Reciprocals of polynomial functions
Learn how to sketch the reciprocal of a polynomial function, \(y=\dfrac{1}{f(x)}\), in Year 11 Specialist Mathematics for Queensland (QCAA). The whole graph is read straight from \(y=f(x)\): a vertical asymptote at each zero of \(f\), and the horizontal asymptote \(y=0\).
You will find asymptotes, intercepts and turning points, match the sign of the reciprocal to the sign of \(f\), and see how a minimum of \(f\) becomes a maximum of \(\dfrac{1}{f}\) — the graph-sketching groundwork for rational and reciprocal trigonometric functions later in the course.
Theory
The graph of \(y=\dfrac{1}{f(x)}\), a reciprocal of a polynomial function, is read straight off the graph of \(y=f(x)\) in Year 11 Specialist Mathematics (QCAA, Queensland). It has a vertical asymptote at every zero of \(f\), the horizontal asymptote \(y=0\), and it keeps the same sign as \(f\). This page shows how to find the asymptotes, intercepts and turning points and sketch the branches.
The reciprocal function of \(f\) is \(y=\dfrac{1}{f(x)}\). Its whole shape is controlled by the graph of \(y=f(x)\), so you never need a table of values — you read the key features from \(f\).
A vertical asymptote occurs at each real zero of \(f\): where \(f(x)=0\), the fraction \(\dfrac{1}{f(x)}\) is undefined and its size blows up. A linear \(f\) has at most one zero, a quadratic \(f\) at most two, so \(y=\dfrac{1}{f(x)}\) has that many vertical asymptotes.
The horizontal asymptote is always \(y=0\). As \(x\to\pm\infty\) the polynomial \(f(x)\) grows without bound, so \(\dfrac{1}{f(x)}\to 0\). The reciprocal graph never actually reaches the \(x\)-axis, because \(\dfrac{1}{f(x)}=0\) has no solution.
Two facts finish the sketch. The reciprocal keeps the same sign as \(f\) (positive where \(f>0\), negative where \(f<0\)), and it swaps size for size: where \(f\) is large \(\dfrac{1}{f}\) is near \(0\), and where \(f\) is close to \(0\), \(\dfrac{1}{f}\) is large. A minimum of \(f\) becomes a maximum of \(\dfrac{1}{f}\) at the same \(x\), and a maximum of \(f\) becomes a minimum of \(\dfrac{1}{f}\).
The reciprocal graph is \(y=\dfrac{1}{f(x)}\). Its vertical asymptotes are the solutions of:
The horizontal asymptote comes from the long-run behaviour:
Key points map through the reciprocal. If \(f\) has a minimum value \(m\ne0\) at \(x=a\), then \(\dfrac{1}{f}\) has a turning point there:
How to sketch \(y=\dfrac{1}{f(x)}\)
- Find the vertical asymptotes: solve \(f(x)=0\). Each real zero of \(f\) gives a dashed vertical line the graph approaches but never meets.
- Add the horizontal asymptote \(y=0\): the graph flattens towards the \(x\)-axis as \(x\to\pm\infty\).
- Mark key points: the \(y\)-intercept \(\dfrac{1}{f(0)}\), and any turning point — a minimum of \(f\) at \((a,m)\) becomes a maximum of \(\dfrac{1}{f}\) at \(\left(a,\dfrac{1}{m}\right)\).
- Match the sign and draw: keep \(\dfrac{1}{f}\) positive where \(f>0\) and negative where \(f<0\), then draw each branch approaching the asymptotes.
Vertical asymptote — solve \(f(x)=0\):
| \(f(x)\) | \(=\) | \(0\) |
| \(x-1\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(1\) |
Horizontal asymptote — behaviour as \(x\to\pm\infty\):
| \(f(x)\) | \(\to\) | \(\pm\infty\) |
| \(\dfrac{1}{f(x)}\) | \(\to\) | \(0\) |
\(y\)-intercept — evaluate at \(x=0\):
| \(f(0)\) | \(=\) | \(0-1\) |
| \(=\) | \(-1\) | |
| \(\dfrac{1}{f(0)}\) | \(=\) | \(\dfrac{1}{-1}\) |
| \(=\) | \(-1\) |
Vertical asymptote \(x=1\), horizontal asymptote \(y=0\), \(y\)-intercept \((0,-1)\).
