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.

 \(p>0\)\(p<0\)
 

\(a<0\)

\(a>0\)

\(a<0\)

\(a>0\)

\(q>0\)

fa0cb90086b72fae7df977c56313712f.pngea96fb1bfb0fc759fe96f1d14938ac73.pnga41390c987133a352ee45ea97e156d27.pngc8a4e43248c5bdc1a671a121353b654e.png

\(q<0\)

b2fddf65f6fc8fb1d438a749fa3c5a35.png6083213c59bb441cb0ecc24e960beec1.png931d7bea3e8ef03c743bd9d52d3c3c93.png84970c5cfc00ebcb31acd0a512dde944.png

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*}

Write the final answer

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*}

a387c7f85b22e8a1a8e19af839e991d6.png

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:

  • quadrants

  • 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

b847d322a356ec2bca6a07678a600720.png

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 \(y = 0\).

20518b728b72942acabe19743103a210.png

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 \(y = 3\) .

0ac3653d6f979bdc0b93825fd71ca6c3.png

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.

4abaeb68c44bd6f80d65d6d9713c6846.png

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 \(y = \cfrac{a}{x + p} + q\).

11eeb3d3a3646f2aa9fd8ce8cea2ed2d.png

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*}

Write the final answer

\[y = -\cfrac{3}{x + 1} + 3\]

[Attributions and Licenses]


This is a lesson from the tutorial, Functions II and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts