = 60, \quad 5(12) = 60, \quad 10(6) = 60$

All equal, so the relationship is inverse variation with $k = 60$.
2

(a) Rule out direct variation as a cross-check

$\tfrac{30}{2} = 15, \quad \tfrac{15}{4} = 3.75$

The ratios differ, so it is certainly not direct variation.
3

(b) Compute the reciprocals

$\tfrac{1}{x} = 0.5, \ 0.25, \ 0.2, \ 0.1$

These become the horizontal coordinates of the new plot.
4

(b) Describe the new graph

$\frac{30}{0.5} = 60, \quad \frac{15}{0.25} = 60, \quad \frac{6}{0.1} = 60$

Every point has the same ratio, so the plot is a straight line through the origin with gradient $60$, which is the constant of variation. The curve has been straightened.

Answer(a) inverse variation with $k = 60$; (b) a straight line through the origin with gradient $60$

Watch Me Solve It · Classifying three graphs

+15 XP per step

PROBLEM

Describe what each graph shows: (a) a straight line through $(0,0)$ and $(3,9)$; (b) a curve through $(1,20)$, $(2,10)$ and $(4,5)$ approaching both axes; (c) a straight line through $(0,6)$ and $(3,0)$.

1

(a) Straight, so check the origin

$k = \tfrac{9}{3} = 3$

It passes through the origin, so it is direct variation with equation $y = 3x$.
2

(b) Curved, so check the products

$1(20) = 20, \quad 2(10) = 20, \quad 4(5) = 20$

Constant, so it is inverse variation with equation $y = \tfrac{20}{x}$. The approach to both axes is consistent with this.
3

(c) Straight, so check the origin again

It crosses the vertical axis at $(0,6)$, not the origin, so it is not direct variation. Its gradient is $\tfrac{0-6}{3-0} = -2$, so its equation is $y = -2x + 6$.
4

State what (c) is

A linear relationship that is neither direct nor inverse variation. It decreases, but by a fixed amount for each unit increase in $x$, whereas inverse variation decreases by a shrinking amount. Its product $xy$ changes: $0$, $4$, $6$, $4$, $0$ at $x = 0,1,2,\ldots$

Answer(a) direct variation, $y = 3x$; (b) inverse variation, $y = \tfrac{20}{x}$; (c) linear but neither

Brain Trainer · 5 problems

Brain Trainer · Read the curve

5 problems

Five items on inverse variation graphs. Work each one, then reveal the answer.

1 How many branches does the graph of $y = \dfrac{5}{x}$ have?

One for positive $x$, one for negative.Two
2 Does the graph of $y = \dfrac{5}{x}$ ever cross the horizontal axis?

Consider whether $\tfrac{5}{x}$ can equal zero.No
3 A curve passes through $(2,15)$ and $(5,6)$. Is it inverse variation?

Check the products: $30$ and $30$.Yes, $k = 30$
4 What shape is the plot of $y$ against $\tfrac{1}{x}$ if $y = \dfrac{8}{x}$?

It becomes direct variation in disguise.A line through the origin, gradient $8$
5 Name the two asymptotes of $y = \dfrac{k}{x}$.

The lines the curve approaches but never meets.Both axes

Complete in your workbook.

Multiple Choice · 5 questions

MC1

The shape

+10 XP

The graph of $y = \dfrac{k}{x}$ with $k > 0$ is:

MC2

The axes

+10 XP

The graph of $y = \dfrac{7}{x}$ meets the horizontal axis:

MC3

The straightening test

+10 XP

To test whether data show inverse variation using a graph, plot:

MC4

Telling them apart

+10 XP

A table gives $x = 1, 2, 4$ with $y = 24, 12, 6$. The relationship is:

MC5

The two branches

+10 XP

The graph of $y = \dfrac{k}{x}$ has two separate branches because:

Short Answer · 3 questions

Sketch and describe

+15 XP

SHORT ANSWER

Consider $y = \dfrac{24}{x}$.
(a) Complete a table of values for $x = 1, 2, 3, 4, 6, 8, 12$.
(b) Describe the shape of the graph for positive $x$, including what happens at each end.
(c) Explain why the graph never meets either axis.
(d) State a symmetry in your table and explain where it comes from.

