# 0.7 Hyperbolic functions and graphs

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## Introduction

In Grade 10, you studied graphs of many different forms. In this chapter, you will learn a little more about the graphs of functions.

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

This form of the hyperbolic function is slightly more complex than the form studied in Grade 10.

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

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

You should have found that the sign of $a$ affects whether the graph is located in the first and third quadrants, or the second and fourth quadrants of Cartesian plane.

You should have also found that the value of $p$ affects whether the $x$ -intercept is negative ( $p>0$ ) or positive ( $p<0$ ).

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 is shown as a dashed line.

 $p<0$ $p>0$ $a>0$ $a<0$ $a>0$ $a<0$ $q>0$ $q<0$

## Domain and range

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

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

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

This shows that the function is undefined at $y=q$ . Therefore the range of $f\left(x\right)=\frac{a}{x+p}+q$ is $\left\{f\left(x\right):f\left(x\right)\in R,f\left(x\right)\ne q$ .

For example, the domain of $g\left(x\right)=\frac{2}{x+1}+2$ is $\left\{x:x\in \mathbb{R},x\ne -1\right\}$ because $g\left(x\right)$ is undefined at $x=-1$ .

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

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

## Domain and range

1. Determine the range of $y=\frac{1}{x}+1$ .
2. Given: $f\left(x\right)=\frac{8}{x-8}+4$ . Write down the domain of $f$ .
3. Determine the domain of $y=-\frac{8}{x+1}+3$

## Intercepts

For functions of the form, $y=\frac{a}{x+p}+q$ , the intercepts with the $x$ and $y$ axis are 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:

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

For example, the $y$ -intercept of $g\left(x\right)=\frac{2}{x+1}+2$ is given by setting $x=0$ to get:

$\begin{array}{ccc}\hfill y& =& \frac{2}{x+1}+2\hfill \\ \hfill {y}_{int}& =& \frac{2}{0+1}+2\hfill \\ & =& \frac{2}{1}+2\hfill \\ & =& 2+2\hfill \\ & =& 4\hfill \end{array}$

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

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

For example, the $x$ -intercept of $g\left(x\right)=\frac{2}{x+1}+2$ is given by setting $x=0$ to get:

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

## Intercepts

1. Given: $h\left(x\right)=\frac{1}{x+4}-2$ . Determine the coordinates of the intercepts of $h$ with the x- and y-axes.
2. Determine the x-intercept of the graph of $y=\frac{5}{x}+2$ . Give the reason why there is no y-intercept for this function.

## Asymptotes

There are two asymptotes for functions of the form $y=\frac{a}{x+p}+q$ . They are determined by examining the domain and range.

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

For example, the domain of $g\left(x\right)=\frac{2}{x+1}+2$ is $\left\{x:x\in \mathbb{R},x\ne -1\right\}$ because $g\left(x\right)$ is undefined at $x=-1$ . 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 $x=-1$ and $y=2$ .

## Asymptotes

1. Given: $h\left(x\right)=\frac{1}{x+4}-2$ .Determine the equations of the asymptotes of $h$ .
2. Write down the equation of the vertical asymptote of the graph $y=\frac{1}{x-1}$ .

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

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

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

For example, sketch the graph of $g\left(x\right)=\frac{2}{x+1}+2$ . Mark the intercepts and asymptotes.

We have determined the domain to be $\left\{x:x\in \mathbb{R},x\ne -1\right\}$ 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 $x=-1$ and $y=2$ .

The $y$ -intercept is ${y}_{int}=4$ and the $x$ -intercept is ${x}_{int}=-2$ .

## Graphs

1. Draw the graph of $y=\frac{1}{x}+2$ . Indicate the horizontal asymptote.
2. Given: $h\left(x\right)=\frac{1}{x+4}-2$ . Sketch the graph of $h$ showing clearly the asymptotes and ALL intercepts with the axes.
3. Draw the graph of $y=\frac{1}{x}$ and $y=-\frac{8}{x+1}+3$ on the same system of axes.
4. Draw the graph of $y=\frac{5}{x-2,5}+2$ . Explain your method.
5. Draw the graph of the function defined by $y=\frac{8}{x-8}+4$ . Indicate the asymptotes and intercepts with the axes.

## End of chapter exercises

1. Plot the graph of the hyperbola defined by $y=\frac{2}{x}$ for $-4\le x\le 4$ . Suppose the hyperbola is shifted 3 units to the right and 1 unit down. What is the new equation then ?
2. Based on the graph of $y=\frac{1}{x}$ , determine the equation of the graph with asymptotes $y=2$ and $x=1$ and passing through the point (2; 3).

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