0.7 Hyperbolic functions and graphs

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 $\{x:x\in \mathbb{R},x\ne -p\}$ .

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

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

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

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

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:

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

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

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

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 $\{x:x\in \mathbb{R},x\ne -1\}$ 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 $\left\{g\right(x):g(x)\in (-\infty ,2)\cup (2,\infty \left)\right\}$ .

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 $\{x:x\in \mathbb{R},x\ne -1\}$ and the range to be $\left\{g\right(x):g(x)\in (-\infty ,2)\cup (2,\infty \left)\right\}$ . 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).