Reciprocal Functions

The reciprocal of a function $f$ is the function $y = \dfrac{1}{f(x)}$. You already know one example: $y = \dfrac{1}{x}$ is the reciprocal of $y = x$. On this page you’ll learn to sketch the reciprocal of any linear or quadratic function straight from the graph of $f$, without plotting dozens of points. It’s the first step into rational functions.

Key ideas

Reciprocal, not inverse

$\dfrac{1}{f(x)}$ means “divide $1$ by the output of $f$”. It is not the inverse $f^{-1}(x)$. For example, if $f(x) = x - 2$, then $\dfrac{1}{f(x)} = \dfrac{1}{x - 2}$, but $f^{-1}(x) = x + 2$.

How the graph of 1/f(x) is built from f

Each output of $\dfrac{1}{f(x)}$ is $1$ divided by the matching output of $f$. That one fact explains every feature:

Where the graph of $f$ …	the graph of $y = \dfrac{1}{f(x)}$ …	Why
has a zero, $f(a) = 0$	has a vertical asymptote $x = a$	you can’t divide by $0$; near $a$, $f$ is tiny, so $\frac{1}{f}$ is huge
gets very large (positive or negative)	gets close to $0$: horizontal asymptote $y = 0$	$1$ divided by a huge number is tiny
is positive / negative	is positive / negative too	$1$ divided by a number keeps its sign
has $y = 1$ or $y = -1$	has the same point (an invariant point)	$\frac{1}{1} = 1$ and $\frac{1}{-1} = -1$
is increasing	is decreasing (and vice versa)	bigger outputs give smaller reciprocals
has a minimum $(a, k)$, $k \ne 0$	has a maximum $\left(a, \frac{1}{k}\right)$ (and vice versa)	the smallest output gives the biggest reciprocal
has $y$-intercept $b$, $b \ne 0$	has $y$-intercept $\frac{1}{b}$	$\frac{1}{f(0)}$

The domain of $y = \dfrac{1}{f(x)}$ is every $x$ except the zeros of $f$. The reciprocal of a linear or quadratic function is never $0$, so it has no $x$-intercepts.

Reciprocals of linear functions

For $f(x) = mx + b$ with $m \ne 0$, the graph of $y = \dfrac{1}{f(x)}$ always has two branches, like $y = \dfrac{1}{x}$: one vertical asymptote at the zero of $f$, and the horizontal asymptote $y = 0$. If $f$ is increasing, both branches decrease; if $f$ is decreasing, both branches increase.

$y = 2x - 4$ (orange) and its reciprocal $y = \dfrac{1}{2x - 4}$ (blue). They meet where $y = \pm 1$.

Reciprocals of quadratic functions

The number of zeros of the quadratic (check the discriminant) decides the number of vertical asymptotes:

Zeros of $f$	Example	Graph of $y = \dfrac{1}{f(x)}$
two	$f(x) = x^2 - 2x - 3$	two vertical asymptotes, three branches; the middle branch has a turning point
one (a double zero)	$f(x) = (x - 3)^2$	one vertical asymptote; both branches go up beside it
none	$f(x) = x^2 + 2x + 5$	no vertical asymptotes; one smooth “hump” with a maximum

The turning point of the reciprocal is always directly above or below the vertex of the parabola (same $x$-value), as long as the vertex isn’t on the $x$-axis.

Steps for sketching y = 1/f(x)

1. Sketch $y = f(x)$ lightly.
2. Draw vertical asymptotes at the zeros of $f$, and the horizontal asymptote $y = 0$.
3. Mark the invariant points, where $f(x) = 1$ or $f(x) = -1$.
4. Plot the reciprocal of the vertex and of the $y$-intercept.
5. Draw each branch on the same side of the $x$-axis as $f$, heading toward the asymptotes.

Worked examples

Example 1: Reciprocal of a linear function

Sketch $y = \dfrac{1}{2x - 4}$. State its domain, range, asymptotes, intercepts, and intervals of increase and decrease.

Solution. Let $f(x) = 2x - 4$.

• Vertical asymptote: $2x - 4 = 0$ when $x = 2$, so $x = 2$.
• Horizontal asymptote: $y = 0$.
• Invariant points: $2x - 4 = 1$ gives $x = 2.5$, so $(2.5, 1)$. $2x - 4 = -1$ gives $x = 1.5$, so $(1.5, -1)$.
• $y$-intercept: $f(0) = -4$, so the reciprocal has $y$-intercept $-\dfrac{1}{4}$. There’s no $x$-intercept.
• Sign: $f$ is negative for $x \lt 2$ and positive for $x \gt 2$, so the reciprocal is too.

