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Year 11 Methods (Unit 1 & 2) A Gallery Of Graphs

Rectangular Hyperbolas

20 practice questions 1 video lesson Theory + worked examples

Understand rectangular hyperbolas for Queensland Year 11 Mathematical Methods (QCAA). A rectangular hyperbola is the graph of a reciprocal such as one over x, a hyperbolic shape with two branches.

You will learn to find the asymptotes and intercepts, describe how the curve behaves for large positive and negative values of x, see the effect of the parameters that shift the curve, and determine its rule from a graph.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1), a rectangular hyperbola is the graph of a reciprocal function such as \(y=\dfrac{1}{x}\) or \(y=\dfrac{a}{x-h}+k\). This page shows how to find its asymptotes, intercepts, domain and range, and how to determine the rule from a graph.

A rectangular hyperbola is the graph of \(y=\dfrac{1}{x}\). It has two branches that sit in opposite quadrants and get closer and closer to the axes without ever touching them.

A line the curve approaches but never meets is an asymptote. For \(y=\dfrac{a}{x-h}+k\) the vertical asymptote is \(x=h\) (the value that makes the denominator zero) and the horizontal asymptote is \(y=k\) (the value \(y\) approaches as \(x\to\pm\infty\)). Here \(a\) is a dilation; a negative \(a\) also reflects the curve.

The domain is every \(x\) except \(x=h\), and the range is every \(y\) except \(y=k\).

Asymptotes first. For \(y=\dfrac{a}{x-h}+k\), read \(x=h\) and \(y=k\) straight from the rule, then find the intercepts.
Graph of y equals one over xThe rectangular hyperbola y=1/x with two branches in opposite quadrants and the axes as asymptotes. x y
\(y=\dfrac{1}{x}\): two branches, with the axes (red dashed) as asymptotes.
Graph of y equals one over x squaredThe curve y=1/x^2 with both branches above the x-axis and the axes as asymptotes. x y
\(y=\dfrac{1}{x^2}\): both branches lie above the \(x\)-axis.

The general reciprocal function and its asymptotes:

\[y=\dfrac{a}{x-h}+k\qquad x=h,\quad y=k\]
y=ax-h+k

Domain and range in set notation:

\[x\in\mathbb{R}\setminus\{h\},\qquad y\in\mathbb{R}\setminus\{k\}\]
x{h}
Intercepts: for the \(y\)-intercept put \(x=0\); for the \(x\)-intercept put \(y=0\) and solve. If \(k\ne 0\) there is exactly one \(x\)-intercept; if \(k=0\) there is none.

How to analyse \(y=\dfrac{a}{x-h}+k\)

  1. Asymptotes: write the vertical asymptote \(x=h\) and the horizontal asymptote \(y=k\) straight from the rule.
  2. Intercepts: substitute \(x=0\) for the \(y\)-intercept, then set \(y=0\) and solve for the \(x\)-intercept.
  3. Domain and range: exclude \(x=h\) and \(y=k\); note whether \(a<0\) reflects the branches.
Example 1 — Reading \(y=\dfrac{6}{x}\)
For \(y=\dfrac{6}{x}\), find \(y\) when \(x=2\) and when \(x=-3\), and state the asymptotes, domain and range.
Solution

Values — substitute each \(x\):

\(y|_{x=2}\)\(=\)\(\dfrac{6}{2}=3\)
\(y|_{x=-3}\)\(=\)\(\dfrac{6}{-3}=-2\)

So the curve passes through \((2,\,3)\) and \((-3,\,-2)\).

Asymptotes — the rule is \(\dfrac{6}{x-0}+0\):

\(x\)\(=\)\(0\)
\(y\)\(=\)\(0\)

The two branches sit in opposite quadrants.

Asymptotes \(x=0,\ y=0\); domain \(x\in\mathbb{R}\setminus\{0\}\); range \(y\in\mathbb{R}\setminus\{0\}\).

Graph of y equals six over xThe hyperbola y=6/x with the points (2,3) and (-3,-2) marked; the axes are asymptotes. x y
x=0,y=0
Example 2 — Asymptotes and intercepts
For \(y=\dfrac{3}{x-2}+1\), find the asymptotes and both intercepts.
Solution

Asymptotes — read \(h\) and \(k\):

\(x\)\(=\)\(2\)
\(y\)\(=\)\(1\)

\(y\)-intercept — put \(x=0\):

\(y\)\(=\)\(\dfrac{3}{0-2}+1\)
\(=\)\(-\dfrac{3}{2}+1\)
\(=\)\(-\dfrac{1}{2}\)

\(x\)-intercept — put \(y=0\):

\(0\)\(=\)\(\dfrac{3}{x-2}+1\)
\(-1\)\(=\)\(\dfrac{3}{x-2}\)
\(-(x-2)\)\(=\)\(3\)
\(x-2\)\(=\)\(-3\)
\(x\)\(=\)\(-1\)

Asymptotes \(x=2,\ y=1\); intercepts \(\left(0,\,-\tfrac{1}{2}\right)\) and \((-1,\,0)\).

