Hyperbolic functions
Functions of the form $y=\frac{1}{x}$
Functions of the general form $y=\frac{a}{x}+q$ are called hyperbolic functions.
Example 1: Plotting a hyperbolic function
Question
Complete the following table for $h\left(x\right)=\frac{1}{x}$ and plot the points on a system of axes.


Join the points with smooth curves.

What happens if $x=0$?

Explain why the graph consists of two separate curves.

What happens to $h\left(x\right)$ as the value of $x$ becomes very small or very large?

The domain of $h\left(x\right)$ is $\{x:x\in \mathbb{R},x\ne 0\}$. Determine the range.

About which two lines is the graph symmetrical?
Answer
Substitute values into the equation

Plot the points and join with two smooth curves
From the table we get the following points: $(3;\frac{1}{3})$, $(2;\frac{1}{2})$, $(1;1)$, $(\frac{1}{2};2)$, $(\frac{1}{4};4)$, $(\frac{1}{4};4)$, $(\frac{1}{2};2)$, $(1;1)$, $(2;\frac{1}{2})$, $(3;\frac{1}{3})$.
For $x=0$ the function $h$ is undefined. This is called a discontinuity at $x=0$.
$y=h\left(x\right)=\frac{1}{x}$ therefore we can write that $x\times y=1$. Since the product of two positive numbers and the product of two negative numbers can be equal to 1, the graph lies in the first and third quadrants.
Determine the asymptotes
As the value of $x$ gets larger, the value of $h\left(x\right)$ gets closer to, but does not equal 0. This is a horizontal asymptote, the line $y=0$. The same happens in the third quadrant; as $x$ gets smaller, $h\left(x\right)$ also approaches the negative $x$axis asymptotically.
We also notice that there is a vertical asymptote, the line $x=0$; as $x$ gets closer to 0, $h\left(x\right)$ approaches the $y$axis asymptotically.
Determine the range
Domain: $\{x:x\in \mathbb{R},x\ne 0\}$
From the graph, we see that $y$ is defined for all values except 0.
Range: $\{y:y\in \mathbb{R},y\ne 0\}$
Determine the lines of symmetry
The graph of $h\left(x\right)$ has two axes of symmetry: the lines $y=x$ and $y=x$. About these two lines, one half of the hyperbola is a mirror image of the other half.
Functions of the form $y=\frac{a}{x}+q$
Investigation 1: The effects of $a$ and $q$ on a hyperbola
On the same set of axes, plot the following graphs:

${y}_{1}={\displaystyle \frac{1}{x}}2$

${y}_{2}={\displaystyle \frac{1}{x}}1$

${y}_{3}={\displaystyle \frac{1}{x}}$

${y}_{4}={\displaystyle \frac{1}{x}}+1$

${y}_{5}={\displaystyle \frac{1}{x}}+2$
Use your results to deduce the effect of $q$.
On the same set of axes, plot the following graphs:

${y}_{6}={\displaystyle \frac{2}{x}}$

${y}_{7}={\displaystyle \frac{1}{x}}$

${y}_{8}={\displaystyle \frac{1}{x}}$

${y}_{9}={\displaystyle \frac{2}{x}}$
Use your results to deduce the effect of $a$.

The effect of $q$
The effect of $q$ is called a vertical shift because all points are moved the same distance in the same direction (it slides the entire graph up or down).

For $q>0$, the graph of $f\left(x\right)$ is shifted vertically upwards by $q$ units.

For $q<0$, the graph of $f\left(x\right)$ is shifted vertically downwards by $q$ units.
The horizontal asymptote is the line $y=q$ and the vertical asymptote is always the $y$axis, the line $x=0$.
The effect of $a$
The sign of $a$ determines the shape of the graph.

If $a>0$, the graph of $f\left(x\right)$ lies in the first and third quadrants.
For $a>1$, the graph of $f\left(x\right)$ will be further away from the axes than $y=\frac{1}{x}$.
For $0<a<1$, as a tends to 0, the graph moves closer to the axes than $y=\frac{1}{x}$.

