Dividing by a very small number gives a very large answer. That single fact is the whole shape of a hyperbola, and it explains both of its asymptotes.
Today's hook, The reciprocal function is the first graph you meet that never touches an axis, no matter how far you follow it. Understanding why is more useful than memorising the shape.
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Recall, your gut answer first
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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Recall, your gut answer first
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Work out $\frac{6}{x}$ for $x = 3$, then $x = 0.1$, then $x = 0.001$. What is happening? Now try $x = 100$ and $x = 1000$. What is happening there?
Before you work it out, what is your instinct? Write it down, then check it against the lesson.
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Small divisor, huge answer
Work through the core explanation before applying it.
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Small divisor, huge answer
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In $y = \dfrac{k}{x}$, as $x$ shrinks towards 0 the quotient grows without bound, so the curve races up the $y$-axis but never reaches it. As $x$ grows large the quotient shrinks towards 0, so the curve flattens towards the $x$-axis without touching it. Those two behaviours are the asymptotes.
$y = \dfrac{k}{x}$, $x \neq 0$ asymptotes $x = 0$ and $y = 0$ domain and range both exclude 0
An asymptote is approached, never reached
The curve gets arbitrarily close to the axis. It never meets it, because $\frac{k}{x}$ is never exactly 0 and $x$ is never 0.
The sign of $k$ picks the quadrants
$k > 0$ puts the two branches in the first and third quadrants; $k < 0$ puts them in the second and fourth.
Both branches, always
A hyperbola has two separate pieces. Sketching only one is an incomplete answer.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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What you'll master
Know
Key facts
$y = \frac{k}{x}$ is undefined at $x = 0$, so its domain is all real $x$ except 0.
It has a vertical asymptote at $x = 0$ and a horizontal asymptote at $y = 0$.
The graph is a hyperbola with two branches, in opposite quadrants.
As $x \to 0$ the magnitude of $y$ grows without bound; as $x \to \pm\infty$, $y \to 0$.
Understand
Concepts
Why dividing by a shrinking number produces an unbounded result.
Why neither asymptote is ever reached rather than merely not drawn.
Why the sign of $k$ decides which pair of quadrants the branches occupy.
Can do
Skills
Sketch $y = \frac{k}{x}$ for positive and negative $k$, showing both asymptotes.
State the domain and range.
Describe the behaviour of the curve near the asymptotes in words.
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Key terms
Reciprocal functionA function of the form $y = \frac{k}{x}$, where the variable is in the denominator. Like this: $y = \frac{6}{x}$ gives $y = 2$ when $x = 3$.
HyperbolaThe two-branch curve that a reciprocal function produces. Like this: $y = \frac{1}{x}$ has one branch in the first quadrant and one in the third.
Vertical asymptoteA vertical line the curve approaches but never meets, here $x = 0$. Like this: as $x$ goes from 0.1 to 0.001, $\frac{6}{x}$ climbs from 60 to 6000 without ever being defined at $x = 0$.
Horizontal asymptoteA horizontal line the curve flattens towards, here $y = 0$. Like this: $\frac{6}{1000} = 0.006$, close to zero but not zero.
BranchOne of the two separate pieces of a hyperbola. Like this: $y = \frac{1}{x}$ has a branch for $x > 0$ and another for $x < 0$, with nothing in between.
Excluded valueAn input the function cannot take, here $x = 0$ because division by zero is undefined. Like this: the domain of $y = \frac{6}{x}$ is all real $x$ except 0.
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Why the curve never meets either axis
Work through the core explanation before applying it.
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Why the curve never meets either axis
core concept
Take $y = \dfrac{6}{x}$. At $x = 3$ it gives 2; at $x = 0.1$ it gives 60; at $x = 0.001$ it gives 6000. The smaller the divisor, the larger the result, with no upper limit.
So as $x$ approaches 0 the curve rises without bound. It never crosses the $y$-axis, because $x = 0$ is not in the domain at all: division by zero is undefined.
Going the other way, $x = 100$ gives 0.06 and $x = 1000$ gives 0.006. The value shrinks towards 0 but is never 0, because a non-zero numerator divided by anything is never zero. So the curve flattens towards the $x$-axis without meeting it.
