Reciprocal graphs and asymptotes

AS · P1 · 11 min

The graph of $y = \dfrac{k}{x}$ is a hyperbola: two separate branches that approach, but never meet, a pair of lines called asymptotes. Translated and stretched versions such as $y = \dfrac{2}{x - 1} + 3$ are favourites in Cambridge functions questions, because they test ranges, inverses and transformations all at once. Once you can read off the asymptotes, sketching and finding the range take a few lines.

The basic curve

For $y = \dfrac{1}{x}$ :

$x = 0$ is not allowed (division by zero), so the graph has no point on the $y$ -axis;
as $x$ gets large and positive, $y$ gets small and positive, approaching $0$ ;
as $x$ approaches $0$ from the right, $y$ becomes very large; from the left, very large and negative.

Definition

An asymptote is a line that a curve approaches more and more closely, without reaching it, as $x$ or $y$ becomes very large. A vertical asymptote $x = p$ occurs where the function is undefined; a horizontal asymptote $y = q$ is the value the function approaches as $x \to \pm\infty$ .

Key result

$y = \dfrac{k}{x}$ has asymptotes $x = 0$ and $y = 0$ .

$k > 0$ : branches in the first and third quadrants.
$k < 0$ : branches in the second and fourth quadrants.

$y = \dfrac{k}{x - p} + q$ is $y = \dfrac{k}{x}$ translated by $\begin{pmatrix} p \\ q \end{pmatrix}$ . Its asymptotes are $x = p$ and $y = q$ .

Domain: $x \neq p$ . Range: $y \neq q$ .

y = 1/x y = -2/x

The graph shows $y = \dfrac{1}{x}$ (first and third quadrants) and $y = -\dfrac{2}{x}$ (second and fourth).

The related curve $y = \dfrac{k}{x^2}$ (with $k > 0$ ) has the same asymptotes, but both branches are above the $x$ -axis, because $x^2 > 0$ .

Sketching a translated hyperbola

Sketching y = k/(x - p) + q

If the function is given as a single fraction, rewrite it in the form $\dfrac{k}{x - p} + q$ (see below).
Draw the asymptotes $x = p$ and $y = q$ as dashed lines.
Use the sign of $k$ to decide which pair of opposite "corners" the branches sit in: top right and bottom left if $k > 0$ ; top left and bottom right if $k < 0$ .
Find the intercepts: put $x = 0$ for the $y$ -intercept, and $y = 0$ for the $x$ -intercept (if they exist).
Draw each branch approaching both asymptotes, through the intercepts.

Sketching from the translated form

Sketch $y = \dfrac{2}{x - 1} + 3$ , stating the equations of the asymptotes and the coordinates of the intercepts.

Solution

Asymptotes: $x = 1$ and $y = 3$ . $k = 2 > 0$ , so the branches are top right and bottom left of the point $(1, 3)$ .

$y$ -intercept: $x = 0$ gives $y = -2 + 3 = 1$ , so $(0, 1)$ .

$x$ -intercept: $\dfrac{2}{x - 1} = -3$ , so $x - 1 = -\dfrac{2}{3}$ , $x = \dfrac{1}{3}$ , giving $\left(\dfrac{1}{3}, 0\right)$ .

y = 2/(x - 1) + 3 x = 1 y = 3 (0, 1) (1/3, 0)

Rewriting a single fraction

A function such as $\dfrac{3x + 1}{x - 2}$ can be rewritten by making the numerator contain a multiple of the denominator:

\frac{3x + 1}{x - 2} = \frac{3(x - 2) + 6 + 1}{x - 2} = 3 + \frac{7}{x - 2}

Now the asymptotes ( $x = 2$ and $y = 3$ ) and the sign of $k$ ( $7 > 0$ ) are visible.

Tip

For $y = \dfrac{ax + b}{cx + d}$ the vertical asymptote is where the denominator is zero, $x = -\dfrac{d}{c}$ , and the horizontal asymptote is the ratio of the $x$ coefficients, $y = \dfrac{a}{c}$ . Use this as a check.

Divide first

Express $\dfrac{3x + 1}{x - 2}$ in the form $a + \dfrac{b}{x - 2}$ , and sketch $y = \dfrac{3x + 1}{x - 2}$ .

Solution

From above, $\dfrac{3x + 1}{x - 2} = 3 + \dfrac{7}{x - 2}$ , so $a = 3$ , $b = 7$ .

Asymptotes $x = 2$ , $y = 3$ ; branches top right and bottom left.

