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# Hyperbolic Functions

## 5.3 Hyperbolic functions (EMBGP)

### Revision (EMBGQ)

#### 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 $$y = -\frac{1}{x}$$.

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

# Success in Maths and Science unlocks opportunities

## Revision

Exercise 5.9

Consider the following hyperbolic functions:

• $$y_1 = \frac{1}{x}$$
• $$y_2 = -\frac{4}{x}$$
• $$y_3 = \frac{4}{x} - 2$$
• $$y_4 = -\frac{4}{x} + 1$$

Complete the table to summarise the properties of the hyperbolic function:

 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ value of $$q$$ $$q = 0$$ effect of $$q$$ no vertical shift value of $$a$$ $$a = 1$$ effect of $$a$$ lies in I and III quad asymptotes $$y$$-axis, $$x = 0$$ $$x$$-axis, $$y = 0$$ axes of symmetry $$y = x$$ $$y = -x$$ domain $$\{x: x \in \mathbb{R}, x \ne 0 \}$$ range $$\{y: y \in \mathbb{R}, y \ne 0 \}$$
 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ value of $$q$$ $$q = 0$$ $$q = 0$$ $$q = -2$$ $$q = 1$$ effect of $$q$$ no vertical shift no vertical shift shift $$\text{2}$$ units down shift $$\text{1}$$ unit up value of $$a$$ $$a = 1$$ $$a = -1$$ $$a = 4$$ $$a = -4$$ effect of $$a$$ lies in I and III quad lies in II and IV quad lies in I and III quad lies in II and IV quad asymptotes $$y$$-axis, $$x = 0$$ $$x$$-axis, $$y = 0$$ $$y$$-axis, $$x = 0$$ $$x$$-axis, $$y = 0$$ $$y$$-axis, $$x = 0$$ $$y = -2$$ $$y$$-axis, $$x = 0$$ $$y = 1$$ axes of symmetry $$y = x$$ $$y = -x$$ $$y = x$$ $$y = -x$$ $$y = x - 2$$ $$y = -x - 2$$ $$y = x + 1$$ $$y = -x + 1$$ domain $$\{x: x \in \mathbb{R}, x \ne 0 \}$$ $$\{x: x \in \mathbb{R}, x \ne 0 \}$$ $$\{x: x \in \mathbb{R}, x \ne 0 \}$$ $$\{x: x \in \mathbb{R}, x \ne 0 \}$$ range $$\{y: y \in \mathbb{R}, y \ne 0 \}$$ $$\{y: y \in \mathbb{R}, y \ne 0 \}$$ $$\{y: y \in \mathbb{R}, y \ne -2 \}$$ $$\{y: y \in \mathbb{R}, y \ne 1 \}$$

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

We now consider hyperbolic functions of the form $$y=\frac{a}{x+p}+q$$ and the effects of parameter $$p$$.

## The effects of $$a$$, $$p$$ and $$q$$ on a hyperbolic graph

1. On the same system of axes, plot the following graphs:

1. $$y_1 = \frac{1}{x}$$
2. $$y_2 = \frac{1}{x-2}$$
3. $$y_3 = \frac{1}{x-1}$$
4. $$y_4 = \frac{1}{x+1}$$

Use your sketches of the functions above to complete the following table:

 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ intercept(s) asymptotes axes of symmetry domain range effect of $$p$$
2. Complete the following sentences for functions of the form $$y = \frac{a}{x + p} + q$$:

1. A change in $$p$$ causes a $$\ldots \ldots$$ shift.
2. If the value of $$p$$ increases, the graph and the vertical asymptote $$\ldots \ldots$$
3. If the value of $$q$$ changes, then the $$\ldots \ldots$$ asymptote of the hyperbola will shift.
4. If the value of $$p$$ decreases, the graph and the vertical asymptote $$\ldots \ldots$$

The effect of the parameters on $$y = \frac{a}{x+p} + q$$

The effect of $$p$$ is a horizontal shift because all points are moved the same distance in the same direction (the entire graph slides to the left or to the right).

• For $$p>0$$, the graph is shifted to the left by $$p$$ units.

• For $$p<0$$, the graph is shifted to the right by $$p$$ units.

The value of $$p$$ also affects the vertical asymptote, the line $$x = -p$$.

The effect of $$q$$ is a vertical shift. The value of $$q$$ also affects the horizontal asymptotes, the line $$y = q$$.

The value of $$a$$ affects the shape of the graph and its position on the Cartesian plane.

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

#### 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 $$y=q$$.

## Worked example 9: Domain and range

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

### Determine the domain

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

### Determine the range

Let $$g(x) = y$$: \begin{align*} y &= \frac{2}{x+1} + 2 \\ y - 2 &= \frac{2}{x+1} \\ (y-2)(x+1) &= 2 \\ x + 1 &= \frac{2}{y-2} \end{align*} Therefore the range is $$\{g(x): g(x) \in \mathbb{R}, g(x) \ne 2 \}$$.