The graph is shown in the figure above.

Domain: $\{x \in \mathbb{R} \mid x \ne 2\}$. Range: $\{y \in \mathbb{R} \mid y \ne 0\}$.

$f$ is increasing everywhere, so the reciprocal is decreasing on both branches: for $x \lt 2$ and for $x \gt 2$.

Example 2: A quadratic with two zeros

Sketch $y = \dfrac{1}{x^2 - 2x - 3}$ and describe its key features.

Solution. Let $f(x) = x^2 - 2x - 3 = (x + 1)(x - 3)$.

• Vertical asymptotes: the zeros of $f$, so $x = -1$ and $x = 3$.
• Horizontal asymptote: $y = 0$.
• Vertex: halfway between the zeros, $x = 1$, and $f(1) = 1 - 2 - 3 = -4$. The vertex $(1, -4)$ is a minimum of $f$, so the reciprocal has a local maximum at $\left(1, -\dfrac{1}{4}\right)$.
• $y$-intercept: $f(0) = -3$, so the reciprocal’s $y$-intercept is $-\dfrac{1}{3}$.
• Sign: $f$ is positive for $x \lt -1$ and $x \gt 3$, and negative for $-1 \lt x \lt 3$. The reciprocal matches.
• Invariant points: $f(x) = 1$ gives $x^2 - 2x - 4 = 0$, so $x = 1 \pm \sqrt{5}$ (about $-1.24$ and $3.24$). $f(x) = -1$ gives $x^2 - 2x - 2 = 0$, so $x = 1 \pm \sqrt{3}$ (about $-0.73$ and $2.73$).

The parabola’s minimum $(1, -4)$ becomes the reciprocal’s local maximum $\left(1, -\frac{1}{4}\right)$.

Domain: $\{x \in \mathbb{R} \mid x \ne -1, 3\}$. Range: $\left\{y \in \mathbb{R} \mid y \gt 0 \text{ or } y \le -\frac{1}{4}\right\}$.

$f$ decreases for $x \lt 1$ and increases for $x \gt 1$. So the reciprocal increases for $x \lt -1$ and for $-1 \lt x \lt 1$, and decreases for $1 \lt x \lt 3$ and for $x \gt 3$.

Example 3: A quadratic with no zeros

Sketch $y = \dfrac{1}{x^2 + 2x + 5}$ and state its domain and range.

Solution. Complete the square: $f(x) = x^2 + 2x + 5 = (x + 1)^2 + 4$. The vertex is the minimum $(-1, 4)$, and $f(x) \ge 4$ for every $x$, so $f$ has no zeros.

• No vertical asymptotes. The horizontal asymptote is $y = 0$.
• $f$ is always positive, so the reciprocal is always positive.
• The minimum $(-1, 4)$ of $f$ becomes a maximum $\left(-1, \dfrac{1}{4}\right)$ of the reciprocal.
• $y$-intercept: $\dfrac{1}{f(0)} = \dfrac{1}{5}$.
• No invariant points, since $f(x)$ is never $1$ or $-1$.

The graph is a single smooth hump: it rises from near $0$ on the far left to the peak $\left(-1, \frac{1}{4}\right)$, then falls back toward $0$. It increases for $x \lt -1$ and decreases for $x \gt -1$.

Domain: $\{x \in \mathbb{R}\}$. Range: $\left\{y \in \mathbb{R} \mid 0 \lt y \le \frac{1}{4}\right\}$.

Example 4: A quadratic with one zero

Describe the graph of $y = \dfrac{1}{(x - 3)^2}$.

Solution. $f(x) = (x - 3)^2$ has a double zero at $x = 3$, so there is one vertical asymptote, $x = 3$, and the horizontal asymptote is $y = 0$.

$f$ is never negative, so the reciprocal is positive on both sides of the asymptote: both branches shoot upward next to $x = 3$.

Invariant points: $(x - 3)^2 = 1$ gives $x = 2$ or $x = 4$, so $(2, 1)$ and $(4, 1)$. $(x - 3)^2 = -1$ has no solutions. The $y$-intercept is $\dfrac{1}{9}$.

$f$ decreases for $x \lt 3$ and increases for $x \gt 3$, so the reciprocal increases for $x \lt 3$ and decreases for $x \gt 3$.

Domain: $\{x \in \mathbb{R} \mid x \ne 3\}$. Range: $\{y \in \mathbb{R} \mid y \gt 0\}$.

