Mathematics » Functions II » Hyperbolic Functions

# Functions of the Form y = a/(x + p) + q

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

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

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

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

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

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

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 $$y = \cfrac{a}{x+p} + q$$

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: $$f(x) = y = \cfrac{a}{x+p} + q$$:

### 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 $y = \cfrac{a}{x+p} + q$ can be re-written as: $y-q = \cfrac{a}{x+p}$ If $$x \ne -p$$ then: \begin{align*} (y-q)(x+p) &= a \\ x + p &= \cfrac{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 $$y=q$$.

## Example

### Question

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

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

### Intercepts

The $$y$$-intercept:

To calculate the $$y$$-intercept we let $$x=0$$. For example, the $$y$$-intercept of $$g(x) = \cfrac{2}{x + 1} + 2$$ is determined by setting $$x=0$$: \begin{align*} g(x) &= \cfrac{2}{x + 1} + 2 \\ g(0) &= \cfrac{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 $$g(x) = \cfrac{2}{x + 1} + 2$$ is determined by setting $$y=0$$: \begin{align*} g(x) &= \cfrac{2}{x + 1} + 2 \\ 0 &= \cfrac{2}{x + 1} + 2 \\ -2 &= \cfrac{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 $$y=\cfrac{a}{x+p}+q$$. 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 $$y =\cfrac{1}{x}$$, we see that if we replace $$x \Rightarrow y$$ and $$y \Rightarrow x$$, we get $$y =\cfrac{1}{x}$$. 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 $$y =\cfrac{a}{x + p} + q$$, 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 $$c_2$$.

## Example

### Question

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

### 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 $$y = \cfrac{2}{x + 1} – 2$$ are the lines \begin{align*} y_1 &= x – 1 \\ y_2 &= -x – 3 \end{align*} ### Sketching graphs of the form $$f(x)=\cfrac{a}{x+p}+q$$

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

• asymptotes

• $$y$$-intercept

• $$x$$-intercept

• domain and range

## Example

### Question

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

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

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 &= \cfrac{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 &= \cfrac{2}{x + 1} + 2\\ -2 &= \cfrac{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: $$\{ y: y \in \mathbb{R}, y \ne 2 \}$$

## Example

### Question

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

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

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

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

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

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 &= \cfrac{1}{0 – 2} + 3 \\ &= 2\cfrac{1}{2} \end{align*} This gives the point $$(0;2\cfrac{1}{2})$$.

### Determine the $$x$$-intercept

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

### Determine the domain and range

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

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

## Example

### Question

Use the graph below to determine the values of $$a$$, $$p$$ and $$q$$ for $$y = \cfrac{a}{x + p} + q$$. ### 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 $$q = 3$$. $y = \cfrac{a}{x + 1} + 3$

### Determine the value of $$a$$

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

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