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Year 12 Maths Standard 2 (2027) Algebraic relationships

Reciprocal Graphs (Hyperbolas)

20 practice questions 0 video lessons Theory + worked examples

Master reciprocal graphs and the hyperbola for NSW Year 12 Mathematics Standard 2. In this topic you recognise inverse variation \(y=\dfrac{k}{x}\) (constant product \(xy=k\)), find the constant \(k\) from a point or a table, and sketch the two branches of the curve.

You will learn how the sign of \(k\) places the branches in quadrants 1 & 3 or 2 & 4, how the \(x\)-axis and \(y\)-axis act as asymptotes the curve approaches but never touches, and why a reciprocal model has real limitations β€” a core Standard 2 skill for modelling quantities whose product stays fixed.

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Theory

A reciprocal relationship is inverse variation \(y=\dfrac{k}{x}\), where the product \(xy=k\) is constant. This Year 12 Standard 2 (NSW) guide shows the hyperbola shape and its two branches, how the sign of \(k\) sets the quadrants, how to find \(k\) from a point, the \(x\)-axis and \(y\)-axis asymptotes, and the limitations of a reciprocal model.

A reciprocal relationship is inverse variation: as one quantity grows the other shrinks so that their product stays constant, \(xy=k\). Rearranged, this is \(y=\dfrac{k}{x}\) with \(k\neq 0\), where \(k\) is the constant of variation.

The graph of \(y=\dfrac{k}{x}\) is a hyperbola β€” two separate smooth branches, not a single line. When \(k>0\) the branches sit in the first and third quadrants; when \(k<0\) they sit in the second and fourth quadrants. A larger \(|k|\) pushes the curve further from the origin.

Both the \(x\)-axis \((y=0)\) and the \(y\)-axis \((x=0)\) are asymptotes: the curve gets ever closer but never touches them, so it has no intercepts. Because \(x=0\) is undefined and \(y\) is never \(0\), a reciprocal model has real limitations β€” it cannot describe a quantity that actually reaches zero.

Reciprocal graphs for positive and negative ky = 4 over x lies in quadrants 1 and 3; y = -4 over x lies in quadrants 2 and 4; both approach the axes. x y y=4/x y=-4/x
Sign of \(k\): \(y=\dfrac{4}{x}\) (quadrants 1 & 3) and \(y=\dfrac{-4}{x}\) (quadrants 2 & 4).
Hyperbola y = 6 over x with its asymptotesThe hyperbola y = 6 over x with the x-axis (y=0) and y-axis (x=0) marked as red dashed asymptotes. x y y=0 x=0
The asymptotes of \(y=\dfrac{6}{x}\) are the \(x\)-axis \((y=0)\) and \(y\)-axis \((x=0)\).

Inverse variation has a constant product, giving the reciprocal equation:

\[xy = k \;\Rightarrow\; y = \dfrac{k}{x}, \qquad k \neq 0\]
y=kx

The constant of variation is the product of the coordinates of any point on the curve, so one known point fixes the whole graph:

\[k = x_1 y_1\]
k=x1y1

Translating the curve moves its asymptotes:

\[y = \dfrac{k}{x} + c \;\;(\text{asymptote } y=c), \qquad y = \dfrac{k}{x-r} \;\;(\text{asymptote } x=r)\]
y=kx-r
Reading a point. If a hyperbola \(y=\dfrac{k}{x}\) passes through \((a,b)\), then \(k=ab\) β€” multiply the coordinates to get the constant.

How to work with a reciprocal graph

  1. Check it is inverse variation: the product \(xy\) is constant, so the equation is \(y=\dfrac{k}{x}\).
  2. Find \(k\): multiply the coordinates of a known point, \(k=xy\) (or read a complete column of a table).
  3. Evaluate: substitute into \(y=\dfrac{k}{x}\) to find \(y\) from \(x\), or rearrange to find \(x\) from \(y\).
  4. Sketch the shape: two branches β€” quadrants 1 & 3 if \(k>0\), quadrants 2 & 4 if \(k<0\).
  5. Mark the asymptotes: the axes for \(y=\dfrac{k}{x}\); shift to \(y=c\) or \(x=r\) for a translated curve. The curve nears them but never touches.
Example 1 β€” Describe the graph
Describe the graph of \(y=\dfrac{-8}{x}\): its shape, which quadrants it lies in, and its asymptotes.
Solution

Use the sign of \(k\) to place the two branches.