Common mistakes

Mixing up the reciprocal and the inverse. $\dfrac{1}{f(x)}$ divides $1$ by the output; $f^{-1}(x)$ undoes $f$. Their graphs look nothing alike.

Putting the reciprocal on the wrong side of the $x$-axis. The reciprocal always has the same sign as $f$. Wherever $f$ is below the axis, so is $\dfrac{1}{f}$. Check each interval between asymptotes.

Putting the turning point at the wrong height. If the vertex of $f$ is $(1, -4)$, the reciprocal’s turning point is $\left(1, -\frac{1}{4}\right)$, not $(1, 4)$ or $(1, -4)$. Take the reciprocal of the $y$-coordinate only.

Forgetting the invariant points. Points where $f(x) = 1$ or $-1$ are on both graphs. They are the easiest points to plot and they help you place each branch.

Letting the graph cross a vertical asymptote or touch $y = 0$. $\dfrac{1}{f(x)}$ is undefined at the zeros of $f$, and it’s never $0$, so the graph never meets these lines.

Drawing vertical asymptotes when $f$ has no zeros. If the discriminant is negative, as in Example 3, the reciprocal has no vertical asymptotes at all: its graph is one unbroken curve.

Practice

1. (Warm-up) Let $f(x) = x + 3$. For $y = \dfrac{1}{f(x)}$, state the asymptotes and the invariant points.

Solution

$f(x) = 0$ at $x = -3$, so the vertical asymptote is $x = -3$. The horizontal asymptote is $y = 0$.

Invariant points: $x + 3 = 1$ gives $x = -2$, so $(-2, 1)$. $x + 3 = -1$ gives $x = -4$, so $(-4, -1)$.

2. (Warm-up) The points $(-2, 4)$, $(1, 0.5)$, $(3, -1)$, and $(5, 0)$ are on the graph of $y = f(x)$. What does each one tell you about the graph of $y = \dfrac{1}{f(x)}$?

Solution

Take the reciprocal of each $y$-coordinate:

• $(-2, 4) \to \left(-2, \frac{1}{4}\right)$
• $(1, 0.5) \to (1, 2)$, since $\frac{1}{0.5} = 2$
• $(3, -1) \to (3, -1)$, an invariant point
• $(5, 0)$: $\frac{1}{0}$ is undefined, so the reciprocal has a vertical asymptote $x = 5$.

3. (Warm-up) How many vertical asymptotes does each graph have?

• (a) $y = \dfrac{1}{x^2 - 9}$
• (b) $y = \dfrac{1}{x^2 + 9}$
• (c) $y = \dfrac{1}{x^2 - 6x + 9}$

Solution

Count the zeros of the denominator.

(a) $x^2 - 9 = (x - 3)(x + 3)$ has two zeros, so two asymptotes: $x = -3$ and $x = 3$.

(b) $x^2 + 9 \ge 9$ is never $0$, so there are none.

(c) $x^2 - 6x + 9 = (x - 3)^2$ has one (double) zero, so one asymptote: $x = 3$.

4. (Core) Sketch $y = \dfrac{1}{3 - x}$. State its domain, range, intercepts, and where it is increasing or decreasing.

Solution

Let $f(x) = 3 - x$, a decreasing line with zero $x = 3$ and $y$-intercept $3$.

• Vertical asymptote $x = 3$; horizontal asymptote $y = 0$.
• Invariant points: $3 - x = 1$ gives $(2, 1)$; $3 - x = -1$ gives $(4, -1)$.
• $y$-intercept $\frac{1}{3}$; no $x$-intercept.
• $f$ is positive for $x \lt 3$ and negative for $x \gt 3$, so the left branch is above the $x$-axis and the right branch is below it.

Domain: $\{x \in \mathbb{R} \mid x \ne 3\}$. Range: $\{y \in \mathbb{R} \mid y \ne 0\}$.

$f$ is decreasing, so the reciprocal is increasing on both branches: for $x \lt 3$ and for $x \gt 3$.

5. (Core) For $y = \dfrac{1}{x^2 - x - 6}$, find the asymptotes, the $y$-intercept, the turning point, the positive and negative intervals, and the range.

Solution

$f(x) = x^2 - x - 6 = (x - 3)(x + 2)$.

