The Hyperbola $y = \dfrac{k}{x}$
An inverse relationship with two curving branches that NEVER touch the axes. Meet asymptotes and learn to spot which way the branches bend.
Printable Worksheets
Print or save as PDF, or build a custom worksheet from any module's questions.
In $y = k/x$, $y$ and $x$ are INVERSELY related: as one grows, the other shrinks. Take $y = 6/x$. Make a quick table for $x = 1, 2, 3, 6$ and for $x = -1, -2, -3, -6$. What happens for $x = 0$? Why?
The hyperbola $y = k/x$ is the shape of an INVERSE relationship: $xy = k$. It has TWO branches, both bending towards the axes but never touching them. Those axes are ASYMPTOTES.
The red curve $y = 6/x$ has $k = 6 > 0$, so its two branches sit in quadrants 1 and 3 (both $x, y$ positive, or both negative). Both branches hug closer and closer to the $x$-axis as $x \to \pm \infty$, and closer to the $y$-axis as $x \to 0$, but they never cross either axis.
Know
- $y = k/x$ describes inverse variation: $xy = k$
- The graph has TWO branches with asymptotes $x = 0$ and $y = 0$
- $k > 0$ gives Q1 + Q3; $k < 0$ gives Q2 + Q4
Understand
- Why $x = 0$ is forbidden (division by zero is undefined)
- Why branches never CROSS the axes, they approach but don't touch
- Why larger $|k|$ pushes branches further from the origin
Can Do
- Build a table of values and plot a hyperbola
- State asymptotes and which quadrants the branches sit in
- Decide whether a given point lies on $y = k/x$
Wrong: Drawing $y = 6/x$ as a single connected curve crossing the axes.
Right: TWO separate branches. Neither touches an axis. There's a gap at $x = 0$.
Wrong: Putting both branches of $y = -4/x$ in quadrants 1 and 3.
Right: $k = -4 < 0$, so the branches sit in Q2 (top-left) and Q4 (bottom-right). $x$ and $y$ have OPPOSITE signs.
The cleanest way to sketch $y = k/x$ is a table of values using divisors of $k$:
For $y = 6/x$: take $x = \pm 1, \pm 2, \pm 3, \pm 6$. Then $y$ values come out as integers: $6, 3, 2, 1, -6, -3, -2, -1$. Plot 8 points, join with TWO smooth curves (one per side), watching them approach the axes but never touch.
Don't pick $x = 0$, it's undefined. Note the symmetry: if $(a, b)$ is on the curve, so is $(-a, -b)$.
Two questions about $k$:
- Sign of $k$: $k > 0$ $\to$ branches in Q1 + Q3 (both coordinates same sign). $k < 0$ $\to$ branches in Q2 + Q4 (opposite signs).
- Size of $|k|$: larger $|k|$ pushes the branches FURTHER from the origin. e.g. $y = 1/x$ passes through $(1, 1)$; $y = 8/x$ passes through $(1, 8)$, further out.
Asymptotes ($x = 0$ and $y = 0$) are the SAME for every $y = k/x$, regardless of $k$.
Watch Me Solve It · 3 examples
- 1Positive $x$ side$x = 1, y = 6$. $x = 2, y = 3$. $x = 3, y = 2$. $x = 6, y = 1$.
- 2Negative $x$ side$x = -1, y = -6$. $x = -2, y = -3$. $x = -3, y = -2$. $x = -6, y = -1$.
- 3Describe sketch$k = 6 > 0$: branches in Q1 and Q3. Both branches approach the axes (asymptotes $x = 0, y = 0$) but never touch.Symmetric: $(a, b)$ and $(-a, -b)$ are mirror images through the origin.
- 1Sign of $k$$k = -4 < 0$, so branches are in Q2 (top-left) and Q4 (bottom-right).
- 2Asymptotes$x = 0$ and $y = 0$ (same as always for $y = k/x$).
- 3Sample points$x = 1: y = -4$ (Q4). $x = -1: y = 4$ (Q2).Opposite-sign coordinates, the trademark of $k < 0$.
- 1Substitute the point$y = k/x \Rightarrow 5 = k/2$.
- 2Solve for $k$Multiply both sides by 2: $k = 10$.
- 3Write the equation$y = 10/x$.Quick check: at $x = 2$, $y = 10/2 = 5$. Matches.