Two different reasons for two asymptotes. The vertical one exists because $x = 0$ is excluded from the domain. The horizontal one exists because $\frac{k}{x}$ can never equal zero. They are not the same argument.
Quick check: what happens to $y = \dfrac{6}{x}$ as $x$ gets closer and closer to 0 from the positive side?
$y = \frac{k}{x}$ has a vertical asymptote at $x = 0$ because that value is excluded from the domain, and a horizontal asymptote at $y = 0$ because a non-zero numerator over anything is never zero. The curve approaches both without ever meeting them.
Pause, copy the two numerical sequences (shrinking $x$ giving growing $y$, growing $x$ giving shrinking $y$), and the two separate reasons for the two asymptotes, into your book.
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Sketching the hyperbola
core concept
We just saw why the curve approaches both axes without touching. That raises a question: what does the whole graph look like, and where exactly do the pieces sit? This card answers it → two branches in opposite quadrants, with the sign of $k$ choosing which pair.
For $k > 0$, both $x$ and $y$ have the same sign, so the branches sit in the **first and third** quadrants. For $y = \frac{6}{x}$, useful points are $(1,6)$, $(2,3)$, $(3,2)$, $(6,1)$ and their negatives.
For $k < 0$ the signs are opposite, so the branches sit in the **second and fourth** quadrants. $y = \frac{-6}{x}$ passes through $(1,-6)$ and $(-1,6)$.
Draw both asymptotes as dashed lines first, then sketch each branch approaching them. A sketch with only one branch is incomplete.
Larger $\lvert k\rvert$ pushes the curve out. $y = \frac{12}{x}$ sits further from the origin than $y = \frac{6}{x}$, because every $y$-value is doubled. The shape is the same; the distance from the axes changes.
Fill the blank: for $y = \dfrac{k}{x}$ with $k$ negative, the branches lie in the second and quadrants.
$k > 0$ puts the branches in the first and third quadrants; $k < 0$ puts them in the second and fourth. Draw both asymptotes dashed, then both branches. Larger $|k|$ moves the curve further from the origin without changing its shape.
Pause, copy the quadrant rule for each sign of $k$, a table of points for $y = \frac{6}{x}$, and the note that both branches must be drawn, into your book.
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Domain, range, and describing the behaviour
core concept
We just saw how to draw the curve. That raises a question: how do you state all of this precisely, in the language a question will ask for? This card answers it → domain and range in interval notation, and the behaviour in words.
The domain excludes the vertical asymptote: all real $x$ except 0, written $(-\infty, 0) \cup (0, \infty)$. The range excludes the horizontal asymptote in the same way: all real $y$ except 0.
The syllabus asks you to **describe the behaviour** as $x \to 0$ and as $x \to \pm\infty$. For $y = \frac{6}{x}$: as $x \to 0^+$, $y \to +\infty$; as $x \to 0^-$, $y \to -\infty$; as $x \to \pm\infty$, $y \to 0$.
Notice the two sides of the vertical asymptote behave oppositely. Approaching 0 from the right sends the curve up; from the left it sends it down. Both must be described.
The interval notation is the same convention you met with inequalities. A round bracket excludes the endpoint, and infinity always takes a round bracket. Excluding a single interior point splits the domain into a union of two intervals.
Which statement about $y = \dfrac{1}{x}$ is FALSE?
Domain and range each exclude 0, written as a union of two intervals. Describe the behaviour on both sides of the vertical asymptote separately, since approaching from the right and the left send the curve in opposite directions, and state that $y \to 0$ as $x \to \pm\infty$.
Pause, copy the domain and range in interval notation, and all three behaviour statements including both sides of the vertical asymptote, into your book.
Worked examples · 3 in a row, reveal as you go
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Work examples end to end
Follow the reasoning through complete worked solutions.
PROBLEM 1 · SKETCHING WITH POSITIVE k
Sketch $y = \dfrac{6}{x}$, showing its asymptotes, and state its domain and range.