Write your working in your book.

Classify from data

+15 XP

SHORT ANSWER

For each data set, decide whether it shows direct variation, inverse variation or neither, showing the calculation you used.
(a) $x = 3, 6, 9$ with $y = 8, 4, \tfrac{8}{3}$
(b) $x = 3, 6, 9$ with $y = 8, 16, 24$
(c) $x = 3, 6, 9$ with $y = 8, 5, 2$
(d) For the set in (a), describe the graph of $y$ against $\tfrac{1}{x}$.

Write your working in your book.

A context on a graph

+15 XP

SHORT ANSWER

A rectangle has a fixed area of $36$ cm$^2$. Its length is $\ell$ cm and its width is $w$ cm.
(a) Write the relationship between $\ell$ and $w$, and state the type of variation.
(b) Complete a table for $\ell = 2, 3, 4, 6, 9, 12$ and describe the graph.
(c) Explain why only one branch of the graph is meaningful here.
(d) The rectangle is a square for exactly one pair of values. Find it, and describe where that point sits on the graph.

Write your working in your book.

Stretch Challenge · Straightening other relationships

+25 XP

CHALLENGE

(a) A set of data is suspected to satisfy $y = \dfrac{k}{x^2}$. Describe a plot that would straighten it, and state what the gradient of that plot would represent.
(b) Explain why the graph of $y = \dfrac{k}{x}$ with $k < 0$ occupies a different pair of quadrants from the case $k > 0$, and describe both.
(c) The graph of $y = \dfrac{k}{x}$ is symmetric about the line $y = x$. Prove this by showing that whenever a point is on the graph, so is the point with its coordinates swapped, and explain what that symmetry means in a context such as the fixed-area rectangle.

(a) Write the relationship as $y = k \times \dfrac{1}{x^2}$, which says $y$ varies directly as $\dfrac{1}{x^2}$. So plotting $y$ against $\dfrac{1}{x^2}$ should give a straight line through the origin, and its gradient would be the constant $k$. The general principle is worth stating: whenever a relationship has the form $y = k \times (\text{some expression in } x)$, plotting $y$ against that expression straightens it, and the gradient is always the constant of variation. That is why linearising is such a common technique with experimental data.

(b) With $k > 0$, a positive $x$ gives a positive $y$ and a negative $x$ gives a negative $y$, so the branches lie where both coordinates share a sign: the upper right and lower left regions. With $k < 0$ the signs are opposite: a positive $x$ gives a negative $y$ and a negative $x$ gives a positive $y$, so the branches lie in the upper left and lower right regions. In both cases the shape of each branch is the same and both axes remain asymptotes; only the position changes. In context, a negative constant is unusual, since the fixed total being shared is normally a positive quantity such as a distance or an amount of work.

(c) Suppose the point $(a,b)$ is on the graph, so $b = \dfrac{k}{a}$, which rearranges to $ab = k$. Now test the swapped point $(b,a)$: it lies on the graph exactly when $a = \dfrac{k}{b}$, that is when $ab = k$, which is the same condition and is already known to hold. So $(b,a)$ is on the graph too. Since swapping the coordinates of a point reflects it in the line $y = x$, the graph is unchanged by that reflection and is therefore symmetric about it. In the fixed-area rectangle this symmetry is the observation that a $4$ cm by $9$ cm rectangle and a $9$ cm by $4$ cm rectangle are the same rectangle turned on its side: interchanging length and width leaves the area unchanged, so both pairs must appear. The point where the graph meets the line $y = x$ is the one case where swapping changes nothing, which is exactly the square.

Quick Review

recap

The shape

A curve in two branches, a hyperbola

The axes

Asymptotes: approached, never met

The test

Constant product, or plot $y$ against $\dfrac{1}{x}$

Straightened

The reciprocal plot has gradient $k$

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