UK Secondary (7-11)

Graphing Hyperbolas (k/x)

Lesson

We can graph a function $\frac{k}{x}$`k``x` by constructing a table of values having first specified a value for the parameter $k$`k`. The shape of the graph will be a hyperbola and the effect of changing $k$`k` is to change the scale of the graph. These properties are illustrated in the following diagram where the graph of $y=\frac{1}{x}$`y`=1`x` is shown in blue, $y=\frac{3}{x}$`y`=3`x` is shown in red and $y=\frac{5}{x}$`y`=5`x` is shown in green.

If we had to draw these graphs by hand, we could construct tables of values like the following. We have restricted $x$`x` to values between $-5$−5 and $5$5.

You should check whether the graphs above really do match the corresponding tables.

$x$x |
$-5$−5 | $-4$−4 | $-3$−3 | $-2$−2 | $-1$−1 | $\frac{1}{2}$12 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
---|---|---|---|---|---|---|---|---|---|---|---|

$\frac{1}{x}$1x |
$-\frac{1}{5}$−15 | $-\frac{1}{4}$−14 | $-\frac{1}{3}$−13 | $-\frac{1}{2}$−12 | $-1$−1 | $2$2 | $1$1 | $\frac{1}{2}$12 | $\frac{1}{3}$13 | $\frac{1}{4}$14 | $\frac{1}{5}$15 |

$\frac{3}{x}$3x |
$-\frac{3}{5}$−35 | $-\frac{3}{4}$−34 | $-1$−1 | $-\frac{3}{2}$−32 | $-3$−3 | $6$6 | $3$3 | $\frac{3}{2}$32 | $1$1 | $\frac{3}{4}$34 | $\frac{3}{5}$35 |

$\frac{5}{x}$5x |
$-1$−1 | $-\frac{5}{4}$−54 | $-\frac{5}{3}$−53 | $-\frac{5}{2}$−52 | $-5$−5 | $10$10 | $5$5 | $\frac{5}{2}$52 | $\frac{5}{3}$53 | $\frac{5}{4}$54 | $1$1 |

To be convinced that, for example, the graph of $f(x)=\frac{2}{x}$`f`(`x`)=2`x` has exactly the same shape as the graph of $g(x)=\frac{1}{x}$`g`(`x`)=1`x`, but with a different scale, we can think of a new variable $u=\frac{x}{2}$`u`=`x`2 or, equivalently, $x=2u$`x`=2`u`. Now, the natural domain of the function $f(x)=\frac{2}{x}$`f`(`x`)=2`x` is the set of real numbers without zero and it is clear that as $x$`x` varies over this domain, $u$`u` must vary over exactly the same set of numbers.

So, with the function $f(x)=\frac{2}{x}$`f`(`x`)=2`x`, we can write $f(x)=\frac{2}{2u}=\frac{1}{u}=g(u)$`f`(`x`)=22`u`=1`u`=`g`(`u`). Thus, we see that $f$`f` and $g$`g` are the same function.

This idea is illustrated in the diagram below.

The values of $g(x)$`g`(`x`) are the same as the values of $f(u)$`f`(`u`).

A glance at all of the hyperbola graphs displayed above suggests that they are symmetrical about the line $y=x$`y`=`x`. We confirm this by noting that the equation $y=\frac{k}{x}$`y`=`k``x` can be written as $k=xy$`k`=`x``y` and it is clear that $x$`x` and $y$`y` can change positions without affecting the relation. We could swap the positions of the $x$`x`- and $y$`y`-axes and the graph should look the same.

Consider the function $y=\frac{2}{x}$`y`=2`x`

Complete the following table of values.

$x$ `x`$-2$−2 $-1$−1 $\frac{-1}{2}$−12 $\frac{1}{2}$12 $1$1 $2$2 $y$ `y`$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Plot the graph.

Loading Graph...In which quadrants does the graph lie?

$3$3

A$2$2

B$1$1

C$4$4

D$3$3

A$2$2

B$1$1

C$4$4

D

Ursula wants to sketch the graph of $y=\frac{7}{x}$`y`=7`x`, but knows that it will look similar to many other hyperbolas.

What can she do to the graph to show that it is the hyperbola $y=\frac{7}{x}$`y`=7`x`, rather than any other hyperbola of the form $y=\frac{k}{x}$`y`=`k``x`?

She can label the axes of symmetry.

AShe can label a point on the graph.

BShe can label the asymptotes.

CShe can label the axes of symmetry.

AShe can label a point on the graph.

BShe can label the asymptotes.

C

A graph of the hyperbola $y=\frac{10}{x}$`y`=10`x` is shown below. Given points $C$`C`$\left(-4,0\right)$(−4,0) and $D$`D`$\left(2,0\right)$(2,0), find the length of interval $AB$`A``B`.

Loading Graph...