Vertical asymptotes — factorise and solve \(f(x)=0\):
| \(x^2-2x\) | \(=\) | \(x(x-2)\) |
| \(x(x-2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(0 \ \text{or}\ x=2\) |
Turning point — find the vertex of \(f\), then reciprocate:
| \(x_v\) | \(=\) | \(\dfrac{0+2}{2}=1\) |
| \(f(1)\) | \(=\) | \((1)^2-2(1)\) |
| \(=\) | \(-1\) | |
| \(\dfrac{1}{f(1)}\) | \(=\) | \(\dfrac{1}{-1}=-1\) |
The vertex is a minimum of \(f\), so this is a maximum of \(\dfrac{1}{f}\).
Vertical asymptotes \(x=0\) and \(x=2\); turning point \((1,-1)\), a maximum.
Vertical asymptotes — look for real zeros by completing the square:
| \(x^2+4x+7\) | \(=\) | \((x+2)^2+3\) |
| \((x+2)^2+3\) | \(=\) | \(0\) |
| \((x+2)^2\) | \(=\) | \(-3\) |
| \(=\) | \(\text{no real solution}\) |
So there are no real zeros, hence \(0\) vertical asymptotes.
Maximum — \(\dfrac{1}{f}\) is greatest where \(f\) is least:
| \(f_{\min}\) | \(=\) | \(3\) |
| \(\text{at } x\) | \(=\) | \(-2\) |
| \(\dfrac{1}{f}\text{ max}\) | \(=\) | \(\dfrac{1}{3}\) |
No vertical asymptotes; maximum point \(\left(-2,\dfrac13\right)\).
Vertical asymptotes — solve \(f(x)=0\):
| \(9-x^2\) | \(=\) | \(0\) |
| \(x^2\) | \(=\) | \(9\) |
| \(x\) | \(=\) | \(3 \ \text{or}\ x=-3\) |
\(y\)-intercept — evaluate at \(x=0\):
| \(f(0)\) | \(=\) | \(9\) |
| \(\dfrac{1}{f(0)}\) | \(=\) | \(\dfrac{1}{9}\) |
Behaviour as \(x\to3^-\) — check the sign of \(f\) just left of \(3\):
| \(x\to 3^-\) | \(\Rightarrow\) | \(f=9-x^2\to 0^+\) |
| \(\dfrac{1}{f}\) | \(\to\) | \(+\infty\) |
Asymptotes \(x=3\) and \(x=-3\); \(y\)-intercept \(\left(0,\dfrac19\right)\); as \(x\to3^-\), \(y\to+\infty\).
Common pitfalls
Frequently asked questions
Where are the vertical asymptotes of \(y=\dfrac{1}{f(x)}\)?
At every real zero of \(f\). Solve \(f(x)=0\): each solution \(x=a\) gives a vertical asymptote \(x=a\). A line has at most one, a quadratic at most two.
Why is the horizontal asymptote always \(y=0\)?
As \(x\to\pm\infty\) the polynomial \(f(x)\) grows without bound, so \(\dfrac{1}{f(x)}\to0\). The graph flattens towards the \(x\)-axis but never reaches it.
Does \(y=\dfrac{1}{f(x)}\) have the same sign as \(f\)?
Yes. \(\dfrac{1}{f(x)}\) is positive exactly where \(f(x)>0\) and negative exactly where \(f(x)<0\), because dividing \(1\) by a number keeps its sign.
What happens to a maximum or minimum of \(f\)?
They swap. A minimum of \(f\) at \((a,m)\) becomes a maximum of \(\dfrac{1}{f}\) at \(\left(a,\dfrac{1}{m}\right)\), and a maximum of \(f\) becomes a minimum of \(\dfrac{1}{f}\).
Can \(y=\dfrac{1}{f(x)}\) ever equal \(0\)?
No. \(\dfrac{1}{f(x)}=0\) would need \(1=0\), which is impossible. The reciprocal graph approaches \(y=0\) but never crosses it.
What if \(f\) has no real zeros?
Then \(y=\dfrac{1}{f(x)}\) has no vertical asymptotes. For example \(f(x)=x^2+4x+7\) is always positive, so its reciprocal is a smooth bump with a maximum where \(f\) is least.