# 1.3 Hyperbolic functions

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## Functions of the form $y=\frac{a}{x}+q$

Functions of the form $y=\frac{a}{x}+q$ are known as hyperbolic functions. The general form of the graph of this function is shown in [link] . General shape and position of the graph of a function of the form f ( x ) = a x + q .

## Investigation : functions of the form $y=\frac{a}{x}+q$

1. On the same set of axes, plot the following graphs:
1. $a\left(x\right)=\frac{-2}{x}+1$
2. $b\left(x\right)=\frac{-1}{x}+1$
3. $c\left(x\right)=\frac{0}{x}+1$
4. $d\left(x\right)=\frac{+1}{x}+1$
5. $e\left(x\right)=\frac{+2}{x}+1$
Use your results to deduce the effect of $a$ .
2. On the same set of axes, plot the following graphs:
1. $f\left(x\right)=\frac{1}{x}-2$
2. $g\left(x\right)=\frac{1}{x}-1$
3. $h\left(x\right)=\frac{1}{x}+0$
4. $j\left(x\right)=\frac{1}{x}+1$
5. $k\left(x\right)=\frac{1}{x}+2$
Use your results to deduce the effect of $q$ .

You should have found that the value of $a$ affects whether the graph is located in the first and third quadrants of Cartesian plane.

You should have also found that the value of $q$ affects whether the graph lies above the $x$ -axis ( $q>0$ ) or below the $x$ -axis ( $q<0$ ).

These different properties are summarised in [link] . The axes of symmetry for each graph are shown as a dashed line.

 $a>0$ $a<0$ $q>0$  $q<0$  ## Domain and range

For $y=\frac{a}{x}+q$ , the function is undefined for $x=0$ . The domain is therefore $\left\{x:x\in \mathbb{R},x\ne 0\right\}$ .

We see that $y=\frac{a}{x}+q$ can be re-written as:

$\begin{array}{ccc}\hfill y& =& \frac{a}{x}+q\hfill \\ \hfill y-q& =& \frac{a}{x}\hfill \\ \hfill \mathrm{If}\phantom{\rule{3pt}{0ex}}x\ne 0\phantom{\rule{3pt}{0ex}}\mathrm{then}:\phantom{\rule{1.em}{0ex}}\left(y-q\right)x& =& a\hfill \\ \hfill x& =& \frac{a}{y-q}\hfill \end{array}$

This shows that the function is undefined at $y=q$ . Therefore the range of $f\left(x\right)=\frac{a}{x}+q$ is $\left\{f\left(x\right):f\left(x\right)\in \left(-\infty ;q\right)\cup \left(q;\infty \right)\right\}$ .

For example, the domain of $g\left(x\right)=\frac{2}{x}+2$ is $\left\{x:x\in \mathbb{R},x\ne 0\right\}$ because $g\left(x\right)$ is undefined at $x=0$ .

$\begin{array}{ccc}\hfill y& =& \frac{2}{x}+2\hfill \\ \hfill \left(y-2\right)& =& \frac{2}{x}\hfill \\ \hfill \mathrm{If}\phantom{\rule{4pt}{0ex}}x\ne 0\phantom{\rule{4pt}{0ex}}\mathrm{then}:\phantom{\rule{1.em}{0ex}}x\left(y-2\right)& =& 2\hfill \\ \hfill x& =& \frac{2}{y-2}\hfill \end{array}$

We see that $g\left(x\right)$ is undefined at $y=2$ . Therefore the range is $\left\{g\left(x\right):g\left(x\right)\in \left(-\infty ;2\right)\cup \left(2;\infty \right)\right\}$ .

## Intercepts

For functions of the form, $y=\frac{a}{x}+q$ , the intercepts with the $x$ and $y$ axis is calculated by setting $x=0$ for the $y$ -intercept and by setting $y=0$ for the $x$ -intercept.

The $y$ -intercept is calculated as follows:

$\begin{array}{ccc}\hfill y& =& \frac{a}{x}+q\hfill \\ \hfill {y}_{int}& =& \frac{a}{0}+q\hfill \end{array}$

which is undefined because we are dividing by 0. Therefore there is no $y$ -intercept.

For example, the $y$ -intercept of $g\left(x\right)=\frac{2}{x}+2$ is given by setting $x=0$ to get:

$\begin{array}{ccc}\hfill y& =& \frac{2}{x}+2\hfill \\ \hfill {y}_{int}& =& \frac{2}{0}+2\hfill \end{array}$

which is undefined.

