Read it off the picture: decreasing within a branch, half-turn and mirror symmetry, and the rectangle of fixed area |k|.
A graph is worth reading, not just drawing. The hyperbola y = k∕x carries four tidy properties you can spot at a glance. On any single branch it always falls (when k > 0) as you move to the right — bigger x, smaller y. It has a beautiful symmetry: a half-turn about the origin carries the whole curve onto itself, and so do reflections across the diagonals y = x and y = −x. And it hides a constant — the rectangle formed by dropping perpendiculars from any point to the two axes always has the same area, |k|, with a bigger |k| pushing the whole curve farther from the origin. Each property comes paired with its picture.
Fix k > 0 and stay on one branch. As x increases, y = k∕x decreases — a bigger x, fed under the same k, gives a smaller y. Said precisely: if x1 < x2 with both points on the same branch, then y1 > y2. So on each branch the function is decreasing.
Watch it on y = 6∕x: at x = 1, 2, 3, 6 the outputs are y = 6, 3, 2, 1 — steadily down. The product x·y stays 6 the whole way, so to keep it fixed a rising x forces a falling y.
On a single branch with k > 0, the inverse-proportion function is decreasing: bigger x ⇒ smaller y. (For k < 0 each branch is increasing instead.)
Here is the catch: you may only compare "bigger y / smaller y" within the same branch. Cross the slit at x = 0 and the rule breaks. On y = 6∕x the point (−1, −6) has a smaller y than (3, 2), even though −1 < 3 — because the two points live on different branches. So "decreasing" is a within-branch statement, never a claim about the whole number line.
The function decreases on each branch separately, not across the gap. Comparing a y from the left branch with a y from the right branch — that is the trap. There is no x = 0 to bridge them.
Spin the whole graph a half-turn (180°) about the origin and it lands exactly on itself. The point (x, y) goes to (−x, −y), and since (−x)(−y) = xy = k, that image is on the curve too. So the hyperbola has central symmetry about O — the origin is its center, and the two branches are half-turn copies of each other.
If (x, y) is on y = k∕x, then (−x, −y) is on it too: a half-turn about O maps the curve onto itself. The center of symmetry is the origin.
The graph is also mirror-symmetric about the diagonal y = x. Swapping x and y leaves the relationship unchanged: if y = k∕x then multiplying by x and dividing by y gives x = k∕y — the very same inverse proportion with the roles traded. So reflecting a point across y = x lands it back on the curve. The same is true across y = −x. These two diagonals are the hyperbola's axes of symmetry.
On y = 6∕x, the point (2, 3) reflects across y = x to (3, 2) — and 3·2 = 6, so it is on the curve. Reflect (2, 3) across y = −x and you reach (−3, −2), also on the curve since (−3)(−2) = 6.
From any point P(x, y) on y = k∕x drop perpendiculars to both axes. The rectangle with corners O, (x, 0), (x, y), (0, y) has area |x·y| = |k| — and that is the same for every point on the curve, because x·y = k everywhere. So the constant k is, quite literally, the area of that rectangle.
On y = 6∕x: the rectangle under (2, 3) is 2 × 3 = 6; the rectangle under (3, 2) is 3 × 2 = 6. Tall and thin, or short and wide — the area never moves off |k|.
The box under any point on y = k∕x has area |k|. Slide the point along the curve and the rectangle reshapes, but its area is locked at |k|.
Compare y = 2∕x, y = 6∕x, and y = 12∕x. The larger |k| is, the farther the whole hyperbola sits from the origin — its rectangle is bigger, so the curve is pushed outward. A smaller |k| pulls the curve in toward O. The shape of each branch is the same; only how far out it rides changes.
