1.2 - Rational functions and asymptotes | Cambridge A-Level Further Mathematics

A rational curve can have a missing point, an infinite branch or a line that it approaches far from the origin. You will distinguish these features, find intercepts and determine horizontal and oblique asymptotes using algebra.

Domain and intercepts

Does cancelling a factor put a missing point back into a graph? Compare $y=(x^2-1)/(x-1)$ with the straight line $y=x+1$ . Their values agree whenever $x\ne1$ , but the fraction has no value at $x=1$ .

Rational function and domain

A rational function is a quotient of polynomials. Its domain contains the real inputs for which its original denominator is nonzero, unless a further restriction is stated.

A hole is a missing point at which a finite limiting value exists. Cancelling a common factor can reveal that value; it does not restore the excluded input.

Factor first and record the exclusions before simplifying. To find an intersection with the $x$ -axis, make the numerator zero and check that the input belongs to the domain. To find an intersection with the $y$ -axis, substitute $x=0$ , if allowed.

Worked example 1: a line with a hole

Question. Find the domain, hole and axis intersections of $f(x)=(x^2-1)/(x-1)$ .

Plan. Keep the denominator restriction while factorising the numerator.

f(x)=\frac{(x-1)(x+1)}{x-1}=x+1,\qquad x\ne1.

The missing input is $1$ , and the simplified expression tends to $2$ there. For $f(x)=0$ , $x+1=0$ , so $x=-1$ . Also $f(0)=1$ .

Answer. Domain: $\mathbb R\setminus\{1\}$ . Hole: $(1,2)$ . Axis intersections: $(-1,0)$ and $(0,1)$ .

Check. Direct substitution into the original fraction gives $f(-1)=0/(-2)=0$ and $f(0)=(-1)/(-1)=1$ . At $x=1$ , it gives $0/0$ , which is undefined.

Move the cancelled factor from $x-1$ to $x-2$ . The line stays the same, but a different point is missing.

Worked example 2: moving the excluded input

Question. Find the same features of $g(x)=(x^2-x-2)/(x-2)$ .

Plan. Factorise, then use the simplified expression only on the original domain.

g(x)=\frac{(x-2)(x+1)}{x-2}=x+1,\qquad x\ne2.

Thus the missing point has height $2+1=3$ . The allowed zero is $x=-1$ , and $g(0)=1$ .

Answer. Domain: $\mathbb R\setminus\{2\}$ . Hole: $(2,3)$ . Axis intersections: $(-1,0)$ and $(0,1)$ .

Check. Polynomial multiplication gives $(x-2)(x+1)=x^2-x-2$ . At $x=3$ , the original fraction is $4$ , agreeing with the line, while at $x=2$ the original denominator is zero.

An open circle records the missing point without suggesting that a whole vertical line is an asymptote.

[DIAGRAM: asset_name: Cancellation leaves a hole; asset_slug: cam9231-rational-holes; caption: The same straight line can have different missing points; description: Two coordinate panels show y equals x plus 1. The left panel has an open circle at (1,2), the right at (2,3). Both show and label the axis intersections (-1,0) and (0,1), distinguishing a removable hole from an infinite branch.]
The same straight line can have different missing points

Use the original denominator again in this check.

Vertical asymptotes

A denominator zero produces a vertical asymptote when an uncancelled denominator factor makes the function unbounded nearby.

Vertical asymptote

The line $x=a$ is a vertical asymptote when $f(x)$ tends to positive or negative infinity as $x$ approaches $a$ from at least one side.

For a rational function, cancel common factors first, keeping the original domain restrictions. A real root of the remaining denominator gives a vertical asymptote. A root removed completely by cancellation gives a hole instead. If only some copies of a repeated factor cancel, the remaining copies can still give an asymptote.

Worked example 3: two uncancelled factors

Question. Find the vertical asymptotes of $f(x)=(x+1)/((x-2)(x+3))$ .

Plan. Check that neither denominator factor cancels.

\begin{aligned} (x-2)(x+3)=0&\quad\Rightarrow\quad x=2\text{ or }x=-3 \\ x=2&:\quad x+1=3\ne0 \\ x=-3&:\quad x+1=-2\ne0. \end{aligned}

The numerator is nonzero at both denominator roots, so neither factor cancels.

Answer. The vertical asymptotes are $x=2$ and $x=-3$ .

Check. Near $2$ , $(x-2)f(x)=(x+1)/(x+3)\to3/5\ne0$ . Near $-3$ , $(x+3)f(x)=(x+1)/(x-2)\to2/5\ne0$ . Each function value therefore grows without bound as the corresponding factor tends to zero.

