Rational Functions Explorer

f(x) = k(x − z)/(x − p): drag the zero and the pole on the graph, or set z, p and k. Every intercept and asymptote is worked out underneath.

Vertical asymptotes	$Q(a)=0,\ P(a)\ne0\ \Rightarrow\ x=a$
Holes	$\frac{(x-a)P_{1}(x)}{(x-a)Q_{1}(x)},\ Q_{1}(a)\ne0\ \Rightarrow\ \left(a,\ \tfrac{P_{1}(a)}{Q_{1}(a)}\right)$
Horizontal asymptote	$\begin{aligned}&\deg P<\deg Q:\ y=0\\&\deg P=\deg Q:\ y=\frac{a_{n}}{b_{n}}\end{aligned}$
Oblique asymptote	$\begin{aligned}&\deg P=\deg Q+1:\ \frac{P}{Q}=mx+c+\frac{R}{Q}\\&y=mx+c\end{aligned}$
This page’s family	$k\,\frac{x-z}{x-p}=k+\frac{k(p-z)}{x-p}$

Worked example — Rational Functions Explorer

x-intercept	1
Vertical asymptote	−2
Horizontal asymptote	1
y-intercept	−0.5
Split form	1 − 3/(x + 2)
\|z − p\|	3

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Rational Functions Explorer

A rational function is a quotient P(x)/Q(x) of two polynomials. It is undefined where Q(x) = 0.

Where the formula comes from

Factorise and cancel. A factor common to top and bottom cancels. Where it was zero and what is left of the denominator is not, the graph has a hole: the simplified function has a value there, the original has none.
$\begin{aligned}\frac{x^{2}-4}{x^{2}-x-2}&=\frac{(x-2)(x+2)}{(x-2)(x+1)}\\&=\frac{x+2}{x+1},\ \ x\ne2\end{aligned}$
Vertical asymptotes are the zeros of the denominator that survive the cancelling. Near one the function grows without bound, with a sign that depends on the side.
$\begin{aligned}&x+1=0\ \Rightarrow\ x=-1\\&f(-1.01)=-99,\qquad f(-0.99)=101\end{aligned}$
Intercepts. The zeros of the cancelled numerator, at points where the original denominator is not zero, are the x-intercepts; f(0) is the y-intercept.
$x+2=0\ \Rightarrow\ x=-2;\qquad f(0)=\frac{2}{1}=2$
Far out, compare degrees. Dividing top and bottom by the highest power of x shows the limit: 0 when the bottom has the higher degree, the ratio of leading coefficients when the degrees are equal.
$\frac{x+2}{x+1}=\frac{1+\frac{2}{x}}{1+\frac{1}{x}}\to1\quad(x\to\pm\infty)$
One degree more on top gives a slant. Divide, and the quotient is the line the curve approaches; the remainder over Q tends to 0.
$\frac{x^{2}+3x+1}{x+1}=x+2-\frac{1}{x+1}\ \to\ y=x+2$

Put the zero on top of the pole and the blow-up vanishes

Drag the zero until it sits exactly on the pole and something odd happens: the curve stops exploding and flattens into a horizontal line. With z = p the factor (x − z) cancels the denominator outright, leaving the constant k everywhere — except at x = p, where the original formula still divides by zero. So the graph is a straight line with one point punched out of it, and this tool switches from reporting an asymptote to reporting a hole at (p, k). A denominator root is a pole only if the numerator does not cancel it; otherwise it is a removable hole, invisible at any zoom level.

A pole is only infinite by choice

Written as (ax+b)/(cx+d), any one-zero one-pole function is a Möbius transformation — the maps that rule complex analysis and hyperbolic geometry. Add a single point at infinity (the Riemann sphere) and the pole stops being a catastrophe: x=p simply maps to that new point, and x=∞ maps to the horizontal asymptote y=k. On the sphere the function is a smooth, reversible bijection — the "explosion" is just an artifact of the flat map.

RATIONAL FUNCTIONS — WHERE DOES THE GRAPH BLOW UP, AND WHERE DOES IT SETTLE?

Which Rational-Function Case Are You In?

f(x) = k(x − z)/(x − p) has only three numbers in it, and each one owns a feature of the graph. The zero z is where the curve crosses the axis. The pole p is where the denominator vanishes and the curve runs off to infinity. The gain k is where it settles far from both. Read those three off the formula and you have sketched the graph before plotting a single point.