Intercepts (easiest from the original fraction): $x = 0$ gives $y = -\tfrac{1}{2}$ ; $y = 0$ gives $3x + 1 = 0$ , $x = -\tfrac{1}{3}$ .

y = (3x + 1)/(x - 2) x = 2 y = 3 (0, -0.5) (-1/3, 0)

Transformations, range and inverse

Describing the transformations

Describe a sequence of transformations that maps $y = \dfrac{1}{x}$ onto $y = \dfrac{2}{x - 1} + 3$ .

Solution

$\dfrac{1}{x} \to \dfrac{2}{x} \to \dfrac{2}{x - 1} \to \dfrac{2}{x - 1} + 3$ :

stretch parallel to the $y$ -axis with scale factor $2$ ;
translation by $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$ .

Range and inverse on half the curve

The function $f$ is defined by $f(x) = 3 - \dfrac{4}{x + 2}$ for $x > -2$ .

(a) State the range of $f$ .

(b) Find $f^{-1}(x)$ and state its domain.

Solution

(a) For $x > -2$ , $x + 2 > 0$ , so $\dfrac{4}{x + 2}$ takes every positive value. Then $3 - \dfrac{4}{x + 2}$ takes every value less than $3$ . Range: $f(x) < 3$ .

(b)

y = 3 - \frac{4}{x + 2} \quad\Rightarrow\quad \frac{4}{x + 2} = 3 - y \quad\Rightarrow\quad x + 2 = \frac{4}{3 - y} \quad\Rightarrow\quad x = \frac{4}{3 - y} - 2

So $f^{-1}(x) = \dfrac{4}{3 - x} - 2$ , with domain $x < 3$ .

y = 3 - 4/(x + 2) + 0 sqrt(x + 2) y = 4/(3 - x) - 2 + 0 sqrt(3 - x) y = x

The two branches are reflections of each other in $y = x$ , as every function and its inverse must be.

Lines meeting hyperbolas

Substituting a line into $y = \dfrac{k}{x}$ and multiplying by $x$ gives a quadratic, so the discriminant decides how many times they meet.

When a line meets a hyperbola

Find the set of values of $c$ for which the line $y = c - x$ meets the curve $y = \dfrac{1}{x}$ at two distinct points, and the values of $c$ for which it is a tangent.

Solution

c - x = \frac{1}{x} \quad\Rightarrow\quad cx - x^2 = 1 \quad\Rightarrow\quad x^2 - cx + 1 = 0

(Multiplying by $x$ is safe because $x = 0$ is not on the curve.)

Discriminant: $c^2 - 4$ .

Two distinct points: $c^2 - 4 > 0$ , so $c < -2$ or $c > 2$ .
Tangent: $c^2 - 4 = 0$ , so $c = \pm 2$ .

y = 1/x y = 2 - x y = -2 - x

The lines $y = 2 - x$ and $y = -2 - x$ touch the two branches at $(1, 1)$ and $(-1, -1)$ .

Watch out

Drawing the curve crossing an asymptote. For $y = \dfrac{k}{x - p} + q$ the curve never meets $x = p$ or $y = q$ .

Wrong sign in the vertical asymptote. $y = \dfrac{5}{x + 4}$ has asymptote $x = -4$ .

Forgetting the range excludes $q$ . On the whole domain the range is $y \neq q$ ; on one branch it is $y > q$ or $y < q$ .

Mixing up quadrants. Check one point: for $y = \dfrac{2}{x - 1} + 3$ at $x = 2$ , $y = 5 > 3$ , so the right branch is above the horizontal asymptote.

Exam tip

On sketches, draw asymptotes as dashed lines and write their equations on the diagram. Label intercepts with coordinates.
Ranges of reciprocal functions are often asked with a domain on one side of the vertical asymptote. Decide whether the fraction is positive or negative there, then build the range step by step.
Inverse questions follow the usual method; the vertical and horizontal asymptotes swap roles in the inverse (here $x = -2, y = 3$ became $x = 3, y = -2$ ).

Summary

$y = \dfrac{k}{x}$ : asymptotes $x = 0$ , $y = 0$ ; branches in quadrants 1 and 3 if $k > 0$ , 2 and 4 if $k < 0$ .
$y = \dfrac{k}{x - p} + q$ is a translation by $\begin{pmatrix} p \\ q \end{pmatrix}$ ; asymptotes $x = p$ , $y = q$ .
Rewrite $\dfrac{ax + b}{cx + d}$ by splitting the numerator into a multiple of the denominator plus a constant.
Domain $x \neq p$ ; range $y \neq q$ , or one side of $q$ on a restricted domain.
Lines meeting hyperbolas: multiply through by $x$ and use the discriminant.