Change only the numerator so that one denominator factor cancels. This changes an infinite branch into a hole.

Worked example 4: one factor cancels

Question. Find the vertical asymptotes and hole of $g(x)=(x-2)/((x-2)(x+3))$ .

Plan. Simplify while retaining both original exclusions.

g(x)=\frac1{x+3},\qquad x\ne2,-3.

At $2$ , the remaining expression tends to $1/5$ , a finite value. At $-3$ , its denominator still tends to zero while its numerator stays $1$ .

Answer. There is a vertical asymptote $x=-3$ and a hole at $(2,1/5)$ .

Check. At $x=2.01$ , the value is approximately $0.200$ , close to $1/5$ . At $x=-2.99$ , the value is $100$ , showing the different behaviour near the asymptote.

The opening question now has a precise answer: cancellation preserves the graph at allowed inputs, but preserves the hole as well.

Watch out

“The denominator is zero, so there is an asymptote” skips the cancellation check. For $(x-2)/(x-2)$ , $x=2$ is excluded but the nearby value is $1$ , not infinity.

Distinguish the two types of exclusion here.

Horizontal and oblique asymptotes

Far from the origin, division separates a rational function into a simple part and a remainder that tends to zero.

For example, $(2x^2+3)/(x^2+1)=2+1/(x^2+1)$ . The second term tends to zero as $x\to\pm\infty$ , so the curve approaches $y=2$ .

Horizontal and oblique asymptotes

A horizontal line $y=b$ is an asymptote if $f(x)-b\to0$ as $x\to\infty$ or $x\to-\infty$ . An oblique asymptote is a line $y=mx+b$ , with $m\ne0$ , for which $f(x)-(mx+b)\to0$ at an infinite end.

With numerator and denominator degrees at most two, these cases guide the algebra:

Relative degrees	Result at infinity
Numerator degree smaller	Horizontal asymptote $y=0$
Equal degrees	Horizontal asymptote: ratio of leading coefficients
Numerator degree one larger	Division gives an oblique asymptote

Always justify the result by a vanishing remainder or division by the highest power. A quadratic polynomial over a constant denominator has no straight-line asymptote.

Worked example 5: an equal-degree fraction

Question. Determine the asymptotes of $f(x)=(2x^2+3)/(x^2+1)$ .

Plan. Check real denominator zeros, then divide.

Since $x^2+1>0$ , there are no vertical asymptotes. Also

f(x)=2+\frac1{x^2+1},\qquad \frac1{x^2+1}\to0\quad(x\to\pm\infty).

Answer. The only asymptote is $y=2$ .

Check. Dividing the original numerator and denominator by $x^2$ gives $(2+3/x^2)/(1+1/x^2)\to2$ , independently confirming the height.

Keep the numerator and lower the denominator degree by one. The quotient is now a sloping line.

Worked example 6: a quadratic divided by a linear factor

Question. Determine the asymptotes of $g(x)=(2x^2+3)/(x+1)$ .

Plan. Use the denominator root and polynomial division.

2x^2+3=(x+1)(2x-2)+5,

g(x)=2x-2+\frac5{x+1}.

The remainder tends to zero at both infinite ends. At $-1$ , the original numerator is $5\ne0$ .

Answer. The asymptotes are $x=-1$ and $y=2x-2$ .

Check. Multiplying back gives $(x+1)(2x-2)+5=2x^2-2+5=2x^2+3$ . This checks the intercept $-2$ as well as the gradient.

The dashed line in each panel represents a difference tending to zero, rather than an extra piece of the curve.

[DIAGRAM: asset_name: Horizontal and oblique approaches; asset_slug: cam9231-rational-asymptotes; caption: Division reveals the line approached at each infinite end; description: Two coordinate panels compare (2x squared plus 3)/(x squared plus 1), approaching the horizontal line y equals 2 from above, with (2x squared plus 3)/(x plus 1), approaching y equals 2x minus 2 and having vertical asymptote x equals minus 1. Solid blue curves and dashed clay asymptotes are labelled.]
Division reveals the line approached at each infinite end

An asymptote can be crossed. For instance, $x/(x^2+1)$ crosses its horizontal asymptote $y=0$ at the origin. The definition concerns behaviour at infinity.

Quick Recap

Record original denominator restrictions before cancelling.
Cancelled factors can leave holes; uncancelled real denominator factors give vertical asymptotes.
Use division to expose a horizontal or oblique asymptote.
An asymptote describes limiting behaviour and may cross the curve.

Cancelling a factor never puts an excluded point back into the original graph.

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1.2 - Rational functions and asymptotes

Domain and intercepts

Vertical asymptotes

Horizontal and oblique asymptotes

Quick Recap

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