Well separated — one vertical asymptote, one horizontal, one crossing x = p, y = k

Pole and zero close together — a violent spike between them |z − p| → 0 ⇒ |f| → ∞

Negative gain — the same skeleton with the branches swapped k < 0 ⇒ f → −f

Pole and zero almost on top of each other — nearly a hole (z − p)/(z − p) = 1, z ≠ p

Well separated — one vertical asymptote, one horizontal, one crossing

What you know: A pole and a zero at a comfortable distance. The vertical asymptote stands at x = p, the horizontal one at y = k, and the curve crosses the axis at x = z.

Feature: x = p, y = k

Worked example: f(x) = (x − 1)/(x + 2): asymptote at x = −2, crossing at x = 1, and f → 1 far out in both directions. At x = 0 the value is −0.5.

Open this case: basic

Well separated — one vertical asymptote, one horizontal, one crossing. The pole and the gain draw the two asymptotes; the zero fixes where the curve crosses. A pole and a zero at a comfortable distance. The vertical asymptote stands at x = p, the horizontal one at y = k, and the curve crosses the axis at x = z. — The pole and the gain draw the two asymptotes; the zero fixes where the curve crosses.

Pole and zero close together — a violent spike between them

What you know: The same shape, but the pole sits near the zero. The curve has to get from the axis to infinity across a short distance, so it climbs steeply.

Feature: |z − p| → 0 ⇒ |f| → ∞

Worked example: f(x) = 1.4(x − 0.5)/(x − 0.2): the pole at 0.2 is only 0.3 from the zero at 0.5, and f(0) is already 3.5 against a far-field value of 1.4

Open this case: steep pole

$Pole and zero close together — a violent spike between them. A short gap between pole and zero forces the curve through a narrow, steep excursion. The same shape, but the pole sits near the zero. The curve has to get from the axis to infinity across a short distance, so it climbs steeply.$

A short gap between pole and zero forces the curve through a narrow, steep excursion.

Negative gain — the same skeleton with the branches swapped

What you know: A negative k flips the whole curve about the horizontal axis. The asymptote positions do not move; the branches change which corner they occupy.

Feature: k < 0 ⇒ f → −f

Worked example: f(x) = −(x + 1.5)/(x − 1.2): asymptote at x = 1.2, crossing at x = −1.5, horizontal asymptote at y = −1, and f(0) = 1.25

Open this case: negative gain

Negative gain — the same skeleton with the branches swapped. The asymptotes are where they were, but each branch has moved to the opposite side. A negative k flips the whole curve about the horizontal axis. The asymptote positions do not move; the branches change which corner they occupy. — The asymptotes are where they were, but each branch has moved to the opposite side.

Pole and zero almost on top of each other — nearly a hole

What you know: When z and p are very close, the factors almost divide out. Away from that neighbourhood the function is close to the constant k, and the pole survives as a needle-thin spike.

Feature: (z − p)/(z − p) = 1, z ≠ p

Worked example: f(x) = (x − 1)/(x − 1.12): only 0.12 apart, so f(0) = 0.893, close to the far-field value of 1 — yet the asymptote at x = 1.12 is still there

Open this case: near cancel

$Pole and zero almost on top of each other — nearly a hole. Almost a constant everywhere, except for the narrow window where the pole still wins. When z and p are very close, the factors almost divide out. Away from that neighbourhood the function is close to the constant k, and the pole survives as a needle-thin spike.$

Almost a constant everywhere, except for the narrow window where the pole still wins.

Problems solved in full

Intercepts and asymptotes 5 steps
For f(x) = 2(x − 3)/(x + 1), find the intercepts and the asymptotes, and say which corners the branches occupy.

Open this problem in the tool
1. Zero: the numerator is 0 at x = 3, where the denominator is not.
  
  $2(x-3)=0\ \Rightarrow\ x=3$
2. Vertical asymptote: the denominator is 0 at x = −1.
  
  $x+1=0\ \Rightarrow\ x=-1$
3. Horizontal asymptote: divide top and bottom by x.
  
  $\frac{2(x-3)}{x+1}=\frac{2-\frac{6}{x}}{1+\frac{1}{x}}\to2\quad(x\to\pm\infty)$
4. y-intercept: put x = 0.
  
  $f(0)=\frac{2(0-3)}{0+1}=-6$
5. Split it to see the branches: the part over x + 1 is negative to the right of −1, so the right branch lies below y = 2 and the left branch above it.
  
