5.3 Hyperbolic functions

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

Functions of the general form $y = \frac{a}{x} + q$ are called hyperbolic functions, where $a$ and $q$ are constants.

The effects of $a$ and $q$ on $f(x) = \frac{a}{x} + q$:

• The effect of $q$ on vertical shift

  • For $q>0$, $f(x)$ is shifted vertically upwards by $q$ units.

  • For $q<0$, $f(x)$ is shifted vertically downwards by $q$ units.

  • The horizontal asymptote is the line $y = q$.

  • The vertical asymptote is the $y$-axis, the line $x = 0$.

• The effect of $a$ on shape and quadrants

  • For $a>0$, $f(x)$ lies in the first and third quadrants.

  • For $a > 1$, $f(x)$ will be further away from both axes than $y = \frac{1}{x}$.

  • For $0<a<1$, as $a$ tends to $\text{0}$, $f(x)$ moves closer to the axes than $y = \frac{1}{x}$.

  • For $a<0$, $f(x)$ lies in the second and fourth quadrants.

  • For $a < -1$, $f(x)$ will be further away from both axes than $y = - \frac{1}{x}$.

  • For $-1<a<0$, as $a$ tends to $\text{0}$, $f(x)$ moves closer to the axes than $y = -\frac{1}{x}$.

	$a<0$	$a>0$
$q>0$
$q=0$
$q<0$

Discovering the characteristics

For functions of the general form: $f(x) = y = \frac{a}{x+p} + q$:

Domain and range

The domain is $\{ x: x \in \mathbb{R}, x \ne -p \}$. If $x = -p$, the dominator is equal to zero and the function is undefined.

We see that $$y = \frac{a}{x+p} + q$$ can be re-written as: $$y-q = \frac{a}{x+p}$$ If $x \ne -p$ then: $$\begin{align*} \left(y-q\right)\left(x+p\right) &= a \\ x + p &= \frac{a}{y-q} \end{align*}$$ The range is therefore $\{ y: y \in \mathbb{R}, y \ne q \}$.

These restrictions on the domain and range determine the vertical asymptote $x=-p$ and the horizontal asymptote $y=q$.

Intercepts

The $y$-intercept:

To calculate the $y$-intercept we let $x=0$. For example, the $y$-intercept of $g(x) = \frac{2}{x + 1} + 2$ is determined by setting $x=0$: $$\begin{align*} g(x) &= \frac{2}{x + 1} + 2 \\ g(0) &= \frac{2}{0 + 1} + 2\\ &= 2 + 2 \\ &= 4 \end{align*}$$ This gives the point $(0;4)$.

The $x$-intercept:

To calculate the $x$-intercept we let $y=0$. For example, the $x$-intercept of $g(x) = \frac{2}{x + 1} + 2$ is determined by setting $y=0$: $$\begin{align*} g(x) &= \frac{2}{x + 1} + 2 \\ 0 &= \frac{2}{x + 1} + 2 \\ -2 &= \frac{2}{x + 1} \\ -2(x + 1) &= 2 \\ -2x - 2 &= 2 \\ -2x &= 4 \\ x &= -2 \end{align*}$$ This gives the point $(-2;0)$.

Asymptotes

There are two asymptotes for functions of the form $y=\frac{a}{x+p}+q$. The asymptotes indicate the values of $x$ for which the function does not exist. In other words, the values that are excluded from the domain and the range. The horizontal asymptote is the line $y=q$ and the vertical asymptote is the line $x=-p$.

Axes of symmetry

There are two lines about which a hyperbola is symmetrical.

For the standard hyperbola $y =\frac{1}{x}$, we see that if we replace $x \Rightarrow y$ and $y \Rightarrow x$, we get $y =\frac{1}{x}$. Similarly, if we replace $x \Rightarrow -y$ and $y \Rightarrow -x$, the function remains the same. Therefore the function is symmetrical about the lines $y = x$ and $y = -x$.

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

To determine the axes of symmetry we define the two straight lines $y_1 = m_1x + c_1$ and $y_2 = m_2x + c_2$. For the standard and shifted hyperbolic function, the gradient of one of the lines of symmetry is $\text{1}$ and the gradient of the other line of symmetry is $-\text{1}$. The axes of symmetry are perpendicular to each other and the product of their gradients equals $-\text{1}$. Therefore we let $y_1 = x + c_1$ and $y_2 = -x + c_2$. We then substitute $(-p;q)$, the point of intersection of the axes of symmetry, into both equations to determine the values of $c_1$ and $c_2$.