You can feel this in the widget above: raise k and the blue curve steps away from the origin, every point's box growing to match. The sign of k still decides the quadrants — k > 0 keeps the branches in Ⅰ and Ⅲ — but the size of k decides the distance.
| Property of y = k∕x | What you see |
|---|---|
| decreasing on each branch (k > 0) | bigger x ⇒ smaller y, within one branch |
| central symmetry about O | (x, y) and (−x, −y) both on the curve |
| mirror symmetry | reflects onto itself across y = x and y = −x |
| rectangle area = |k| | the box under any point is always |k| |
| bigger |k| | the whole curve sits farther from O |
"Decreasing" and "increasing" are per-branch facts; the symmetry holds for every k ≠ 0; and a curve gets farther from O as |k| grows, not closer. Keep the sign of k for the quadrants and the size of k for the distance — two separate jobs.
The hyperbola y = k∕x wears its properties on its face. On each branch it is decreasing when k > 0 (increasing when k < 0) — but that comparison is only valid within one branch; never compare y-values across the gap at x = 0. The whole curve has central symmetry about O: (x, y) on the curve guarantees (−x, −y) is too. It also has mirror symmetry across the two diagonals y = x and y = −x. The rectangle dropped from any point to the axes has the fixed area |k| (because x·y = k), and a bigger |k| pushes the whole curve farther from the origin.
For y = 8∕x with x > 0, as x increases does y increase or decrease?
It decreases. With k = 8 > 0, the function is decreasing on each branch — a bigger x under the same k gives a smaller y (e.g. x = 2 → y = 4, x = 4 → y = 2).
The point (3, 4) is on y = 12∕x. Use central symmetry to name another point on the curve.
A half-turn about O sends (x, y) → (−x, −y), so (−3, −4) is on the curve. Check: (−3)(−4) = 12 ✓.
The rectangle dropped from the point (5, 2) to the two axes on a hyperbola has what area, and what is k?
The box is 5 wide × 2 tall, so its area is 10. Since the rectangle's area equals |k| and x·y = k = 5·2 = 10, we have k = 10 (the curve is y = 10∕x).
Which curve sits farther from the origin, y = 2∕x or y = 9∕x?
y = 9∕x — the larger |k| (9 > 2) means a bigger rectangle under every point, so the whole curve is pushed farther from O.
Name the two lines of mirror symmetry of y = k∕x.
The diagonals y = x and y = −x. (Swapping x and y turns y = k∕x into x = k∕y — the same relationship — so reflecting across y = x lands the curve back on itself; likewise for y = −x.)
On y = −6∕x, compare y at x = 2 and x = 3. Is the branch increasing or decreasing?
At x = 2, y = −6∕2 = −3; at x = 3, y = −6∕3 = −2. Since −3 < −2, y increases as x grows — each branch of a k < 0 hyperbola is increasing.
Six questions to lock it in. Tap the answer you think is right.
This lesson moves from building the inverse-proportion graph to reading it. The four properties — monotonicity within a branch, central symmetry, mirror symmetry across the diagonals, and the fixed rectangle area |k| — each connect a piece of algebra (x·y = k) to a feature you can point at on the picture. The widgets let students test rather than memorize: walk two points up a branch and watch y fall, reflect a point across the diagonals and see it land on the curve, slide a point and confirm the rectangle's area never moves.
The misconception to watch is the over-generalized "decreasing." Students who saw a falling curve will say the function "decreases everywhere," but it decreases on each branch separately — comparing a y on the left branch with a y on the right is meaningless because there is no path between them across x = 0. A second slip is assuming only k > 0 has symmetry; in fact every k ≠ 0 hyperbola has the full central and mirror symmetry, only the quadrants change. Encourage students to always ask "same branch?" before comparing, and to verify k = x·y as the rectangle's area.
Common Core. This lesson supports F-IF.B.4 / F-IF.B.6 (describe where a function is increasing or decreasing; interpret the behavior of a graph), F-BF.B.3 / G-CO.A.3 (symmetry of a graph — central and reflective), and A-SSE.A.1 (interpret k = xy as the area of the rectangle under a point).