  $\frac{2(x-3)}{x+1}=\frac{2(x+1)-8}{x+1}=2-\frac{8}{x+1}$
Answer

Zero x = 3, y-intercept −6, asymptotes x = −1 and y = 2; f(x) = 2 − 8/(x + 1), branches top left and bottom right.

$f(x)=2-\frac{8}{x+1}:\quad x=3,\ \ y=-6,\ \ x=-1,\ \ y=2$
A hole 3 steps
f(x) = (x² − 4)/(x² − x − 2). Find any holes, the asymptotes and the intercepts.
Show solution
1. Factorise both. x − 2 is common.
  
  $f(x)=\frac{(x-2)(x+2)}{(x-2)(x+1)}$
2. Cancel it. The original is undefined at x = 2; the simplified form gives the height of the hole.
  
  $=\frac{x+2}{x+1},\ \ x\ne2;\qquad\frac{2+2}{2+1}=\frac43$
3. What is left: the vertical asymptote from x + 1, the zero from x + 2, equal degrees for the horizontal asymptote.
  
  $\begin{aligned}&x+1=0\Rightarrow x=-1;\qquad x+2=0\Rightarrow x=-2\\&y\to\frac11=1;\qquad f(0)=\frac{-4}{-2}=2\end{aligned}$
Answer

Hole at (2, 4/3); asymptotes x = −1 and y = 1; zero x = −2; y-intercept 2.

$\left(2,\tfrac43\right);\quad x=-1,\ y=1;\quad x=-2;\quad f(0)=2$
Horizontal asymptotes by degree 3 steps
Find the horizontal asymptote, if any, of (a) (3x + 1)/(x² + 1), (b) (6x² − x)/(2x² + 5), (c) x³/(x² + 1).
Show solution
1. (a) The bottom has the higher degree.
  
  $\frac{3x+1}{x^{2}+1}=\frac{\frac{3}{x}+\frac{1}{x^{2}}}{1+\frac{1}{x^{2}}}\to0$
2. (b) Equal degrees: the ratio of the leading coefficients.
  
  $\frac{6x^{2}-x}{2x^{2}+5}=\frac{6-\frac1x}{2+\frac{5}{x^{2}}}\to\frac62=3$
3. (c) The top has the higher degree, by one, so there is no horizontal asymptote; division gives a slant one.
  
  $\frac{x^{3}}{x^{2}+1}=x-\frac{x}{x^{2}+1}$
Answer

(a) y = 0, (b) y = 3, (c) none: the curve approaches y = x.

$y=0;\qquad y=3;\qquad y=x$
An oblique asymptote 3 steps
Find the asymptotes of f(x) = (x² + 3x + 1)/(x + 1).
Show solution
1. The top has degree one more than the bottom. Divide.
  
  $x^{2}+3x+1=(x+1)(x+2)-1\ \Rightarrow\ f(x)=x+2-\frac{1}{x+1}$
2. The remainder term tends to 0 as x grows either way.
  
  $\frac{1}{x+1}\to0\quad(x\to\pm\infty)$
3. The denominator is 0 at x = −1, and the numerator is not (it is −1 there).
  
  $x+1=0\Rightarrow x=-1,\qquad(-1)^{2}+3(-1)+1=-1\ne0$
Answer

Vertical x = −1; oblique y = x + 2.

$x=-1;\qquad y=x+2$
Crossing the horizontal asymptote 2 steps
Does f(x) = (x² − 2x)/(x² + 1) cross its horizontal asymptote? If so, where?
Show solution
1. Equal degrees, so the horizontal asymptote is y = 1.
  
  $\frac{x^{2}-2x}{x^{2}+1}\to\frac11=1$
2. Set f(x) = 1 and solve. The x² terms cancel and leave a linear equation.
  
  $\frac{x^{2}-2x}{x^{2}+1}=1\iff x^{2}-2x=x^{2}+1\iff x=-\frac12$
Answer

Yes, once, at (−1/2, 1). A horizontal asymptote describes the far ends only.

$\left(-\tfrac12,\ 1\right)$
A rational inequality 3 steps
Solve (x − 3)/(x + 1) > 0.

Open this problem in the tool
Show solution
1. The sign can change only at the zero, 3, and at the pole, −1. They split the line into three intervals.
  
  $x<-1,\qquad-1<x<3,\qquad x>3$
2. Test one point in each.
  
  $\begin{aligned}&f(-2)=\frac{-5}{-1}=5>0\\&f(0)=\frac{-3}{1}=-3<0\\&f(4)=\frac15>0\end{aligned}$
3. x = 3 gives 0, which is not greater than 0, and x = −1 is not in the domain.
  