Worked example 10: Axes of symmetry

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

Determine the point of intersection $(-p;q)$

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

Define two straight line equations

$$\begin{align*} y_1 &= x + c_1 \\ y_2 &= -x + c_2 \end{align*}$$

Solve for $c_1$ and $c_2$

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

$$\begin{align*} y_1 &= x + c_1 \\ -2 &= -1 + c_1 \\ -1 &= c_1 \end{align*}$$

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

$$\begin{align*} y_2 &= -x + c_2 \\ -2 &= -(-1) + c_2 \\ -3 &= c_2 \end{align*}$$

Write the final answer

The axes of symmetry for $y = \frac{2}{x + 1} - 2$ are the lines $$\begin{align*} y_1 &= x - 1 \\ y_2 &= -x - 3 \end{align*}$$

Axes of symmetry

Textbook Exercise 5.13

Complete the following for \(f(x)\) and \(g(x)\):

• Sketch the graph.
• Determine \((-p;q)\).
• Find the axes of symmetry.

Compare \(f(x)\) and \(g(x)\) and also their axes of symmetry. What do you notice?

\(f(x) = \frac{2}{x}\)

\(g(x) = \frac{2}{x} + 1\)

\(f(x) = -\frac{3}{x}\)

\(g(x) = -\frac{3}{x + 1}\)

\(f(x) = \frac{5}{x}\)

\(g(x) = \frac{5}{x - 1} - 1\)

A hyperbola of the form \(k(x) = \frac{a}{x +p} + q\) passes through the point \((4;3)\). If the axes of symmetry intersect at \((-1;2)\), determine the equation of \(k(x)\).

Try the textbook questions

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

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

• quadrants

• asymptotes

• $y$-intercept

• $x$-intercept

• domain and range

Worked example 11: Sketching a hyperbola

Sketch the graph of $y = \frac{2}{x + 1} + 2$. Determine the intercepts, asymptotes and axes of symmetry. State the domain and range of the function.

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

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

Determine the asymptotes

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

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

Determine the $y$-intercept

The $y$-intercept is obtained by letting $x = 0$: $$\begin{align*} y &= \frac{2}{0 + 1} + 2 \\ &= 4 \end{align*}$$ This gives the point $(0;4)$.

Determine the $x$-intercept

The $x$-intercept is obtained by letting $y = 0$: $$\begin{align*} 0 &= \frac{2}{x + 1} + 2\\ -2 &= \frac{2}{x + 1} \\ -2(x +1) &= 2 \\ -2x -2 &= 2 \\ -2x &= 4 \\ x &= -2 \end{align*}$$ This gives the point $(-2;0)$.

Determine the axes of symmetry

Using $(-1;2)$ to solve for $c_1$: $$\begin{align*} y_1 &= x + c_1 \\ 2 &= -1 + c_1 \\ 3 &= c_1 \end{align*}$$ $$\begin{align*} y_2 &= -x + c_2 \\ 2 &= -(-1) + c_2 \\ 1 &= c_2 \end{align*}$$ Therefore the axes of symmetry are $y = x + 3$ and $y = -x + 1$.

Plot the points and sketch the graph

State the domain and range

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

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

Worked example 12: Sketching a hyperbola

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

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

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

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

Start with a sketch of the standard hyperbola $g(x) = \frac{1}{x}$.

The vertical asymptote is $x = 0$ and the horizontal asymptote is $y = 0$.

Determine the vertical shift

From the equation we see that $q = 3$, which means $g(x)$ must shifted $\text{3}$ units up.

The horizontal asymptote is also shifted $\text{3}$ units up to $y = 3$ .

Determine the horizontal shift

From the equation we see that $p = -2$, which means $g(x)$ must shifted $\text{2}$ units to the right.

The vertical asymptote is also shifted $\text{2}$ units to the right.

Determine the $y$-intercept

The $y$-intercept is obtained by letting $x = 0$: $$\begin{align*} y &= \frac{1}{0 - 2} + 3 \\ &= 2\frac{1}{2} \end{align*}$$ This gives the point $(0;2\frac{1}{2})$.