Practice questions

Question

State the equations of the asymptotes of $y = \dfrac{5}{x + 4} - 2$ and find the coordinates of its intercepts.
Express $\dfrac{2x - 5}{x + 1}$ in the form $a + \dfrac{b}{x + 1}$ , and state the range of $y = \dfrac{2x - 5}{x + 1}$ for $x \neq -1$ .
Find the range of $f(x) = \dfrac{4}{x - 3} + 1$ for $x < 3$ .
Find the range of $f(x) = \dfrac{6}{x + 1}$ for $1 \le x \le 5$ .
$f(x) = \dfrac{2}{x - 1} + 3$ for $x > 1$ . Find $f^{-1}(x)$ and state its domain.
Describe a sequence of transformations mapping $y = \dfrac{1}{x}$ onto $y = -\dfrac{3}{x + 2}$ .
Sketch $y = \dfrac{1}{(x - 2)^2}$ , giving its asymptotes and $y$ -intercept, and state its range.
Find the points where the line $y = 2x + 1$ meets the curve $y = \dfrac{3}{x}$ .
Find the value of $m$ for which the line $y = mx + 2$ is a tangent to the curve $y = -\dfrac{1}{x}$ , and find the point of contact.
$f(x) = \dfrac{2x + 3}{x - 1}$ for $x > 1$ . (a) Express $f(x)$ in the form $a + \dfrac{b}{x - 1}$ . (b) State the range of $f$ . (c) Find $f^{-1}(x)$ and its domain.

Answers

Asymptotes $x = -4$ , $y = -2$ . $y$ -intercept: $\tfrac{5}{4} - 2 = -\tfrac{3}{4}$ , so $\left(0, -\tfrac{3}{4}\right)$ . $x$ -intercept: $\dfrac{5}{x + 4} = 2$ , $x + 4 = \tfrac{5}{2}$ , $x = -\tfrac{3}{2}$ , so $\left(-\tfrac{3}{2}, 0\right)$ .
$\dfrac{2(x + 1) - 7}{x + 1} = 2 - \dfrac{7}{x + 1}$ . Range $y \neq 2$ .
For $x < 3$ , $x - 3 < 0$ , so $\dfrac{4}{x - 3}$ takes every negative value. Range $f(x) < 1$ .
$f$ is decreasing on this domain: $f(1) = 3$ , $f(5) = 1$ . Range $1 \le f(x) \le 3$ .
Range of $f$ is $f(x) > 3$ . $y - 3 = \dfrac{2}{x - 1}$ , $x = 1 + \dfrac{2}{y - 3}$ . $f^{-1}(x) = 1 + \dfrac{2}{x - 3}$ for $x > 3$ .
Stretch parallel to the $y$ -axis with scale factor $3$ ; reflection in the $x$ -axis; translation by $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$ . (The translation acts in the $x$ -direction, so it can come at any point in the sequence.)
Asymptotes $x = 2$ and $y = 0$ . Both branches above the $x$ -axis. $y$ -intercept $\tfrac{1}{4}$ , so $\left(0, \tfrac{1}{4}\right)$ . Range $y > 0$ .
$2x + 1 = \dfrac{3}{x}$ gives $2x^2 + x - 3 = 0$ , $(2x + 3)(x - 1) = 0$ . Points $(1, 3)$ and $\left(-\tfrac{3}{2}, -2\right)$ .
$mx + 2 = -\dfrac{1}{x}$ gives $mx^2 + 2x + 1 = 0$ . For a tangent (with $m \neq 0$ ), $4 - 4m = 0$ , so $m = 1$ . Then $x^2 + 2x + 1 = 0$ , $x = -1$ , $y = 1$ . Point of contact $(-1, 1)$ . (If $m = 0$ the equation is linear, $2x + 1 = 0$ : the line $y = 2$ crosses the curve once at $x = -\tfrac{1}{2}$ , but it crosses rather than touches, so it is not a tangent.)
(a) $\dfrac{2(x - 1) + 5}{x - 1} = 2 + \dfrac{5}{x - 1}$ . (b) For $x > 1$ , $\dfrac{5}{x - 1} > 0$ , so $f(x) > 2$ . (c) $y - 2 = \dfrac{5}{x - 1}$ , $x = 1 + \dfrac{5}{y - 2}$ . $f^{-1}(x) = 1 + \dfrac{5}{x - 2}$ for $x > 2$ (equivalently $\dfrac{x + 3}{x - 2}$ ).

The basic curve

Sketching a translated hyperbola

Rewriting a single fraction

Transformations, range and inverse

Lines meeting hyperbolas

Practice questions

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