Functions of the Form <em>y = <sup>1</sup>/<sub>x</sub></em>
Functions of the form \(y=\dfrac{1}{x}\)
Functions of the general form \(y=\dfrac{a}{x}+q\) are called hyperbolic functions.
Example
Question
Complete the following table for \(h(x)=\dfrac{1}{x}\) and plot the points on a system of axes.
|
\(x\) |
\(-\text{3}\) |
\(-\text{2}\) |
\(-\text{1}\) |
\(-\dfrac{1}{2}\) |
\(-\dfrac{1}{4}\) |
0 |
\(\dfrac{1}{4}\) |
\(\dfrac{1}{2}\) |
\(\text{1}\) |
\(\text{2}\) |
\(\text{3}\) |
|
\(h(x)\) |
\(-\dfrac{1}{3}\) |
-
Join the points with smooth curves.
-
What happens if \(x = 0\)?
-
Explain why the graph consists of two separate curves.
-
What happens to \(h(x)\) as the value of \(x\) becomes very small or very large?
-
The domain of \(h(x)\) is \(\{x:x\in \mathbb{R},x\ne 0\}\). Determine the range.
-
About which two lines is the graph symmetrical?
Substitute values into the equation
\begin{align*} h(x)& = \cfrac{1}{x} \\ h(-3)& = \cfrac{1}{-3} = -\cfrac{1}{3} \\ h(-2)& = \cfrac{1}{-2} = -\cfrac{1}{2} \\ h(-1)& = \cfrac{1}{-1} = -1 \\ h(-\cfrac{1}{2})& = \cfrac{1}{-\cfrac{1}{2}} = -2 \\ h(-\cfrac{1}{4})& = \cfrac{1}{-\cfrac{1}{4}} = -4 \\ h(0)& = \cfrac{1}{0} = \text{undefined} \\ h\left(\cfrac{1}{4}\right)& = \cfrac{1}{\cfrac{1}{4}} = 4 \\ h\left(\cfrac{1}{2}\right)& = \cfrac{1}{\cfrac{1}{2}} = 2 \\ h(1)& = \cfrac{1}{1} = 1 \\ h(2)& = \cfrac{1}{2} = \cfrac{1}{2} \\ h(3)& = \cfrac{1}{3} = \cfrac{1}{3} \end{align*}|
\(x\) |
\(-\text{3}\) |
\(-\text{2}\) |
\(-\text{1}\) |
\(-\dfrac{1}{2}\) |
\(-\dfrac{1}{4}\) |
\(\text{0}\) |
\(\dfrac{1}{4}\) |
\(\dfrac{1}{2}\) |
\(\text{1}\) |
\(\text{2}\) |
\(\text{3}\) |
|
\(h(x)\) |
\(-\dfrac{1}{3}\) |
\(-\dfrac{1}{2}\) |
\(-\text{1}\) |
\(-\text{2}\) |
\(-\text{4}\) |
undefined |
\(\text{4}\) |
\(\text{2}\) |
\(\text{1}\) |
\(\dfrac{1}{2}\) |
\(\dfrac{1}{3}\) |
Plot the points and join with two smooth curves
From the table we get the following points: \((-3;-\cfrac{1}{3})\), \((-2;-\cfrac{1}{2})\), \((-1;-1)\), \((-\cfrac{1}{2};-2)\), \((-\cfrac{1}{4};-4)\), \(\left(\cfrac{1}{4};4)\), \((\cfrac{1}{2};2)\), \((1;1)\), \((2;\cfrac{1}{2}\right)\), \((3;\cfrac{1}{3})\)
For \(x=0\) the function \(h\) is undefined. This is called a discontinuity at \(x=0\).
\(y=h(x)=\dfrac{1}{x}\) therefore we can write that \(x\times y=1\). Since the product of two positive numbers and the product of two negative numbers can be equal to \(\text{1}\), the graph lies in the first and third quadrants.
Determine the asymptotes
As the value of \(x\) gets larger, the value of \(h(x)\) gets closer to, but does not equal \(\text{0}\). This is a horizontal asymptote, the line \(y=0\). The same happens in the third quadrant; as \(x\) gets smaller \(h(x)\) also approaches the negative \(x\)-axis asymptotically.
We also notice that there is a vertical asymptote, the line \(x=0\); as \(x\) gets closer to \(\text{0}\), \(h(x)\) approaches the \(y\)-axis asymptotically.
Determine the range
Domain: \(\{x:x\in \mathbb{R},x\ne 0\}\)
From the graph, we see that \(y\) is defined for all values except \(\text{0}\).
Range: \(\{y:y\in \mathbb{R},y\ne 0\}\)
Determine the lines of symmetry
The graph of \(h(x)\) has two axes of symmetry: the lines \(y=x\) and \(y=-x\). About these two lines, one half of the hyperbola is a mirror image of the other half.
This lesson is part of:
Functions I