Graphs of rational functions
Master graphs of rational functions in Year 12 VCE Specialist Mathematics. The graph of \(y=\dfrac{P(x)}{Q(x)}\) is shaped by its asymptotes and key features — vertical, horizontal and oblique (slant) asymptotes, holes (removable discontinuities) and stationary points. It sits in the Functions, relations and graphs area of study of the VCE Mathematics Study Design (VCAA), within the Rational functions topic of Unit 3.
You will learn to factorise and cancel to locate holes, read the end behaviour from the degrees of \(P\) and \(Q\), find stationary points with the quotient rule, and sketch curves such as \(y=\dfrac{ax+b}{cx+d}\) and \(y=\dfrac{\text{quadratic}}{\text{linear}}\) — core graph-sketching skills for the Rational functions topic.
Theory
The graph of a rational function \(y=\dfrac{P(x)}{Q(x)}\) is shaped by its asymptotes and key features. In Year 12 Specialist Mathematics you locate vertical, horizontal and oblique (slant) asymptotes, spot holes (removable discontinuities), find stationary points with the quotient rule, and sketch curves such as \(y=\dfrac{ax+b}{cx+d}\), \(y=\dfrac{1}{\text{quadratic}}\) and \(y=\dfrac{\text{quadratic}}{\text{linear}}\).
A rational function is a quotient of polynomials, \(y=\dfrac{P(x)}{Q(x)}\) with \(Q(x)\neq 0\). Its graph is controlled by where the denominator vanishes and by how the degrees of \(P\) and \(Q\) compare.
A vertical asymptote is a line \(x=a\) where the denominator is zero (and the numerator is not): near it the curve shoots to \(\pm\infty\). A horizontal asymptote is a line \(y=c\) that the curve approaches as \(x\to\pm\infty\); it exists when \(\deg P \le \deg Q\). An oblique (slant) asymptote is a line \(y=mx+k\) approached as \(x\to\pm\infty\); it occurs when \(\deg P\) is exactly one more than \(\deg Q\), and is found by polynomial division.
A hole (removable discontinuity) appears where a factor cancels from top and bottom: the function equals the simplified form everywhere except that one \(x\), where a single point is missing from the curve.
A stationary point is where \(\dfrac{dy}{dx}=0\). For a rational function you differentiate with the quotient rule \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\), solve \(y'=0\), and use the sign of \(y''\) (or a sign test) to classify each point as a maximum or minimum.
Read a rational function's asymptotes straight from its structure.
Vertical asymptotes occur where the denominator is zero (after cancelling any common factors):
The end behaviour (horizontal or oblique asymptote) depends on the degrees of \(P\) and \(Q\):
| Degrees | End behaviour |
|---|---|
| \(\deg P < \deg Q\) | horizontal asymptote \(y=0\) |
| \(\deg P = \deg Q\) | horizontal asymptote \(y=\dfrac{\text{lead }P}{\text{lead }Q}\) |
| \(\deg P = \deg Q + 1\) | oblique asymptote (divide to find it) |
An oblique asymptote comes from polynomial division; the remainder term vanishes at infinity:
Stationary points use the quotient rule:
How to sketch a rational function
- Factorise and cancel \(\dfrac{P(x)}{Q(x)}\). A cancelled factor marks a hole; substitute its \(x\)-value into the simplified form for the hole's coordinates.
- Vertical asymptotes: set the simplified denominator to zero, \(Q(x)=0\).
- End behaviour: compare degrees for a horizontal asymptote (\(\deg P\le\deg Q\)) or, when \(\deg P=\deg Q+1\), divide for the oblique asymptote.
- Intercepts: \(y\)-intercept at \(x=0\); \(x\)-intercepts where the numerator is zero.
- Stationary points: differentiate with the quotient rule, solve \(y'=0\), and classify each with \(y''\) or a sign test; then sketch the branches between the asymptotes.
Vertical asymptote — denominator zero:
| \(x+1\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(-1\) |
Horizontal asymptote — leading coefficients (equal degree):
| \(y\) | \(=\) | \(\dfrac{3x-6}{x+1}\) |
| \(\to\) | \(\dfrac{3}{1} \quad (x\to\pm\infty)\) | |
| \(=\) | \(3\) |
Intercepts:
| \(3x-6=0\) | \(\Rightarrow\) | \((2,0)\) |
| \(x=0:\ y=\dfrac{-6}{1}\) | \(\Rightarrow\) | \((0,-6)\) |
Vertical asymptote \(x=-1\), horizontal asymptote \(y=3\); intercepts \((2,0)\) and \((0,-6)\).
Polynomial-divide the numerator by the denominator:
| \(x^2+3x+4\) | \(=\) | \((x+1)(x+2)+2\) |
| \(\dfrac{x^2+3x+4}{x+1}\) | \(=\) | \(x+2+\dfrac{2}{x+1}\) |
Let \(x\to\pm\infty\) so the remainder term vanishes:
| \(\dfrac{2}{x+1}\) | \(\to\) | \(0\) |
| \(y\) | \(\to\) | \(x+2\) |
The oblique asymptote is \(y=x+2\) (with a vertical asymptote at \(x=-1\)).
Factorise and cancel the common factor:
| \(\dfrac{x^2-4}{x-2}\) | \(=\) | \(\dfrac{(x-2)(x+2)}{x-2}\) |
| \(=\) | \(x+2, \quad x\neq 2\) |
The cancelled point \(x=2\) is the hole; substitute:
| \(y\) | \(=\) | \(2+2\) |
| \(=\) | \(4\) |
There is a hole at \((2,4)\); elsewhere the graph is the line \(y=x+2\).
Differentiate with the quotient rule:
| \(y'\) | \(=\) | \(\dfrac{2x(x^2+1)-(x^2-1)(2x)}{(x^2+1)^2}\) |
| \(=\) | \(\dfrac{4x}{(x^2+1)^2}\) |
Solve \(y'=0\) and evaluate:
| \(4x\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(0\) |
| \(y(0)\) | \(=\) | \(\dfrac{-1}{1} = -1\) |
For \(x<0\), \(y'<0\); for \(x>0\), \(y'>0\) — the gradient turns from negative to positive, so it is a minimum. The horizontal asymptote is \(y=1\).
Minimum turning point \((0,-1)\); horizontal asymptote \(y=1\).
Common pitfalls
Frequently asked questions
How do you find the vertical asymptotes of a rational function?
Factorise and cancel, then set the simplified denominator to zero. Each solution \(x=a\) (where the numerator is not also zero) is a vertical asymptote \(x=a\).
When does a rational function have an oblique (slant) asymptote?
When the numerator's degree is exactly one more than the denominator's. Divide the numerator by the denominator; the quotient \(y=mx+k\) is the oblique asymptote.
What is a hole in a rational graph?
A hole (removable discontinuity) occurs where a factor cancels from the numerator and denominator. The graph is the simplified curve with a single point missing at that \(x\)-value.
How do you find the horizontal asymptote?
Compare degrees. If \(\deg P<\deg Q\) the horizontal asymptote is \(y=0\); if \(\deg P=\deg Q\) it is the ratio of the leading coefficients; if \(\deg P>\deg Q\) there is none (an oblique one instead when the difference is 1).
How do you find stationary points of a rational function?
Differentiate with the quotient rule \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\), set \(y'=0\) (the numerator zero), solve for \(x\), then classify each point with the second derivative or a sign test.
What is the difference between a hole and a vertical asymptote?
A vertical asymptote comes from a denominator zero that does not cancel — the curve races to \(\pm\infty\) there. A hole comes from a factor that cancels — the curve is continuous through that region apart from one missing point.