NZ Level 7 (NZC) Level 2 (NCEA) Transformations of Hyperbolas
Lesson

## Principles of translation

The hyperbola given by $y=\frac{a}{x-h}+k$y=axh+k can be thought of as the basic rectangular  hyperbola $y=\frac{a}{x}$y=ax 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=12x3+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=12x 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=12x is translated to horizontally and vertically to become the transformed function $y=\frac{12}{x-3}+7$y=12x3+7

We can also note that as $x\rightarrow3,y\rightarrow\infty$x3,yand as $x\rightarrow\infty,y\rightarrow7$x,y7

### Finding points

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=12x3+7 so that $y=\frac{12}{9-3}+7$y=1293+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=12x3+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).

### Applet

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

#### Worked Examples

##### QUESTION 1

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

1. What would be the new equation if the graph was shifted upwards by $4$4 units?

2. What would be the new equation if the graph was shifted to the right by $7$7 units?

##### QUESTION 2

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

1. How do we shift the graph of $y=\frac{1}{x}$y=1x to get the graph of $y=\frac{1}{x}+3$y=1x+3?

Move the graph $3$3 units to the left.

A

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

B

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

C

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

D

Move the graph $3$3 units to the left.

A

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

B

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

C

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

D
2. Hence plot $y=\frac{1}{x}+3$y=1x+3 on the same graph as $y=\frac{1}{x}$y=1x.

##### QUESTION 3

1. Consider $y=\frac{-1}{x}$y=1x. What value cannot be substituted for $x$x?

2. In which quadrants does $y=\frac{-1}{x}$y=1x lie?

1

A

2

B

4

C

3

D

1

A

2

B

4

C

3

D
3. Consider $y=\frac{-1}{x-4}$y=1x4. What value cannot be substituted for $x$x?

4. In which quadrants does $y=\frac{-1}{x-4}$y=1x4 lie?

4

A

1

B

3

C

2

D

4

A

1

B

3

C

2

D
5. How can the graph of $y=\frac{-1}{x}$y=1x be altered to create the graph of $y=\frac{-1}{x-4}$y=1x4?

translated $4$4 units right

A

reflected about $x$x-axis

B

steepened

C

translated $4$4 units down

D

translated $4$4 units right

A

reflected about $x$x-axis

B

steepened

C

translated $4$4 units down

D

### Outcomes

#### M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

#### 91257

Apply graphical methods in solving problems