If $a<0$, the graph of $f\left(x\right)$ lies in the second and fourth quadrants.
For $a<1$, the graph of $f\left(x\right)$ will be further away from the axes than $y=\frac{1}{x}$.
For $1<a<0$, as a tends to 0, the graph moves closer to the axes than $y=\frac{1}{x}$.
Discovering the characteristics
The standard form of a hyperbola is the equation $y=\frac{a}{x}+q$.
Domain and range
For $y=\frac{a}{x}+q$, the function is undefined for $x=0$. The domain is therefore $\{x:x\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}x\ne 0\}$.
We see that $y=\frac{a}{x}+q$ can be rewritten as:
This shows that the function is undefined only at $y=q$.
Therefore the range is $\left\{f\right(x):f(x)\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}f(x)\ne q\}$.
Example 2: Domain and range of a hyperbola
Question
If $g\left(x\right)=\frac{2}{x}+2$, determine the domain and range of the function.
Answer
Determine the domain
The domain is $\{x:x\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}x\ne 0\}$ because $g\left(x\right)$ is undefined only at $x=0$.
Determine the range
We see that $g\left(x\right)$ is undefined only at $y=2$. Therefore the range is $\left\{g\right(x):g(x)\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}g(x)\ne 2\}$.
Intercepts
The $y$intercept:
Every point on the $y$axis has an $x$coordinate of 0, therefore to calculate the $y$intercept, let $x=0$.
For example, the $y$intercept of $g\left(x\right)=\frac{2}{x}+2$ is given by setting $x=0$:
which is undefined, therefore there is no $y$intercept.
The $x$intercept:
Every point on the $x$axis has a $y$coordinate of 0, therefore to calculate the $x$intercept, let $y=0$.
For example, the $x$intercept of $g\left(x\right)=\frac{2}{x}+2$ is given by setting $y=0$:
This gives the point $(1;0)$.
Asymptotes
There are two asymptotes for functions of the form $y=\frac{a}{x}+q$.
The horizontal asymptote is the line $y=q$ and the vertical asymptote is always the $y$axis, the line $x=0$.
Axes of symmetry
There are two lines about which a hyperbola is symmetrical: $y=x+q$ and $y=x+q$.
Sketching graphs of the form $y=\frac{a}{x}+q$
In order to sketch graphs of functions of the form, $y=f\left(x\right)=\frac{a}{x}+q$, we need to determine four characteristics:

sign of $a$

$y$intercept

$x$intercept

asymptotes
Example 3: Sketching a hyperbola
Question
Sketch the graph of $g\left(x\right)=\frac{2}{x}+2$. Mark the intercepts and the asymptotes.
Answer
Examine the standard form of the equation
We notice that $a>0$ therefore the graph of $g\left(x\right)$ lies in the first and third quadrant.
Calculate the intercepts
For the $y$intercept, let $x=0$:
This is undefined, therefore there is no $y$intercept.
For the $x$intercept, let $y=0$:
This gives the point $(1;0)$.
Determine the asymptotes
The horizontal asymptote is the line $y=2$. The vertical asymptote is the line $x=0$.
Sketch the graph
Domain: $\{x:x\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}x\ne 0\}$.
Range: $\{y:y\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}y\ne 2\}$.
Example 4: Sketching a hyperbola
Question
Sketch the graph of $y=\frac{4}{x}+7$.
Answer
Examine the standard form of the equation
We see that $a<0$ therefore the graph lies in the second and fourth quadrants.
Calculate the intercepts
For the $y$intercept, let $x=0$:
This is undefined, therefore there is no $y$intercept.
For the $x$intercept, let $y=0$:
This gives the point $\left(\frac{4}{7};0\right)$.
Determine the asymptotes
The horizontal asymptote is the line $y=7$. The vertical asymptote is the line $x=0$.
Sketch the graph
Domain: $\{x:x\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}x\ne 0\}$
Range: $\{y:y\in \mathbb{R},\phantom{\rule{3.33333pt}{0ex}}y\ne 7\}$
Axis of symmetry: $y=x+7$ and $y=x+7$
Exercise 1:
Draw the graph of $xy=6$.

Does the point $(2;3)$ lie on the graph? Give a reason for your answer.

If the $x$value of a point on the drawn graph is 0,25 what is the corresponding $y$value?

What happens to the $y$values as the $x$values become very large?

Give the equations of the asymptotes.

With the line $y=x$ as line of symmetry, what is the point symmetrical to $(2;3)$?
Draw the graph of $h\left(x\right)=\frac{8}{x}$.

How would the graph $g\left(x\right)=\frac{8}{x}+3$ compare with that of $h\left(x\right)=\frac{8}{x}$? Explain your answer fully.

Draw the graph of $y=\frac{8}{x}+3$ on the same set of axes, showing asymptotes, axes of symmetry and the coordinates of one point on the graph.
Todo.
Todo.