The $x$ -intercepts are calculated by setting $y=0$ as follows:

$\begin{array}{ccc}\hfill y& =& \frac{a}{x}+q\hfill \\ \hfill 0& =& \frac{a}{{x}_{int}}+q\hfill \\ \hfill \frac{a}{{x}_{int}}& =& -q\hfill \\ \hfill a& =& -q\left({x}_{int}\right)\hfill \\ \hfill {x}_{int}& =& \frac{a}{-q}\hfill \end{array}$

For example, the $x$ -intercept of $g\left(x\right)=\frac{2}{x}+2$ is given by setting $x=0$ to get:

$\begin{array}{ccc}\hfill y& =& \frac{2}{x}+2\hfill \\ \hfill 0& =& \frac{2}{{x}_{int}}+2\hfill \\ \hfill -2& =& \frac{2}{{x}_{int}}\hfill \\ \hfill -2\left({x}_{int}\right)& =& 2\hfill \\ \hfill {x}_{int}& =& \frac{2}{-2}\hfill \\ \hfill {x}_{int}& =& -1\hfill \end{array}$

## Asymptotes

There are two asymptotes for functions of the form $y=\frac{a}{x}+q$ . Just a reminder, an asymptote is a straight or curved line, which the graph of a function will approach, but never touch. They are determined by examining the domain and range.

We saw that the function was undefined at $x=0$ and for $y=q$ . Therefore the asymptotes are $x=0$ and $y=q$ .

For example, the domain of $g\left(x\right)=\frac{2}{x}+2$ is $\left\{x:x\in \mathbb{R},x\ne 0\right\}$ because $g\left(x\right)$ is undefined at $x=0$ . We also see that $g\left(x\right)$ is undefined at $y=2$ . Therefore the range is $\left\{g\left(x\right):g\left(x\right)\in \left(-\infty ;2\right)\cup \left(2;\infty \right)\right\}$ .

From this we deduce that the asymptotes are at $x=0$ and $y=2$ .

## Sketching graphs of the form $f\left(x\right)=\frac{a}{x}+q$

In order to sketch graphs of functions of the form, $f\left(x\right)=\frac{a}{x}+q$ , we need to determine four characteristics:

1. domain and range
2. asymptotes
3. $y$ -intercept
4. $x$ -intercept

For example, sketch the graph of $g\left(x\right)=\frac{2}{x}+2$ . Mark the intercepts and asymptotes.

We have determined the domain to be $\left\{x:x\in \mathbb{R},x\ne 0\right\}$ and the range to be $\left\{g\left(x\right):g\left(x\right)\in \left(-\infty ;2\right)\cup \left(2;\infty \right)\right\}$ . Therefore the asymptotes are at $x=0$ and $y=2$ .

There is no $y$ -intercept and the $x$ -intercept is ${x}_{int}=-1$ . Graph of g ( x ) = 2 x + 2 .

Draw the graph of $y=\frac{-4}{x}+7$ .

1. The domain is: $\left\{x:x\in \mathbb{R},x\ne 0\right\}$ and the range is: $\left\{f\left(x\right):f\left(x\right)\in \left(-\infty ;7\right)\cup \left(7;\infty \right)\right\}$ .
2. We look at the domain and range to determine where the asymptotes lie. From the domain we see that the function is undefined when $x=0$ , so there is one asymptote at $x=0$ . The other asymptote is found from the range. The function is undefined at $y=q$ and so the second asymptote is at $y=7$
3. There is no y-intercept for graphs of this form.
4. The x-intercept occurs when $y=0$ . Calculating the x-intercept gives:
$\begin{array}{ccc}\hfill y& =& \frac{-4}{x}+7\hfill \\ \hfill 0& =& \frac{-4}{x}+7\hfill \\ \hfill -7& =& \frac{-4}{x}\hfill \\ \hfill {x}_{\mathrm{int}}& =& \frac{4}{7}\hfill \end{array}$
So there is one x-intercept at $\left(\frac{4}{7},0\right)$ .
5. Putting all this together gives us the following graph:

## Graphs

1. Using graph (grid) paper, draw the graph of $xy=-6$ .
1. Does the point (-2; 3) lie on the graph ? Give a reason for your answer.
2. Why is the point (-2; -3) not on the graph ?
3. If the $x$ -value of a point on the drawn graph is 0,25, what is the corresponding $y$ -value ?
4. What happens to the $y$ -values as the $x$ -values become very large ?
5. With the line $y=-x$ as line of symmetry, what is the point symmetrical to (-2; 3) ?
2. Draw the graph of $xy=8$ .
1. How would the graph $y=\frac{8}{3}+3$ compare with that of $xy=8$ ? Explain your answer fully.
2. Draw the graph of $y=\frac{8}{3}+3$ on the same set of axes.

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