Graph of y = -8 over xA hyperbola y = -8 over x with two branches in the second and fourth quadrants. x y
\(k\)\(=\)\(-8 < 0\)
\(x>0\)\(\Rightarrow\)\(y<0\ \text{(quadrant 4)}\)
\(x<0\)\(\Rightarrow\)\(y>0\ \text{(quadrant 2)}\)

A hyperbola with two branches in the second and fourth quadrants, approaching but never touching the \(x\)- and \(y\)-axes.

Example 2 β€” Find k from a point
The hyperbola \(y=\dfrac{k}{x}\) passes through \((3,5)\). Find \(k\), write the equation, and find \(y\) when \(x=5\).
Solution

The constant \(k\) is the product of the coordinates.

Hyperbola y = k over x through (3,5)A hyperbola in quadrants 1 and 3 passing through the marked point (3,5). x y 3 5
\(k\)\(=\)\(xy = 3\times 5 = 15\)
\(y\)\(=\)\(\dfrac{15}{x}\)
\(\text{At } x=5:\ y\)\(=\)\(\dfrac{15}{5} = 3\)

\(k=15\), so \(y=\dfrac{15}{x}\), and \(y=3\) when \(x=5\).

k=15
Example 3 β€” Inverse-variation table
\(y\) varies inversely with \(x\). Find \(k\), the missing value, and whether \((10,2)\) lies on the curve.
\(x\)1245
\(y\)20105?
Solution

For inverse variation the product \(xy\) is constant.

\(k\)\(=\)\(xy = 1\times 20 = 20\)
\(\text{At } x=5:\ y\)\(=\)\(\dfrac{20}{5} = 4\)
\((10,2):\ 10\times 2\)\(=\)\(20 = k\)

\(k=20\); the missing value is 4; and \((10,2)\) does lie on the curve since \(10\times 2=20\).

Example 4 β€” Asymptotes of a shifted curve
State the asymptotes of \(y=\dfrac{4}{x}+1\) and explain why the curve never reaches them.
Solution

Let \(x\) grow large, then let \(x\) approach 0.

Hyperbola y = 4 over x plus 1The hyperbola y = 4 over x + 1 with a red dashed horizontal asymptote at y = 1. x y y=1
\(x\to\pm\infty:\ \dfrac{4}{x}\)\(\to\)\(0,\ \text{so } y\to 1\)
\(x\to 0:\ \dfrac{4}{x}\)\(\to\)\(\pm\infty\)

Horizontal asymptote \(y=1\), vertical asymptote \(x=0\). The curve gets ever closer, but \(\dfrac{4}{x}\) is never exactly \(0\) and \(x=0\) is undefined.

Common pitfalls

You cannot divide by zero. \(x=0\) is not allowed, so the curve has no point on the \(y\)-axis, and \(y\) is never \(0\) β€” there are no intercepts.
Two separate branches. A hyperbola is not one line; never join the branches through the origin.
Reciprocal models have limits. Since \(y=\dfrac{k}{x}\) never equals \(0\) and is undefined at \(x=0\), it cannot model a quantity that truly reaches zero.

Frequently asked questions

What is a reciprocal relationship?

It is inverse variation: two quantities whose product is constant, xy equals k, which rearranges to y equals k over x. As one quantity increases the other decreases so that the product stays the same.

What shape is the graph of y = k/x?

A hyperbola. It has two separate smooth branches, not a single straight line. Each branch curves toward the axes without ever touching them.

How does the sign of k change the graph?

When k is positive the two branches lie in the first and third quadrants. When k is negative they lie in the second and fourth quadrants. A larger size of k pushes the curve further from the origin.

What are the asymptotes of y = k/x?

The x-axis (y = 0) and the y-axis (x = 0). The curve approaches both lines as x or y become very large or very small, but it never actually touches them, so there are no intercepts.

How do you find k from a point on the curve?

Multiply the coordinates of the point. If y equals k over x passes through the point (a, b), then k equals a times b, and the equation is y equals that value over x.

Why does a reciprocal model have limitations?

Because y equals k over x is undefined at x = 0 and can never give y = 0, the model cannot describe a situation where a quantity reaches exactly zero, and it breaks down near x = 0 where the value grows without bound.