Hyperbolic Functions Revision Notes for Grade 11 NSC Matric Mathematics

Hyperbolic Functions

Introduction to hyperbolic functions

Hyperbolic functions are functions that have the general form $y = \frac{a}{x} + q$ , where $a$ and $q$ are constants. These functions create curved graphs that have two separate branches and approach certain lines (called asymptotes) but never actually touch them.

The most basic hyperbolic function is $y = \frac{1}{x}$ , which you can think of as the "parent function" for all hyperbolic functions.

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About Asymptotes: Asymptotes are invisible boundary lines that the graph approaches infinitely close to, but never actually touches or crosses. Think of them as "forbidden zones" for the function's graph.

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

Effects of parameter $q$ (vertical shifts)

The parameter $q$ controls the vertical shift of the hyperbola:

When $q > 0$ , the graph shifts upwards by $q$ units
When $q < 0$ , the graph shifts downwards by $|q|$ units
The horizontal asymptote becomes the line $y = q$
The vertical asymptote remains the line $x = 0$ (the y-axis)

Effects of parameter $a$ (shape and quadrants)

The parameter $a$ affects both the shape and position of the hyperbola:

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Understanding Parameter $a$

When $a > 0$ :

The hyperbola lies in the first and third quadrants
As $|a|$ increases, the branches move further away from both axes
As $a$ approaches 0, the branches move closer to the axes

When $a < 0$ :

The hyperbola lies in the second and fourth quadrants
As $|a|$ increases, the branches move further away from both axes
As $a$ approaches 0, the branches move closer to the axes

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

The extended form includes parameter $p$ , which creates horizontal shifts.

Effects of parameter $p$ (horizontal shifts)

The parameter $p$ causes a horizontal shift of the entire graph:

When $p > 0$ , the graph shifts left by $p$ units
When $p < 0$ , the graph shifts right by $|p|$ units
The vertical asymptote becomes the line $x = -p$
The horizontal asymptote remains $y = q$

chatImportant

Critical Point About Horizontal Shifts: Notice that when $p > 0$ , the graph shifts LEFT, not right! This is because the function is $\frac{a}{x+p}$ , so to make the denominator zero, we need $x = -p$ . This is a common source of confusion.

Combined effects of all parameters

The diagram above shows how different combinations of $a$ , $p$ , and $q$ values affect the hyperbola's position and orientation.

Key characteristics of hyperbolic functions

Domain and range

For a function $f(x) = \frac{a}{x+p} + q$ :

Domain: $\{x : x \in \mathbb{R}, x \neq -p\}$

The domain excludes $x = -p$ because this makes the denominator zero, causing the function to be undefined

Range: $\{y : y \in \mathbb{R}, y \neq q\}$

The range excludes $y = q$ because the horizontal asymptote can never be reached

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Worked Example 1: Domain and Range

Question: Determine the domain and range for $g(x) = \frac{2}{x+1} + 2$

Solution:

Step 1: Determine the domain

The domain is $\{x : x \in \mathbb{R}, x \neq -1\}$ since $g(x)$ is undefined for $x = -1$ .

Step 2: Determine the range

Let $g(x) = y$ :

$y = \frac{2}{x+1} + 2$
$y - 2 = \frac{2}{x+1}$
$(y-2)(x+1) = 2$
$x + 1 = \frac{2}{y-2}$

Therefore the range is $\{g(x) : g(x) \in \mathbb{R}, g(x) \neq 2\}$ .

Intercepts

Understanding how to find intercepts is essential for graphing hyperbolic functions.

Finding the y-intercept:

Set $x = 0$ and solve for $y$ .

For example, if $g(x) = \frac{2}{x+1} + 2$ :

$g(0) = \frac{2}{0+1} + 2 = \frac{2}{1} + 2 = 4$
This gives the y-intercept $(0; 4)$ .

Finding the x-intercept:

Set $y = 0$ and solve for $x$ .

Using the same example:

$0 = \frac{2}{x+1} + 2$
$-2 = \frac{2}{x+1}$
$-2(x+1) = 2$
$-2x - 2 = 2$
$-2x = 4$
$x = -2$

This gives the x-intercept $(-2; 0)$ .

Asymptotes

Asymptotes are lines that the graph approaches but never touches.

For $y = \frac{a}{x+p} + q$ :

Horizontal asymptote: $y = q$
Vertical asymptote: $x = -p$

These asymptotes represent the values excluded from the range and domain respectively.

Axes of symmetry

Hyperbolic functions have two axes of symmetry that intersect at the point $(-p; q)$ .

For the standard hyperbola $y = \frac{1}{x}$ , the axes of symmetry are:

$y = x$
$y = -x$

For the shifted hyperbola $y = \frac{a}{x+p} + q$ , the axes of symmetry intersect at $(-p; q)$ .

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Worked Example 2: Axes of Symmetry

Question: Determine the axes of symmetry for $y = \frac{2}{x+1} - 2$

Solution:

Step 1: Determine the intersection point $(-p; q)$

From the equation we see that $p = 1$ and $q = -2$ . So the axes of symmetry will intersect at $(-1; -2)$ .