• Vertical asymptotes $x = -2$ and $x = 3$; horizontal asymptote $y = 0$.
• $y$-intercept: $\frac{1}{f(0)} = -\frac{1}{6}$.
• Vertex of $f$: $x = \frac{-2 + 3}{2} = \frac{1}{2}$, and $f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} - 6 = -\frac{25}{4}$. So the reciprocal has a local maximum at $\left(\frac{1}{2}, -\frac{4}{25}\right)$.
• Positive for $x \lt -2$ or $x \gt 3$; negative for $-2 \lt x \lt 3$.

Range: $\left\{y \in \mathbb{R} \mid y \gt 0 \text{ or } y \le -\frac{4}{25}\right\}$.

6. (Core) Sketch $y = \dfrac{1}{x^2 + 4x + 8}$, and state its maximum value, domain, and range.

Solution

$f(x) = x^2 + 4x + 8 = (x + 2)^2 + 4$, with minimum $(-2, 4)$ and no zeros.

The reciprocal has no vertical asymptotes, a horizontal asymptote $y = 0$, and is always positive. Its maximum is at $\left(-2, \frac{1}{4}\right)$, so the maximum value is $\frac{1}{4}$. Its $y$-intercept is $\frac{1}{8}$. It increases for $x \lt -2$ and decreases for $x \gt -2$: one hump centred on $x = -2$.

Domain: $\{x \in \mathbb{R}\}$. Range: $\left\{y \in \mathbb{R} \mid 0 \lt y \le \frac{1}{4}\right\}$.

7. (Core) The graph of $y = \dfrac{1}{ax + b}$ has a vertical asymptote $x = -\dfrac{1}{2}$ and a $y$-intercept of $1$. Find $a$ and $b$.

Solution

The $y$-intercept is $\frac{1}{b} = 1$, so $b = 1$.

The asymptote is the zero of $ax + 1$: $a\left(-\frac{1}{2}\right) + 1 = 0$, so $a = 2$.

The function is $y = \dfrac{1}{2x + 1}$. Check: at $x = 0$, $y = 1$ ✓, and $2x + 1 = 0$ at $x = -\frac{1}{2}$ ✓.

8. (Challenge) Consider the family $y = \dfrac{1}{x^2 - 2x + n}$, where $n$ is a constant. Describe how the vertical asymptotes and the turning point depend on $n$. Consider $n \lt 1$, $n = 1$, and $n \gt 1$.

Solution

Complete the square: $f(x) = x^2 - 2x + n = (x - 1)^2 + (n - 1)$, so the parabola’s vertex is $(1, n - 1)$.

• $n \lt 1$: the vertex is below the axis, so $f$ has two zeros, $x = 1 \pm \sqrt{1 - n}$, and the reciprocal has two vertical asymptotes. The middle branch is negative with a local maximum at $\left(1, \frac{1}{n - 1}\right)$. For example, $n = -3$ gives $f(x) = (x - 3)(x + 1)$, asymptotes $x = -1$ and $x = 3$, and a turning point $\left(1, -\frac{1}{4}\right)$.
• $n = 1$: $y = \dfrac{1}{(x - 1)^2}$ has one vertical asymptote, $x = 1$, with both branches above the axis. There’s no turning point, since the vertex of $f$ is on the axis.
• $n \gt 1$: $f$ has no zeros, so there are no vertical asymptotes. The graph is a positive hump with maximum $\left(1, \frac{1}{n - 1}\right)$. The bigger $n$ is, the lower and flatter the hump.

In every case the horizontal asymptote is $y = 0$.

9. (Challenge) The graph of $y = \dfrac{1}{f(x)}$, where $f$ is a quadratic function, has vertical asymptotes $x = 1$ and $x = 5$ and passes through $\left(3, -\frac{1}{2}\right)$. Find $f(x)$, and describe the turning point of the reciprocal.

Solution

The zeros of $f$ are $1$ and $5$, so $f(x) = a(x - 1)(x - 5)$.

At $x = 3$ the reciprocal is $-\frac{1}{2}$, so $f(3) = -2$:

$$a(3 - 1)(3 - 5) = -2 \quad\Rightarrow\quad -4a = -2 \quad\Rightarrow\quad a = \frac{1}{2}$$$$f(x) = \frac{1}{2}(x - 1)(x - 5)$$

The vertex of $f$ is halfway between the zeros, at $x = 3$, so $(3, -2)$ is the minimum of $f$. That makes $\left(3, -\frac{1}{2}\right)$ a local maximum of the reciprocal.

Check: $\dfrac{1}{f(3)} = \dfrac{1}{\frac{1}{2}(2)(-2)} = -\dfrac{1}{2}$ ✓