Hong Kong

Stage 4 - Stage 5

Lesson

The hyperbola given by $y=\frac{a}{x-h}+k$`y`=`a``x`−`h`+`k` can be thought of as the basic rectangular hyperbola $y=\frac{a}{x}$`y`=`a``x` translated horizontally (parallel to the $x$`x` axis) a distance of $h$`h` units and translated vertically (parallel to the $y$`y` axis) a distance of $k$`k` units. We note that under this type of translation:

The centre will move to the point $\left(h,k\right)$(`h`,`k`).

The orientation of the hyperbola will remain unaltered.

The asymptotes will become the straight lines $x=h$`x`=`h` and $y=k$`y`=`k`.

For example, to sketch the hyperbola $y=\frac{12}{x-3}+7$`y`=12`x`−3+7, first place the centre at $\left(3,7\right)$(3,7). Then draw in the two orthogonal asymptotes (orthogonal means at right angles) given by $x=3$`x`=3 and $y=7$`y`=7. Finally, draw the hyperbola as if it were the basic hyperbola $y=\frac{12}{x}$`y`=12`x` but now centred on the point $\left(3,7\right)$(3,7).

Note that the domain includes all values of $x$`x` not equal to $3$3 and the range includes all values of $y$`y` not equal to $7$7. Here is a graph showing how the basic function $y=\frac{12}{x}$`y`=12`x` is translated to horizontally and vertically to become the transformed function $y=\frac{12}{x-3}+7$`y`=12`x`−3+7.

We can also note that as $x\rightarrow3,y\rightarrow\infty$`x`→3,`y`→∞and as $x\rightarrow\infty,y\rightarrow7$`x`→∞,`y`→7.

Continuing with our example, to find the point where $x=9$`x`=9, we simply substitute $x=9$`x`=9 into $y=\frac{12}{x-3}+7$`y`=12`x`−3+7 so that $y=\frac{12}{9-3}+7$`y`=129−3+7 or when simplified $y=9$`y`=9. Thus the point $\left(9,9\right)$(9,9) lies on the hyperbola.

To find the point where $y=13$`y`=13, set $13=\frac{12}{x-3}+7$13=12`x`−3+7 and solve for $x$`x`, so that:

$13$13 | $=$= | $\frac{12}{x-3}+7$12x−3+7 |

$6$6 | $=$= | $\frac{12}{x-3}$12x−3 |

$6\left(x-3\right)$6(x−3) |
$=$= | $12$12 |

$6x-18$6x−18 |
$=$= | $12$12 |

$6x$6x |
$=$= | $30$30 |

$x$x |
$=$= | $5$5 |

Thus another point on the hyperbola is $\left(5,13\right)$(5,13).

Use the applet to understand how the translation of the basic function works. Try positive and negative values of $a$`a`.

This is a graph of the hyperbola $y=\frac{1}{x}$`y`=1`x`.

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What would be the new equation if the graph of $y=\frac{1}{x}$

`y`=1`x` was shifted upwards by $4$4 units?What would be the new equation if the graph of $y=\frac{1}{x}$

`y`=1`x` was shifted to the right by $7$7 units?

This is a graph of $y=\frac{1}{x}$`y`=1`x`.

How do we shift the graph of $y=\frac{1}{x}$

`y`=1`x` to get the graph of $y=\frac{1}{x}+3$`y`=1`x`+3?Move the graph $3$3 units to the left.

AMove the graph upwards by $3$3 unit(s).

BMove the graph downwards by $3$3 unit(s).

CMove the graph $3$3 units to the right.

DHence plot $y=\frac{1}{x}+3$

`y`=1`x`+3 on the same graph as $y=\frac{1}{x}$`y`=1`x`.

Answer the following.

Consider $y=\frac{-1}{x}$

`y`=−1`x`. What value cannot be substituted for $x$`x`?In which two quadrants does the graph of $y=\frac{-1}{x}$

`y`=−1`x` lie?1

A2

B4

C3

DConsider $y=\frac{-1}{x-4}$

`y`=−1`x`−4. What value cannot be substituted for $x$`x`?In which three quadrants does the graph of $y=\frac{-1}{x-4}$

`y`=−1`x`−4 lie?4

A1

B3

C2

DHow can the graph of $y=\frac{-1}{x}$

`y`=−1`x` be altered to create the graph of $y=\frac{-1}{x-4}$`y`=−1`x`−4?translated $4$4 units right

Areflected about $x$

`x`-axisBsteepened

Ctranslated $4$4 units down

D