Hong Kong

Stage 4 - Stage 5

Lesson

The graph of every hyperbola has two separate arcs and these arcs in their extremities move toward straight lines called asymptotes.

Every hyperbola, just like every circle, has a centre but, just like a circle, this centre is not part of the curve itself. One important feature of every hyperbola is that if we rotate any one of the arcs by $180^{\circ}$180∘ about the hyperbola's centre, that arc will fall directly onto the other arc.

The centre of the general rectangular hyperbola given by $y=\frac{a}{x-h}+k$`y`=`a``x`−`h`+`k` is the point $\left(h,k\right)$(`h`,`k`). It provides both the key to the asymptotes and to the efficient plotting of the curve.

The asymptotes of a rectangular hyperbola are given by the vertical line $x=h$`x`=`h` (this is the line passing through the point $\left(h,0\right)$(`h`,0) and parallel to the $y$`y` axis) and the horizontal line $y=k$`y`=`k` (this is the line passing through the point $\left(0,k\right)$(0,`k`) and parallel to the $x$`x` axis).

To ensure a a good plot of the hyperbola we should select a range $x$`x` values on either side of the asymptote $x=h$`x`=`h`.

As an example, suppose we wish to construct a plot of the hyperbola $y=\frac{12}{x}$`y`=12`x`. Here, $h=0$`h`=0 and $k=0$`k`=0, so the asymptotes are the vertical line given by $x=0$`x`=0 (this happens to be the $y$`y` axis), and the horizontal line $y=0$`y`=0 (this happens to be the $x$`x` axis).

We might choose $4$4 equally spaced values of $x$`x` to the left of the line $x=0$`x`=0 and four values to the right, and then use the hyperbola's equation to determine $6$6 values of $y$`y`:

$x$x |
$-4$−4 | $-3$−3 | $-2$−2 | $-1$−1 | $1$1 | $2$2 | $3$3 | $4$4 |
---|---|---|---|---|---|---|---|---|

$y$y |
$-3$−3 | $-4$−4 | $-6$−6 | $-12$−12 | $12$12 | $6$6 | $4$4 | $3$3 |

We plot these points, and use our knowledge of the features of a hyperbola to imagine the location of the two arcs (shown as dotted lines) and the asymptotes (the axes in this case) that the arcs are moving toward.

We need to keep in mind the rotational symmetry of the two arcs about the centre $\left(0,0\right)$(0,0).

One improvement we could make to our strategy is to choose $x$`x` values that, while not equally separated, give rise to integer $y$`y` values whenever possible.

For example suppose we wish to plot the graph of $y=\frac{12}{x-3}+7$`y`=12`x`−3+7. We note that the centre is $\left(3,7\right)$(3,7), and the two mutually perpendicular asymptotes are the straight lines $x=3$`x`=3 and $y=7$`y`=7.

As in the first example, we choose an equal number of $x$`x` values on either side of the line $x=3$`x`=3. However this time we will choose the $x$`x` values so that the quantity $\left(x-3\right)$(`x`−3) is a factor of $12$12. This ensures that the $y$`y` values will be integers.

For example we might choose $x=4$`x`=4 because $\frac{12}{4-3}+7$124−3+7 is the integer $19$19. Similarly the five values $x=5$`x`=5, $x=6$`x`=6, $x=7$`x`=7, $x=9$`x`=9, $x=15$`x`=15 reveal $y$`y` values of $y=13$`y`=13, $y=11$`y`=11, $y=10$`y`=10, $y=9$`y`=9 and $y=8$`y`=8.

On the left hand side, moving away from the asymptote $x=3$`x`=3 , the values $x=2$`x`=2, $x=1$`x`=1, $x=0$`x`=0, $x=-1$`x`=−1, $x=-3$`x`=−3, $x=-9$`x`=−9 reveal integer $y$`y` values of $y=-5$`y`=−5, $y=1$`y`=1, $y=3$`y`=3, $y=4$`y`=4, $y=5$`y`=5, $y=6$`y`=6.

The dilation factor $a=12$`a`=12 has many factors, and we were lucky to pick up $6$6 integer points because of it. In general however the dilation factor will not be as abundant in factors, and so sometimes we are forced to plot a number of non-integer points.

Here is the table of values, using just $10$10 of the above points, split up to show the points on each side of the asymptote:

Left side $x$x |
$-9$−9 | $-3$−3 | $-1$−1 | $0$0 | $2$2 |
---|---|---|---|---|---|

$y$y |
$6$6 | $5$5 | $4$4 | $3$3 | $-5$−5 |

:

Right side $x$x |
$4$4 | $5$5 | $7$7 | $9$9 | $15$15 |
---|---|---|---|---|---|

$y$y |
$19$19 | $13$13 | $10$10 | $9$9 | $8$8 |

The $x$`x` values chosen gives as wide a range as possible so that the imagined arcs might be easier to draw after the points have been plotted.

Here is the plot.

Consider the function $y=\frac{-1}{4x}$`y`=−14`x`

a) Complete the table of values

b) Plot the graph

c) In what quadrants does the graph lie?