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.

y = 6/x has two branches (Q1 and Q3) and approaches the axes x = 0 and y = 0 as asymptotes.

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.