The Trucus
Explore the truncus for Victorian Year 11 Mathematical Methods (VCAA) — the graph of one over x squared and its transformations, a smooth curve above a horizontal asymptote with a vertical asymptote.
You will learn to find its asymptotes, intercepts, domain and range, sketch it after dilations, reflections and translations, and build its rule from a graph — key skills in the study of functions.
Theory
In Year 11 Mathematical Methods, Unit 1, a truncus is the power function \(y=\dfrac{a}{(x-h)^2}+k\) — the graph of \(y=\dfrac{1}{x^2}\) dilated by factor \(a\) and translated so its asymptotes become \(x=h\) and \(y=k\). This page shows how to read off the asymptotes, domain and range, find intercepts, and build the rule from given information.
A truncus is a member of the family of power functions studied in Unit 1. Its rule is \(y=\dfrac{a}{(x-h)^2}+k\), a transformation of the base graph \(y=\dfrac{1}{x^2}\).
Because the denominator is squared, \((x-h)^2>0\) for every \(x\neq h\). So the fraction \(\dfrac{a}{(x-h)^2}\) keeps the sign of \(a\): the whole graph sits on one side of its horizontal asymptote. Both branches point the same way — up if \(a>0\), down if \(a<0\).
The three constants are read straight from the rule:
- \(a\) is the dilation factor from the \(x\)-axis (and reflects in the \(x\)-axis when \(a<0\));
- \(h\) is the horizontal translation, giving the vertical asymptote \(x=h\);
- \(k\) is the vertical translation, giving the horizontal asymptote \(y=k\).
The standard form of a truncus and the features you read from it:
| Feature | Result |
|---|---|
| Vertical asymptote | \(x=h\) |
| Horizontal asymptote | \(y=k\) |
| Maximal domain | \(\mathbb{R}\setminus\{h\}\) |
| Range \((a>0)\) | \((k,\,\infty)\) |
| Range \((a<0)\) | \((-\infty,\,k)\) |
Reading a truncus \(y=\dfrac{a}{(x-h)^2}+k\)
- Asymptotes: the vertical asymptote is \(x=h\) (where the denominator is zero); the horizontal asymptote is \(y=k\).
- Domain: all reals except the value at the vertical asymptote, \(\mathbb{R}\setminus\{h\}\).
- Range: check the sign of \(a\). If \(a>0\) the range is \((k,\,\infty)\); if \(a<0\) it is \((-\infty,\,k)\). The value \(y=k\) is never reached.
- Intercepts: \(y\)-intercept at \(x=0\); \(x\)-intercepts by solving \(y=0\).
Building the rule
- Read \(h\) from the vertical asymptote and \(k\) from the horizontal asymptote.
- Substitute a known point \((x_1,\,y_1)\) into \(y=\dfrac{a}{(x-h)^2}+k\) and solve for \(a\).
Vertical asymptote — the rule is undefined where the denominator is zero:
| \((x-3)^2\) | \(=\) | \(0\) |
| \(x-3\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(3\) |
So the vertical asymptote is \(x=3\).
Horizontal asymptote — as \(x\to\pm\infty\), the fraction \(\to 0\), so \(y\to k\):
| \(y\) | \(=\) | \(\dfrac{1}{(x-3)^2}+2\) |
| \(y\) | \(\to\) | \(0+2\) |
| \(y\) | \(=\) | \(2\) |
So the horizontal asymptote is \(y=2\).
Domain — exclude the value at the vertical asymptote: the rule exists for every real \(x\) except \(x=3\).
Vertical asymptote \(x=3\); horizontal asymptote \(y=2\); domain \(=\mathbb{R}\setminus\{3\}\).
The \(y\)-intercept is where \(x=0\) — substitute:
| \(y\) | \(=\) | \(\dfrac{8}{(0-2)^2}-1\) |
| \(y\) | \(=\) | \(\dfrac{8}{(-2)^2}-1\) |
| \(y\) | \(=\) | \(\dfrac{8}{4}-1\) |
| \(y\) | \(=\) | \(2-1\) |
| \(y\) | \(=\) | \(1\) |
The \(y\)-intercept is \((0,\,1)\).
Sign of the fraction — here \(a=-3<0\) and \((x+2)^2>0\) for all \(x\neq-2\), so:
| \(\dfrac{-3}{(x+2)^2}\) | \(<\) | \(0\) |
Add the vertical translation \(k=-1\):
| \(\dfrac{-3}{(x+2)^2}-1\) | \(<\) | \(-1\) |
| \(y\) | \(<\) | \(-1\) |
The horizontal asymptote \(y=-1\) is never reached, so it is excluded.
Range \(=(-\infty,\,-1)\).
\(h\) from the vertical asymptote \(x=h\):
| \(h\) | \(=\) | \(2\) |
\(k\) from the horizontal asymptote \(y=k\):
| \(k\) | \(=\) | \(-3\) |
\(a\) — substitute the point \((3,\,1)\) into \(y=\dfrac{a}{(x-2)^2}-3\):
| \(1\) | \(=\) | \(\dfrac{a}{(3-2)^2}-3\) |
| \(1\) | \(=\) | \(\dfrac{a}{1}-3\) |
| \(1\) | \(=\) | \(a-3\) |
| \(a\) | \(=\) | \(4\) |
\(h=2,\ k=-3,\ a=4\); the rule is \(y=\dfrac{4}{(x-2)^2}-3\).
Common pitfalls
Frequently asked questions
What is a truncus?
A truncus is the power function \(y=\dfrac{a}{(x-h)^2}+k\), a transformation of \(y=\dfrac{1}{x^2}\). The squared denominator means both branches lie on the same side of the horizontal asymptote.
How do you find the asymptotes of a truncus?
The vertical asymptote is \(x=h\) (where \((x-h)^2=0\)); the horizontal asymptote is \(y=k\), the value \(y\) approaches as \(x\to\pm\infty\).
What is the domain and range of a truncus?
The maximal domain is \(\mathbb{R}\setminus\{h\}\). The range is \((k,\infty)\) when \(a>0\) and \((-\infty,k)\) when \(a<0\); \(y=k\) is never reached.
How is a truncus different from a hyperbola?
A hyperbola \(y=\dfrac{a}{x-h}+k\) has a linear denominator, so its two branches sit on opposite sides of \(y=k\). A truncus has a squared denominator, so both branches sit on the same side.
How do you find the rule of a truncus?
Read \(h\) from the vertical asymptote and \(k\) from the horizontal asymptote, then substitute a known point into \(y=\dfrac{a}{(x-h)^2}+k\) and solve for \(a\).
Does a truncus always have an x-intercept?
No. Put \(y=0\): there are two \(x\)-intercepts only when \(a\) and \(k\) have opposite signs; if they share a sign the graph never crosses the \(x\)-axis.