  $f(3)=0\not>0,\qquad x=-1\notin\operatorname{dom}f$
Answer

x < −1 or x > 3

$x<-1\ \ \lor\ \ x>3$
A function from its features 2 steps
Find f(x) = k(x − z)/(x − p) with vertical asymptote x = 2, horizontal asymptote y = 3 and y-intercept −3.

Open this problem in the tool
Show solution
1. The vertical asymptote gives p, and the horizontal one gives k.
  
  $x=2\Rightarrow p=2;\qquad y=3\Rightarrow k=3$
2. The y-intercept fixes z.
  
  $f(0)=\frac{3(0-z)}{0-2}=\frac{3z}{2}=-3\ \Rightarrow\ z=-2$
Answer

f(x) = 3(x + 2)/(x − 2)

$f(x)=\frac{3(x+2)}{x-2}$
The curve reaching its horizontal asymptote at y = 1 5 steps
f(x) = (x − 1)/(x + 2) has a horizontal asymptote at y = 1. Work out whether the curve ever actually reaches it.

Open this problem in the tool
Show solution
1. The zero and the pole read straight off the factored form: the numerator vanishes at 1, the denominator at −2.
  
  $f(x)=0\iff x=1;\qquad x+2=0\iff x=-2\ \text{(vertical asymptote)}$
2. For large |x| both top and bottom are dominated by their x terms, so the ratio tends to the ratio of leading coefficients. That is the horizontal asymptote, and here it is 1.
  
  $\lim_{x\to\pm\infty}\frac{x-1}{x+2}=\frac{1}{1}=1\ \text{(horizontal asymptote)}$
3. Now ask the question directly instead of eyeballing the graph: set the function equal to its asymptote and solve.
4. The x cancels and leaves −1 = 2, which is false for every x. So the curve never touches y = 1 — not far out, not anywhere.
  
  $\frac{x-1}{x+2}=1\iff x-1=x+2\iff -1=2\quad\text{— no solution}$
5. The vertical asymptote behaves oppositely. Just right of −2, at −1.99, the function is −299; just left of it, at −2.01, it is +301. It does not approach a value from both sides, it diverges in opposite directions.
  
  $f(-1.99)=-299,\qquad f(-2.01)=+301$
Answer

The tool prints an x-intercept of 1, a vertical asymptote at −2 and a horizontal one at 1. The result worth keeping is that a horizontal asymptote is not a barrier — this one is never met, but that is a fact about this function and not a rule. Change the numerator to x − 1 + something that vanishes and a curve can cross its horizontal asymptote freely, even repeatedly. Vertical asymptotes are the strict ones: the function is undefined there, so it can never cross. Confusing the two is the most common mistake on this topic, and one line of algebra settles it every time.

$\text{the curve approaches}\ y=1\ \text{and never reaches it};\ \text{the pole flips sign}$

Supplementary problems, answers only

Find the intercepts and asymptotes of f(x) = (3x − 6)/(x + 2).

Show answer

x-intercept 2, y-intercept −3; asymptotes x = −2 and y = 3

$x=2,\ \ y=-3;\qquad x=-2,\ \ y=3$
Find the hole and the asymptotes of f(x) = (x² + x − 6)/(x² − 9).

Show answer

Hole at (−3, 5/6); asymptotes x = 3 and y = 1

$\left(-3,\tfrac56\right);\qquad x=3,\ \ y=1$
Find the oblique asymptote of f(x) = (2x² + x − 1)/(x − 1).

Show answer

y = 2x + 3

$y=2x+3$
Solve (x + 2)/(x − 1) ≤ 0.

Show answer

−2 ≤ x < 1

$-2\le x<1$
Find the horizontal asymptote of f(x) = (4x³ − x)/(2x³ + 7).

Show answer

y = 2

$y=2$

Learning path

Beyond the quadratic

References (2)

Why k(x−z)/(x−p) is a Möbius map, and why circles and lines are the only shapes it produces: T. Needham, Visual Complex Analysis, Chapter 3: Möbius Transformations and Inversion. Oxford University Press, 1997.
The asymptotes as ordinary behaviour at a point, once infinity is a point: L. V. Ahlfors, Complex Analysis, 3rd edition. McGraw-Hill, 1979 — rational functions on the extended plane.

By the 7 Raven Room team · Reviewed against published sources. Corrections welcome.