Determine the $x$-intercept

The $x$-intercept is obtained by letting $y = 0$: $$\begin{align*} 0 &= \frac{1}{x - 2} + 3\\ -3 &= \frac{1}{x -2} \\ -3(x - 2) &= 1 \\ -3x + 6 &= 1 \\ -3x &= -5 \\ x &= \frac{5}{3} \end{align*}$$ This gives the point $(\frac{5}{3};0)$.

Determine the domain and range

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

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

Worked example 13: Finding the equation of a hyperbola from the graph

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

Examine the graph and deduce the sign of $a$

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

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

Determine the value of $a$

To determine the value of $a$ we substitute a point on the graph, namely $(0;0)$: $$\begin{align*} y &= \frac{a}{x + 1} + 3 \\ 0 &= \frac{a}{0 + 1} + 3 \\ \therefore -3 &= a \end{align*}$$

Write the final answer

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

Interactive Exercise 5.12

Sketching graphs

Textbook Exercise 5.14

Draw the graphs of the following functions and indicate:

• asymptotes
• intercepts, where applicable
• axes of symmetry
• domain and range

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

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

\(y = -\frac{1}{x + 1} + 3\)

\(y = -\frac{5}{x - 2\frac{1}{2}} - 2\)

\(y = \frac{8}{x - 8} + 4\)

Given the graph of the hyperbola of the form \(y = \frac{1}{x + p} + q\), determine the values of \(p\) and \(q\).

\begin{align*} y &= \frac{1}{x + p} + q \\ \text{From graph } \quad p &= 2 \\ q &= -1 \\ \therefore y &= \frac{1}{x + 2} - 1 \end{align*}

Given a sketch of the function of the form \(y = \frac{a}{x + p} + q\), determine the values of \(a\), \(p\) and \(q\).

\begin{align*} y &= \frac{a}{x + p} + q \\ \text{From graph } \quad p &= 0 \\ q &= 2 \\ \therefore y &= \frac{a}{x} + 2 \\ \text{Subst. } (2;0) \quad 0 &= \frac{a}{2} + 2 \\ -2 &= \frac{a}{2} \\ -2(2) &= a \\ \therefore a &= -4 \\ y &= -\frac{4}{x} + 2 \end{align*}

Draw the graph of \(f(x) = -\frac{3}{x}\), \(x > 0\).

Determine the average gradient of the graph between \(x=1\) and \(x=3\).

\begin{align*} \text{Average gradient } &= \frac{f(3) - f(1)}{3 - 1} \\ &= \dfrac{-\frac{3}{3} - \left( -\frac{3}{1} \right)}{3-1} \\ &= \dfrac{-1 +3}{2} \\ &= \dfrac{2}{2} \\ &= 1 \end{align*}

Average gradient between \(x=1\) and \(x=3\) is \(1\).

Is the gradient at \((\frac{1}{2};-6)\) less than or greater than the average gradient between \(x=1\) and \(x=3\)? Illustrate this on your graph.

\begin{align*} \text{Average gradient } &= \frac{f(a+h) - f(a)}{(a+h) - a} \\ &= \dfrac{-\frac{3}{a+h} - \left( -\frac{3}{a} \right)}{(a+h)-a} \\ &= \dfrac{-\frac{3}{a+h} + \frac{3}{a}}{h} \\ &= \dfrac{\frac{-3(a) + 3(a+h)}{a(a+h)}}{h} \\ &= \dfrac{\frac{-3a + 3a+3h}{a^2+ah}}{h} \\ &= \dfrac{\frac{3h}{a^2+ah}}{h} \\ &= \frac{3h}{a^2+ah} \times \frac{1}{h} \\ &= \frac{3}{a^2+ah} \\ \therefore \text{At } (\frac{1}{2}; -6) \qquad a &= \frac{1}{2} \\ \therefore \text{And } \qquad h &= 0 \\ \therefore \text{Average gradient } &= \frac{3}{\left( \frac{1}{2} \right)^2+\frac{1}{2}(0)} \\ &= 3 \times \frac{4}{1} \\ &= 12 \end{align*}

Average gradient at \((\frac{1}{2};-6)\) is greater than the average gradient between \(x=1\) and \(x=3\).

Try the textbook questions