We now consider hyperbolic functions of the form \(y=\frac{a}{x+p}+q\) and the effects of parameter \(p\).
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).
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.
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 \(\{g(x): g(x) \in \mathbb{R}, g(x) \ne 2 \}\).
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Sign up hereDomain and range
Exercise 5.10
\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*}
\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
\(f(x) = \frac{1}{x + 4} - 2\)
\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
\(y = \frac{1}{x + 4} - 2\)
\begin{align*}
\text{Vertical asymptote: } y &= -2 \\
\text{Horizontal asymptote: } x &= -4
\end{align*}
\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*}
\begin{align*}
\text{Vertical asymptote: } y &= -8 \\
\text{Horizontal asymptote: } x &= 0
\end{align*}
\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*}
Axes of symmetry
Exercise 5.13
\(f(x) = \frac{2}{x}\)
\(g(x) = \frac{2}{x} + 1\)
\(f(x) = -\frac{3}{x}\)
\(g(x) = -\frac{3}{x + 1}\)
\(f(x) = \frac{5}{x}\)
\(g(x) = \frac{5}{x - 1} - 1\)
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 \(k(x)\).