Graph of y equals three over x minus two plus oneThe hyperbola y=3/(x-2)+1 with vertical asymptote x=2 and horizontal asymptote y=1. x y
x=2,y=1
Example 3 — Determine the rule
A hyperbola has asymptotes \(x=-1\) and \(y=2\) and passes through \((0,\,5)\). Find its rule in the form \(y=\dfrac{a}{x-h}+k\).
Solution

Frame — the asymptotes give \(h\) and \(k\):

\(h\)\(=\)\(-1\)
\(k\)\(=\)\(2\)

So \(y=\dfrac{a}{x+1}+2\).

Find \(a\) — substitute the point \((0,\,5)\):

\(5\)\(=\)\(\dfrac{a}{0+1}+2\)
\(5\)\(=\)\(a+2\)
\(a\)\(=\)\(3\)

Rule: \(y=\dfrac{3}{x+1}+2\).

Hyperbola with vertical asymptote x equals minus oneThe hyperbola y=3/(x+1)+2 through the point (0,5); asymptotes x=-1 and y=2. x y
y=3x+1+2
Example 4 — A reflected hyperbola
For \(y=-\dfrac{4}{x+2}+3\), find the asymptotes, both intercepts, and the domain and range.
Solution

Asymptotes:

\(x\)\(=\)\(-2\)
\(y\)\(=\)\(3\)

\(y\)-intercept — put \(x=0\):

\(y\)\(=\)\(-\dfrac{4}{0+2}+3\)
\(=\)\(-2+3\)
\(=\)\(1\)

\(x\)-intercept — put \(y=0\):

\(0\)\(=\)\(-\dfrac{4}{x+2}+3\)
\(\dfrac{4}{x+2}\)\(=\)\(3\)
\(4\)\(=\)\(3(x+2)\)
\(4\)\(=\)\(3x+6\)
\(3x\)\(=\)\(-2\)
\(x\)\(=\)\(-\dfrac{2}{3}\)

Asymptotes \(x=-2,\ y=3\); intercepts \((0,\,1)\) and \(\left(-\tfrac{2}{3},\,0\right)\); domain \(\mathbb{R}\setminus\{-2\}\), range \(\mathbb{R}\setminus\{3\}\).

Reflected hyperbola with asymptotes x equals minus two and y equals threeThe hyperbola y=-4/(x+2)+3 with a reflection; asymptotes x=-2 and y=3. x y
x=-2,y=3

Common pitfalls

Sign of the vertical asymptote. In \(y=\dfrac{a}{x-h}+k\) the asymptote is \(x=h\); for \(\dfrac{3}{x+1}\) that means \(x=-1\), not \(x=1\).
Forgetting the horizontal asymptote. The \(+k\) shifts the whole curve up by \(k\), so the horizontal asymptote is \(y=k\), not \(y=0\).
Claiming an intercept that is not there. When \(k=0\) the curve never reaches the \(x\)-axis, so \(y=\dfrac{a}{x}\) has no \(x\)-intercept.

Frequently asked questions

What is a rectangular hyperbola?

It is the graph of a reciprocal function such as \(y=\dfrac{1}{x}\): two smooth branches in opposite quadrants that approach the axes without touching them.

How do you find the asymptotes of a hyperbola?

For \(y=\dfrac{a}{x-h}+k\), the vertical asymptote is \(x=h\) (denominator zero) and the horizontal asymptote is \(y=k\).

What are the domain and range of a hyperbola?

The domain is all real \(x\) except \(x=h\), and the range is all real \(y\) except \(y=k\).

How do you find the intercepts of a rectangular hyperbola?

Substitute \(x=0\) for the \(y\)-intercept and set \(y=0\), then solve, for the \(x\)-intercept.

What is the difference between the graphs of one over x and one over x squared?

\(y=\dfrac{1}{x}\) has branches in opposite quadrants (one positive, one negative), while \(y=\dfrac{1}{x^2}\) has both branches above the \(x\)-axis.