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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 \(y = -\frac{1}{x}\).

\(a<0\)

\(a>0\)

\(q>0\)

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\(q=0\)

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\(q<0\)

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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 shiftno vertical shiftshift \(\text{2}\) units downshift \(\text{1}\) unit up
value of \(a\)\(a = 1\)\(a = -1\)\(a = 4\)\(a = -4\)
effect of \(a\)lies in I and III quadlies in II and IV quadlies in I and III quadlies 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.

\(p>0\)\(p<0\)

\(a<0\)

\(a>0\)

\(a<0\)

\(a>0\)

\(q>0\)

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\(q<0\)

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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 \(\{g(x): g(x) \in \mathbb{R}, g(x) \ne 2 \}\).

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*}

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*}

a387c7f85b22e8a1a8e19af839e991d6.png

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?

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

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

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

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

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

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

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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 \(k(x)\).

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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

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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 \(y = 0\).

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

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 \(q = 3\). \[y = \frac{a}{x + 1} + 3\]

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

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

Sketching graphs

Exercise 5.14

Draw the graphs of the following functions and indicate:

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

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

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

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

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

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

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

08e3852eb2827733028d15f136d79fc2.png
\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 \(q\).

1a2f293601e4c3cb63f144a4ea604f2e.png
\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 \(f(x) = -\frac{3}{x}\), \(x > 0\).

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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 \(x=3\).