What does the graph of y = k/x look like for positive k?

It is a reciprocal curve (hyperbola), not a straight line. For positive x, the curve falls steeply near the y-axis and flattens as x increases. It never touches either axis because y can never be zero (k divided by x is zero only when x is infinite) and x can never be zero (division by zero is undefined). Two exam features to state: the curve falls steeply then flattens, and it never touches either axis. A straight line through the origin would mean y = kx (direct proportion), which is a different relationship entirely.

How do you verify an inverse proportion relationship from a table of values?

Use the multiplicative test: compute x times y at each row. If the product is the same constant k at every row, the relationship is y = k/x. For example, E = 36/D means E times D = 36 at every point. Check D = 4: E = 9, product = 36. Check D = 9: E = 4, product = 36. The trap is using an additive test (checking whether E decreases by a fixed amount), which would apply to a linear relationship, not an inverse one.

GCSE Maths Higher · AQA · Proportion

Constant of proportionality and graph shape: k = x times y in inverse proportion

Q: y is inversely proportional to x. The point (6, 21) lies on the curve. How do you find k?

Multiply x and y together: k = 6 times 21 = 126. The formula for inverse proportion is y = k/x, so rearranging gives k = x times y. The trap is using the direct-proportion formula k = y/x = 21/6 = 3.5, which gives completely wrong values at every subsequent x. With the correct k = 126, y at x = 12 is 126 divided by 12 = 10.5. Product check: 12 times 10.5 = 126. This is the exact structure of AQA Higher JUN24 Paper 2 Q24a.

The most common error when finding $k$ in $y = k/x$ : students apply the direct-proportion formula $k = y/x$ instead of the correct inverse formula $k = xy$ . From point $(6, 21)$ , the trap gives $k = 3.5$ and then inflates $y$ to 42 at $x = 12$ . The correct $k = 6 \times 21 = 126$ gives $y = 10.5$ .

A second error is sketching the graph of $y = k/x$ as a straight line. It is a reciprocal curve that falls steeply near the $y$ -axis and flattens as $x$ grows, never touching either axis.

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How to spot it in your own work

You wrote $k = y/x = 3.5$ for the point $(6, 21)$ on $y = k/x$ , giving $y = 42$ at $x = 12$ instead of 10.5 (JUN24 Paper 2 Q24a).
You sketched the graph of $s = 60/t$ as a straight line rather than a reciprocal curve that never touches either axis (NOV24 Paper 3 Q10).
You failed the multiplicative verification test for $E = 36/D$ : checking constant differences instead of constant products $E \times D = 36$ (JUN23 Paper 3 Q20a).
You read $k$ in $D = k/b$ as “the number of desserts” rather than the fixed total $D \times b$ (JUN24 Paper 2 Q3a).

An exam question that triggers it

Here is JUN24 Paper 2 Question 24a, reproduced in structure:

$y$ is inversely proportional to $x$ .

The point $(6, 21)$ lies on the curve $y = k/x$ .

Find the value of $y$ when $x = 12$ .

The trap: $k = 21/6 = 3.5$ (direct formula), giving $y = 3.5 \times 12 = 42$ . The fix: $k = 6 \times 21 = 126$ , so $y = 126/12 = 10.5$ . Product check: $12 \times 10.5 = 126$ .

Why students fall for this

The default model for a proportionality constant is the ratio: in $y = kx$ , $k = y/x$ . That reflex carries over to $y = k/x$ without adjustment. But rearranging $y = k/x$ by multiplying both sides by $x$ gives $k = xy$ — the product, not the ratio.

The graph error follows from the same overextension: students expect all proportion graphs to be straight lines. For $y = kx$ that is correct. For $y = k/x$ , the curve shoots to infinity as $x \to 0$ and decays to zero as $x \to \infty$ — the opposite of a line.

A third version arises with tables: students apply an additive test (constant differences) when the structure demands a multiplicative test (constant products). Verifying $E \times D = 36$ at each row is faster and more reliable than computing $E = 36/D$ for every entry.

The fix: Multiply x and y to find k; use the product test to verify

Step 1: rearrange to find the formula for $k$ . Start from $y = k/x$ . Multiply both sides by $x$ : $k = xy$ . This is the inverse formula. The direct formula $k = y/x$ only applies to $y = kx$ .

Step 2: compute k from the given point. Multiply the $x$ -value by the $y$ -value. For $(6, 21)$ : $k = 6 \times 21 = 126$ .

Step 3: find y at the new x. Divide $k$ by the new $x$ . At $x = 12$ : $y = 126/12 = 10.5$ .

Step 4: product check. Multiply the new $x$ by your answer and confirm it equals $k$ . $12 \times 10.5 = 126$ . If it does not equal $k$ , recheck.

Worked example

JUN24·2·24a: $y \propto 1/x$ , point $(6, 21)$ .

Find k.
$k = x \times y = 6 \times 21 = 126$
Find y at x = 12.
$y = \frac{126}{12} = 10.5$
Product check.
$12 \times 10.5 = 126 \checkmark$
Trap: $k = 21/6 = 3.5$ (direct formula), so $y = 3.5 \times 12 = 42$ . Product check fails: $12 \times 42 = 504 \ne 126$ .

NOV24·3·10: $s = 60/t$ , sketch speed against time.

At $t = 1$ : $s = 60$ .
At $t = 4$ : $s = 15$ .
At $t = 60$ : $s = 1$ .
The curve falls steeply near the $s$ -axis and flattens as $t$ grows. It never touches either axis. Trap: a straight line with negative gradient.

JUN23·3·20a: $E = 36/D$ . Verify with product test.

At $D = 4$ , $E = 9$ : $4 \times 9 = 36$ .
At $D = 9$ , $E = 4$ : $9 \times 4 = 36$ .
Product is fixed at 36 at every row. Relationship confirmed.

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Common questions

$y = k/x$ . The point $(6, 21)$ lies on the curve. Find $k$ .: $k = 6 \times 21 = 126$ . Rearrange $y = k/x$ to get $k = xy$ . The trap $k = 21/6 = 3.5$ applies the direct-proportion formula $k = y/x$ , which only works for $y = kx$ . Product check: $6 \times 21 = 126$ , and at $x = 12$ : $y = 126/12 = 10.5$ , verified by $12 \times 10.5 = 126$ .
What does the graph of $y = k/x$ look like for positive $k$ and positive $x$ ?: A reciprocal curve that falls steeply near the $y$ -axis and flattens as $x$ increases. It never touches either axis: as $x \to 0^+$ , $y \to \infty$ ; as $x \to \infty$ , $y \to 0$ but never reaches it. The trap is a straight line with negative gradient, which would mean $y = -mx + c$ , an entirely different relationship.
How do you verify that $E = 36/D$ from a table of values?: Compute $E \times D$ at each row and confirm it equals 36. For example, if the table shows $D = 4, E = 9$ : product $4 \times 9 = 36$ . If $D = 9, E = 4$ : product $9 \times 4 = 36$ . The trap is checking constant differences, which tests linearity, not inverse proportion.

Related misconceptions

Inverse proportion treated as directWriting y = kx (or scaling y up as x rises) for a y proportional to 1/x relationship. The product xy is fixed, not the ratio.
Proportion to a power scales linearlyFor y proportional to x squared, doubling x multiplies y by 4, not 2. The scale factor carries the power.

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