Common Pitfalls
Equation
- $y = k/x$
- $xy = k$ (inverse)
- $x \neq 0$
Asymptotes
- $x = 0$ (y-axis)
- $y = 0$ (x-axis)
- Never touched
Branches
- $k > 0$: Q1, Q3
- $k < 0$: Q2, Q4
- Symmetric thru origin
Table tips
- Use divisors of $k$
- Both signs of $x$
- Skip $x = 0$
How are you completing this lesson?
Brain Trainer · 4 problems
Four quick hyperbola questions.
1 For $y = 12/x$, find $y$ when $x = 4$.
$y = 12/4$.$y = 3$2 State the asymptotes of $y = -8/x$.
Same for every $y = k/x$.$x = 0$ and $y = 0$3 Which quadrants contain the branches of $y = -10/x$?
$k = -10 < 0$.Q2 and Q44 A hyperbola $y = k/x$ passes through $(3, 4)$. Find $k$.
$4 = k/3 \Rightarrow k = 12$.$y = 12/x$
Quick Check · 5 questions
Show Your Working · 3 questions
Q6. Complete the table for $y = 12/x$ using $x = -6, -3, -2, -1, 1, 2, 3, 6$. State which quadrants contain branches and the asymptotes.
Q7. A hyperbola $y = k/x$ passes through $(-2, 6)$. (a) Find $k$. (b) State the equation. (c) Which quadrants do the branches sit in?
Q8. Two hyperbolas: $y = 4/x$ and $y = 16/x$. (a) Do they share asymptotes? Justify. (b) Compare the points at $x = 2$ for each, which is further from the $x$-axis? (c) Hence describe how increasing $|k|$ changes the position of the branches.
Quick Check
1. B asymptotes are $x = 0$ and $y = 0$.
2. D$k < 0$: branches in Q2 and Q4.
3. A$y = 8/2 = 4$.
4. C undefined (division by zero).
5. B$5 = k/3 \Rightarrow k = 15$.
Show Your Working Model Answers
Q6 (3 marks): Table: $(-6, -2), (-3, -4), (-2, -6), (-1, -12), (1, 12), (2, 6), (3, 4), (6, 2)$ [1]. $k = 12 > 0$ so branches in Q1 and Q3 [1]. Asymptotes $x = 0$ and $y = 0$ [1].
Q7 (3 marks): (a) $6 = k/(-2) \Rightarrow k = -12$ [1]. (b) $y = -12/x$ [1]. (c) $k < 0$, so branches in Q2 and Q4 [1].
Q8 (3 marks): (a) Both have asymptotes $x = 0, y = 0$ (true for every $y = k/x$) [1]. (b) At $x = 2$: $y = 4/2 = 2$ vs $y = 16/2 = 8$. So $y = 16/x$ is further from the $x$-axis [1]. (c) Larger $|k|$ pushes the branches further from the origin / from the asymptotes [1].
Where Hyperbola Meets Line
Find any intersection points of $y = 6/x$ and $y = x$. (a) Set them equal: $6/x = x$. (b) Multiply both sides by $x$ (noting $x \neq 0$): $6 = x^2$. (c) Solve: $x = \pm\sqrt{6}$. (d) Find matching $y$ values. (e) Explain what symmetry of the answers tells you about the geometry of the two curves.
Reveal solution
(a) $6/x = x$. (b) $x^2 = 6$ (with $x \neq 0$). (c) $x = \pm\sqrt{6}$. (d) Matching $y = x$: $y = \pm\sqrt{6}$. Points $(\sqrt{6}, \sqrt{6})$ and $(-\sqrt{6}, -\sqrt{6})$. (e) Both points are symmetric about the origin, consistent with both $y = x$ (line through origin) and $y = 6/x$ (hyperbola symmetric about origin) sharing that symmetry.
Equation
$y = k/x$
Asymptotes
$x = 0$, $y = 0$
$k > 0$
Q1 and Q3
$k < 0$
Q2 and Q4
$x = 0$
Undefined
Symmetry
$(a, b) \leftrightarrow (-a, -b)$
Your Badges
0 of 6Mark lesson as complete
Tick when you've finished Learn, Practice and the Stretch. Earns +85 XP and +25 coins.