5.3 Hyperbolic functions

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] .

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 $\{x:x\in \mathbb{R},x\ne 0\}$ .

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

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

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

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

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:

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:

which is undefined.

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

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

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 $\{x:x\in \mathbb{R},x\ne 0\}$ 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 $\left\{g\right(x):g(x)\in (-\infty ;2)\cup (2;\infty \left)\right\}$ .

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

There is no $y$ -intercept and the $x$ -intercept is $x_{int}=-1$ .

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.