Mathematics • Year 9 • Unit 2 • Lesson 13

The Hyperbola $y = \dfrac{k}{x}$

Build the table, plot, and quadrant-spot habit for hyperbolas. Watch a worked example for $y = 6/x$, fill in a guided one for $y = -4/x$, then run eight independent problems from quick value reads to finding $k$ from a point.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. The hyperbola $y = k/x$ has TWO branches, never touches the axes, and is undefined at $x = 0$.

Problem. Build a table of values for $y = 6/x$ at $x = -6, -3, -2, -1, 1, 2, 3, 6$. State the asymptotes and which quadrants contain the branches.

Step 1 — Positive $x$ side.

$x = 1: y = 6/1 = 6$. $x = 2: y = 3$. $x = 3: y = 2$. $x = 6: y = 1$.

Reason: as $x$ grows, $y = 6/x$ shrinks. The product $xy = 6$ stays constant — that's inverse variation.

Step 2 — Negative $x$ side.

$x = -1: y = -6$. $x = -2: y = -3$. $x = -3: y = -2$. $x = -6: y = -1$.

Reason: dividing a positive by a negative gives a negative. Mirror image of the positive side, through the origin.

Step 3 — Quadrants.

Positive-$x$ points have $y > 0$: that's Q1. Negative-$x$ points have $y < 0$: that's Q3.

Reason: $k = 6 > 0$, so $xy > 0$ — same-sign coordinates — Q1 and Q3.

Step 4 — Asymptotes.

$x = 0$ (vertical) and $y = 0$ (horizontal). Branches approach these lines but never touch.

Answer: Branches in Q1 and Q3. Asymptotes $x = 0$ and $y = 0$.

Stuck? Revisit lesson § "Plotting from a Table" — build both sides of the table, then sketch each branch.

2. We do — fill in the missing steps

Same structure as Section 1, but with the working faded. Fill in each blank. 4 marks

Problem. For $y = -4/x$, build a quick table at $x = -4, -2, -1, 1, 2, 4$, state the asymptotes, and name the quadrants of the branches.

Step 1 — Positive $x$ side: $x = 1$: $y = $ ______ . $x = 2$: $y = $ ______ . $x = 4$: $y = $ ______ .

Step 2 — Negative $x$ side: $x = -1$: $y = $ ______ . $x = -2$: $y = $ ______ . $x = -4$: $y = $ ______ .

Step 3 — Sign of $k$: $k = -4$, so $k$ is __________________ (positive / negative). That puts the branches in quadrants __________________ and __________________ .

Step 4 — Asymptotes: for ALL hyperbolas $y = k/x$ the asymptotes are $x = $ ______ and $y = $ ______ .

Stuck? Revisit lesson § "Watch Me Solve It · Effect of negative $k$" — same approach.

3. You do — independent practice

Show your working under each problem. 3.1–3.4 are foundation (one value, one quadrant call). 3.5–3.6 are standard (mini-tables and asymptotes). 3.7–3.8 are extension (find $k$ from a given point).

Foundation — quick reads

3.1 For $y = 12/x$, find $y$ when $x = 4$. 1 mark

3.2 For $y = 12/x$, find $y$ when $x = -3$. 1 mark

3.3 Name the asymptotes of $y = -8/x$. 1 mark

3.4 Which quadrants contain the branches of $y = -10/x$? 1 mark

Standard — tables and features

3.5 Build a table of values for $y = 8/x$ at $x = -8, -4, -2, -1, 1, 2, 4, 8$. State the asymptotes and the two quadrants containing branches. 2 marks

3.6 Build a table for $y = -6/x$ at $x = -6, -3, -1, 1, 3, 6$. State the asymptotes and the two quadrants of the branches. 2 marks

Extension — find $k$ from a point

3.7 A hyperbola $y = k/x$ passes through $(2, 5)$. Find $k$, write the equation, and verify by substituting the point back in. 2 marks

3.8 A hyperbola $y = k/x$ passes through $(-2, 6)$. (a) Find $k$. (b) State the equation. (c) Which two quadrants contain the branches? 2 marks

Stuck on 3.7 or 3.8? Sub the point into $y = k/x$ and solve: $5 = k/2 \Rightarrow k = 10$. For 3.8, $6 = k/(-2) \Rightarrow k = -12$.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (faded $y = -4/x$)

Step 1: $y(1) = \mathbf{-4}$, $y(2) = \mathbf{-2}$, $y(4) = \mathbf{-1}$.
Step 2: $y(-1) = \mathbf{4}$, $y(-2) = \mathbf{2}$, $y(-4) = \mathbf{1}$.
Step 3: $k = -4$ is negative, so branches sit in quadrants 2 and 4.
Step 4: asymptotes $x = \mathbf{0}$ and $y = \mathbf{0}$.

3.1 — $y = 12/x$ at $x = 4$

$y = 12/4 = \mathbf{3}$.

3.2 — $y = 12/x$ at $x = -3$

$y = 12/(-3) = \mathbf{-4}$.

3.3 — Asymptotes of $y = -8/x$

$x = 0$ (vertical) and $y = 0$ (horizontal). Same for every $y = k/x$.

3.4 — Quadrants of $y = -10/x$

$k = -10 < 0$, so branches sit in Q2 and Q4.

3.5 — Table for $y = 8/x$

$y$ values: $-1, -2, -4, -8, 8, 4, 2, 1$.
Asymptotes: $x = 0$, $y = 0$. Branches in Q1 and Q3 ($k > 0$).

3.6 — Table for $y = -6/x$

$y$ values at $x = -6, -3, -1, 1, 3, 6$: $1, 2, 6, -6, -2, -1$.
Asymptotes: $x = 0$, $y = 0$. Branches in Q2 and Q4 ($k < 0$).

3.7 — $y = k/x$ through $(2, 5)$

$5 = k/2 \Rightarrow k = 10$. Equation: $\mathbf{y = 10/x}$. Check: at $x = 2$, $y = 10/2 = 5$ ✓

3.8 — $y = k/x$ through $(-2, 6)$

(a) $6 = k/(-2) \Rightarrow k = -12$.
(b) Equation: $\mathbf{y = -12/x}$.
(c) $k = -12 < 0$, so branches in Q2 and Q4.