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Asymptotes: $x = 0$ and $y = 0$; draw both dashed
Always mark them before the curve.
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Points: $(1,6)$, $(2,3)$, $(3,2)$, $(6,1)$ and the negatives $(-1,-6)$, $(-2,-3)$
$k > 0$, so branches in the first and third quadrants.
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Domain all real $x \neq 0$; range all real $y \neq 0$
Each excludes the value its asymptote sits at.
PROBLEM 2 · NEGATIVE k
Describe how the graph of $y = \dfrac{-4}{x}$ differs from $y = \dfrac{4}{x}$.
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Both have the same asymptotes, $x = 0$ and $y = 0$
The value of $k$ does not move the asymptotes.
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$y = \frac{4}{x}$ has branches in quadrants 1 and 3; $y = \frac{-4}{x}$ has them in 2 and 4
A negative $k$ makes $y$ take the opposite sign to $x$.
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$y = -\sqrt[3]{x}$ is $y = \sqrt[3]{x}$ reflected in either axis
Reflecting in the $x$-axis or the $y$-axis gives the same result here.
PROBLEM 3 · DESCRIBING BEHAVIOUR
Describe the behaviour of $y = \dfrac{3}{x}$ as $x \to 0$ and as $x \to \pm\infty$.
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As $x \to 0^+$: the divisor shrinks while staying positive, so $y \to +\infty$
Approaching from the right.
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As $x \to 0^-$: the divisor shrinks while staying negative, so $y \to -\infty$
Approaching from the left gives the opposite direction.
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As $x \to +\infty$ and as $x \to -\infty$: $y \to 0$
The curve flattens towards the horizontal asymptote from above and from below.
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Quick-fire practice
Work through the core explanation before applying it.
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Quick-fire practice
+10 XP
State the asymptotes of $y = \dfrac{5}{x}$.
In which quadrants do the branches of $y = \dfrac{-2}{x}$ lie?
State the domain of $y = \dfrac{7}{x}$ in interval notation.
What happens to $y = \dfrac{1}{x}$ as $x$ becomes very large?
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Revisit your table of values
Run the quick drill and copy the summary into your book.
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Revisit your table of values
At the start you evaluated $\frac{6}{x}$ at $x = 3$, $0.1$, $0.001$, then at $100$ and $1000$. Name the asymptote each sequence is heading towards, and explain why the curve never actually arrives at either.
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Multiple choice
Answer the drill bank and rate your confidence.
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Multiple choice
+5 XP per correct · +25 XP all-correct
Pick your answer, then rate your confidence, that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 43 marks
Q1. Sketch $y = \dfrac{-8}{x}$, showing both asymptotes and at least two points on each branch. (3 marks)
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UnderstandBand 43 marks
Q2. Explain why the graph of $y = \dfrac{k}{x}$ never crosses either axis, giving a separate reason for each axis. (3 marks)
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UnderstandBand 32 marks
Q3. State the domain and range of $y = \dfrac{4}{x}$ in interval notation. (2 marks)
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📖 Comprehensive answers (click to reveal)
Practice 1: $x = 0$ and $y = 0$. Practice 2: second and fourth. Practice 3: $(-\infty, 0) \cup (0, \infty)$. Practice 4: $y$ approaches 0 without reaching it.
Q1 (3 marks): Asymptotes $x = 0$ and $y = 0$ drawn dashed [1]. $k = -8 < 0$, so branches in the second and fourth quadrants [1]. Points such as $(1,-8)$, $(2,-4)$, $(-1,8)$, $(-2,4)$ plotted, both branches drawn [1].
Q2 (3 marks): It never crosses the $y$-axis because that would require $x = 0$, and $\frac{k}{0}$ is undefined, so $x = 0$ is excluded from the domain [1]. It never crosses the $x$-axis because that would require $y = 0$, and a non-zero $k$ divided by any number is never zero [1]. The two reasons are different: one is a domain exclusion, the other is a property of division [1].
Take the module quiz if you are ready, then mark the lesson complete.
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Boss battle · Asymptote Attack
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Read asymptotes, quadrants and behaviour off reciprocal functions at speed. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.