Graphing Reciprocal Functions

Graphing Reciprocal Functions — Moosa Academy

Put the variable underneath, in the denominator, and the graph stops being a single connected curve. Because division by zero is undefined, one x-value has to be removed — and around that gap the curve splits into two branches that race off toward invisible guide lines they never reach.

Concept The parent reciprocal function

The simplest reciprocal function is

f(x) = \dfrac{1}{x}

. Its graph is a hyperbola: two separate branches sitting in opposite corners of the plane. Feed it a large

x

and the output shrinks toward zero; feed it a value near zero and the output blows up.

Domain	All real numbers except $0$
Range	All real numbers except $0$
Vertical asymptote	$x = 0$
Horizontal asymptote	$y = 0$
Intercepts	None — the curve meets neither axis

Theorem The general form

For $f(x) = \dfrac{a}{x - b} + c$ , the vertical asymptote is $x = b$ and the horizontal asymptote is $y = c$ .

$a$ — stretches the branches; a negative value flips them into the opposite pair of quadrants

$b$ — slides the graph sideways, carrying the vertical asymptote with it

$c$ — slides the graph up or down, carrying the horizontal asymptote with it

Notice the sign trap in $b$ . Written $x - b$ , a graph shifted to $x = 3$ has denominator $x - 3$ ; a graph shifted to $x = -1$ has denominator $x + 1$ . The asymptote sits where the denominator vanishes, not at the number you see printed.

Concept Finding the two asymptotes

Vertical: set the denominator equal to zero and solve for

x

. That value is excluded from the domain.

Horizontal: read the constant added on the end. That value is excluded from the range.

Both lines are guides, not part of the graph. The curve approaches each one without limit and never touches it, which is exactly why those two values are missing from the domain and the range.

Example Where is the function undefined?

Find the value of $x$ that makes $f(x) = \dfrac{1}{x - 3}$ undefined.

Set the denominator to zero:

x - 3 = 0

Solve:

x = 3

Undefined at $x = 3$ ; the vertical asymptote is $x = 3$

The domain is every real number except $3$ . Nothing else about the function matters for this question — only the denominator can break it.

Example Reading a shifted graph

State the asymptotes, domain and range of $f(x) = \dfrac{2}{x + 1} - 4$ .

Denominator zero:

x + 1 = 0

, so

x = -1

Constant term:

-4

Vertical asymptote

x = -1

, horizontal asymptote

y = -4

Domain: $x \neq -1$ Range: $y \neq -4$

The whole parent hyperbola has simply been picked up and moved one unit left and four units down. Its shape is unchanged.

Example A reciprocal in real life

A hot-air balloon basket has $20$ square feet of floor, shared equally. The space each passenger gets is $f(x) = \dfrac{20}{x}$ , where $x$ is the number of people.

People $x$	2	4	5	10
Space $f(x)$	10	5	4	2

Mathematically the asymptotes are still $x = 0$ and $y = 0$ , but a balloon cannot hold zero or a negative number of passengers. The practical domain is $x > 0$ , so only the first-quadrant branch is drawn. Add more people and everyone's share slides closer to the horizontal axis — small, but never zero.

Summary

A reciprocal function has the variable in the denominator; its graph is a hyperbola of two branches.
The parent $f(x) = \dfrac{1}{x}$ has asymptotes $x = 0$ and $y = 0$ , and no intercepts.
In $f(x) = \dfrac{a}{x - b} + c$ , the asymptotes are $x = b$ and $y = c$ .
Find the vertical asymptote by setting the denominator to zero; the horizontal one is the constant added on the end.
The excluded $x$ -value is missing from the domain; the excluded $y$ -value is missing from the range.
In a real-world model, physical sense can restrict the domain further — often to $x > 0$ .