# Success in Maths and Science unlocks opportunities

## Domain and range

Exercise 5.10

Determine the domain and range for each of the following functions:

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

\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq 0 \right \} \\ \text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 1 \right \} \end{align*}

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

\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq 8 \right \} \\ \text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 4 \right \} \end{align*}

$$y = -\frac{4}{x + 1} -3$$

\begin{align*} \text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq -1 \right \} \\ \text{Range: } & \left \{ y: y \in \mathbb{R},y \neq -3 \right \} \end{align*}

$$x = \frac{2}{3 - y} + 5$$

\begin{align*} x &= \frac{2}{3 - y} + 5 \\ x -5 &= \frac{2}{3 - y} \\ (x -5)(3 - y) &= 2\\ 3 - y &= \frac{2}{3x - 5} \\ - y &= \frac{2}{x - 5} - 3 \\ \therefore y &= -\frac{2}{x - 5} + 3 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq 5 \right \} \\ \text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 3 \right \} \end{align*}

$$(y - 2)(x + 2) = 3$$

\begin{align*} (y - 2)(x + 2) &= 3\\ y - 2 &= \frac{3}{x + 2} \\ \therefore y &= \frac{3}{x + 2} + 2 \\ \text{Domain: } & \left \{ x: x \in \mathbb{R},x \neq -2 \right \} \\ \text{Range: } & \left \{ y: y \in \mathbb{R},y \neq 2 \right \} \end{align*}

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 $$(-2;0)$$.

## Intercepts

Exercise 5.11

Determine the $$x$$- and $$y$$-intercepts for each of the following functions:

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

\begin{align*} f(x) &= \frac{1}{x + 4} - 2 \\ \text{Let } x & = 0 \\ f(0) &= \frac{1}{4} - 2 \\ \therefore y_\text{int} &=(0;-1\frac{3}{4}) \\ \text{Let } y & = 0 \\ 0 &= \frac{1}{x + 4} - 2 \\ 2 &= \frac{1}{x + 4} \\ 2(x + 4) &= 1 \\ 2x + 8 &= 1 \\ 2x &= -7 \\ \therefore x &= -\frac{7}{2} \\ \therefore x_\text{int} &= \left( -3\frac{1}{2}; 0 \right) \end{align*}

$$g(x) = -\frac{5}{x} + 2$$

\begin{align*} g(x) &= -\frac{5}{x} + 2 \\ \text{Let } x & = 0 \\ \therefore g(x) & \text{is undefined } \\ \therefore \text{no } & x-\text{intercepts} \\ \text{Let } y & = 0 \\ 0 &= -\frac{5}{x} + 2 \\ -2 &= -\frac{5}{x} \\ 2x &= 5 \\ \therefore x &= \frac{5}{2} \\ \therefore x_\text{int} &= \left( \frac{5}{2} ; 0 \right) \end{align*}

$$j(x) = \frac{2}{x - 1} + 3$$

\begin{align*} j(x) &= \frac{2}{x - 1} + 3 \\ \text{Let } x & = 0 \\ j(0) &= \frac{2}{- 1} + 3 \\ &= 1 \\ \therefore y_\text{int} &=(0;1) \\ \text{Let } y & = 0 \\ 0 &= \frac{2}{x - 1} + 3 \\ -3 &= \frac{2}{x - 1} \\ -3(x-1) &= 2 \\ -3x + 3 &= 2 \\ -3x &= -1 \\ \therefore x &= \frac{1}{3} \\ \therefore x_\text{int} &= \left( \frac{1}{3} ; 0 \right) \end{align*}

$$h(x) = \frac{3}{6 - x} + 1$$

\begin{align*} h(x) &= \frac{3}{6 - x} + 1 \\ \text{Let } x & = 0 \\ h(0) &= \frac{3}{6} + 1 \\ &= \frac{3}{2} \\ \therefore y_\text{int} &= \left(0;\frac{3}{2} \right) \\ \text{Let } y & = 0 \\ 0 &= \frac{3}{6 - x} + 1 \\ -1 &= \frac{3}{6 - x} \\ -(6 - x) &= 3 \\ -6 + x &= 3 \\ -3x &= -1 \\ \therefore x &= \frac{1}{3} \\ \therefore x_\text{int} &= \left( \frac{1}{3} ; 0 \right) \end{align*}

$$k(x) = \frac{5}{x + 2} - \frac{1}{2}$$

\begin{align*} k(x) &= \frac{5}{x + 2} - \frac{1}{2} \\ \text{Let } x & = 0 \\ k(0) &= \frac{5}{2} - \frac{1}{2} \\ &= 2 \\ \therefore y_\text{int} &= \left(0;2 \right) \\ \text{Let } y & = 0 \\ 0 &= \frac{5}{x + 2} - \frac{1}{2} \\ \frac{1}{2} &= \frac{5}{x + 2} \\ x + 2 &= 5(2) \\ x &= 10 - 2 \\ \therefore x &= 8 \\ \therefore x_\text{int} &= \left( 8 ; 0 \right) \end{align*}

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 $$x=-p$$.

## Asymptotes

Exercise 5.12

Determine the asymptotes for each of the following functions:

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

\begin{align*} \text{Vertical asymptote: } y &= -2 \\ \text{Horizontal asymptote: } x &= -4 \end{align*}

$$y = -\frac{5}{x}$$

\begin{align*} \text{Vertical asymptote: } y &= 0 \\ \text{Horizontal asymptote: } x &= 0 \end{align*}

$$y = \frac{3}{2 -x} + 1$$

\begin{align*} y &= \frac{3}{2 -x} + 1 \\ &= \frac{3}{-(x - 2)} + 1 \\ &= -\frac{3}{x - 2} + 1 \\ \text{Vertical asymptote: } y &= 1 \\ \text{Horizontal asymptote: } x &= 2 \end{align*}

$$y = \frac{1}{x} -8$$

\begin{align*} \text{Vertical asymptote: } y &= -8 \\ \text{Horizontal asymptote: } x &= 0 \end{align*}

$$y = -\frac{2}{x - 2}$$

\begin{align*} \text{Vertical asymptote: } y &= 0 \\ \text{Horizontal asymptote: } x &= 2 \end{align*}

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*}

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

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 $$k(x)$$.

#### 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:

• 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 $$y = -x + 1$$.

### 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*}

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

## Sketching graphs

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 $$x=3$$.