Step 2: Define two straight line equations

$y_1 = x + c_1$ and $y_2 = -x + c_2$

Step 3: Solve for $c_1$ and $c_2$

Using $(-1; -2)$ to solve for $c_1$ :

$y_1 = x + c_1$
$-2 = -1 + c_1$
$c_1 = -1$

Using $(-1; -2)$ to solve for $c_2$ :

$y_2 = -x + c_2$
$-2 = -(-1) + c_2$
$-2 = 1 + c_2$
$c_2 = -3$

Step 4: Write the final answer

The axes of symmetry are: $y_1 = x - 1$ and $y_2 = -x - 3$

Sketching hyperbolic graphs

To sketch a hyperbola of the form $f(x) = \frac{a}{x+p} + q$ , follow these systematic steps:

Identify the quadrants (based on the sign of $a$ )
Find the asymptotes ( $x = -p$ and $y = q$ )
Calculate the y-intercept (let $x = 0$ )
Calculate the x-intercept (let $y = 0$ )
Determine domain and range

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Worked Example 3: Sketching a Hyperbola

Question: Sketch the graph of $y = \frac{2}{x+1} + 2$ . Determine the intercepts, asymptotes and axes of symmetry.

Solution:

Step 1: Examine the equation

We notice that $a > 0$ , therefore the graph will lie in the first and third quadrants.

Step 2: Determine the asymptotes

From the equation we know that $p = 1$ and $q = 2$ .

Therefore the horizontal asymptote is $y = 2$ and the vertical asymptote is $x = -1$ .

Step 3: Determine the y-intercept

The y-intercept is obtained by letting $x = 0$ :

$y = \frac{2}{0+1} + 2 = 4$
This gives the point $(0; 4)$ .

Step 4: Determine the x-intercept

The x-intercept is obtained by letting $y = 0$ :

$0 = \frac{2}{x+1} + 2$
$-2 = \frac{2}{x+1}$
$-2(x+1) = 2$
$x = -2$

This gives the point $(-2; 0)$ .

Step 5: Determine the axes of symmetry

Using $(-1; 2)$ to solve for the axes of symmetry:

$y_1 = x + 3$ and $y_2 = -x + 1$

Step 6: Plot the points and sketch the graph

Step 7: State the domain and range

Domain: $\{x : x \in \mathbb{R}, x \neq -1\}$

Range: $\{y : y \in \mathbb{R}, y \neq 2\}$

Using transformations to sketch graphs

An alternative approach to sketching hyperbolic functions is to use transformations of the parent function $y = \frac{1}{x}$ .

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Worked Example 4: Using Transformations

Question: Use horizontal and vertical shifts to sketch the graph of $f(x) = \frac{1}{x-2} + 3$

Solution:

Step 1: Examine the equation

We notice that $a > 0$ , therefore the graph will lie in the first and third quadrants.

Step 2: Sketch the standard hyperbola $y = \frac{1}{x}$

Start with the standard hyperbola. The vertical asymptote is $x = 0$ and the horizontal asymptote is $y = 0$ .

Step 3: Determine the vertical shift

From the equation we see that $q = 3$ , which means the graph shifts 3 units up. The horizontal asymptote shifts up to $y = 3$ .

Step 4: Determine the horizontal shift

From the equation we see that $p = -2$ , which means the graph shifts 2 units to the right. The vertical asymptote shifts right to $x = 2$ .

Domain: $\{x : x \in \mathbb{R}, x \neq 2\}$

Range: $\{y : y \in \mathbb{R}, y \neq 3\}$

Finding equations from graphs

Sometimes you need to work backwards from a graph to determine the function's equation. This requires careful observation of the graph's key features.

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Worked Example 5: Finding Equation from Graph

Question: Use the graph below to determine the values of $a$ , $p$ and $q$ for $y = \frac{a}{x+p} + q$

Solution:

Step 1: Examine the graph and deduce the sign of $a$

We notice that the graph lies in the second and fourth quadrants, therefore $a < 0$ .

Step 2: Determine the asymptotes

From the graph we see that the vertical asymptote is $x = -1$ , therefore $p = 1$ . The horizontal asymptote is $y = 3$ , and therefore $q = 3$ .

$y = \frac{a}{x+1} + 3$

Step 3: Determine the value of $a$

To determine the value of $a$ we substitute a point on the graph, namely $(0; 0)$ :

$y = \frac{a}{x+1} + 3$
$0 = \frac{a}{0+1} + 3$
$∴ -3 = a$

Step 4: Write the final answer

$y = \frac{-3}{x+1} + 3$

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Key Points to Remember:

Hyperbolic functions have the form $y = \frac{a}{x+p} + q$ where $a$ , $p$ , and $q$ are constants
Parameter effects: $a$ determines quadrants and shape, $p$ creates horizontal shifts, $q$ creates vertical shifts
Domain excludes values where the denominator equals zero: $\{x : x \in \mathbb{R}, x \neq -p\}$
Asymptotes are lines the graph approaches: horizontal asymptote $y = q$ , vertical asymptote $x = -p$
Intercepts are found by setting $x = 0$ (y-intercept) or $y = 0$ (x-intercept) and solving

Hyperbolic Functions (Grade 11 NSC Matric Mathematics): Revision Notes