When linearising data, it may be necessary to determine the equation for the transformation so other values can be predicted. This often requires being able to determine an equation for these types of functions from a set of points or features from the data.
Worked examples
Example 1
Consider the data:
| $x$x | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|---|
| $y$y | $1$1 | $3$3 | $9$9 | $19$19 | $33$33 | $51$51 |
Using an appropriate linearising transformation, find the equation of the line for the transformed data.
Think: A graph of the data will help identify the type of transformation.
Do: Plot the data of $y$y against $x$x:
This data looks parabolic. The transformed data can be expressed in the following table:
| $x$x | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|---|
| $x^2$x2 | $0$0 | $1$1 | $4$4 | $9$9 | $16$16 | $25$25 |
| $y$y | $1$1 | $3$3 | $9$9 | $19$19 | $33$33 | $51$51 |
Plotting $y$y against $x^2$x2 leads to the linear graph:
To determine the equation of the line $y=m\left(x^2\right)+c$y=m(x2)+c, from the table of values, when $x^2=0$x2=0, $y=1$y=1. So the $y$y-intercept is $c=1$c=1. Using the coordinate $(4,9)$(4,9) from the table, the gradient $m$m can be found:
| $9$9 | $=$= | $m\times4+1$m×4+1 |
| $m$m | $=$= | $2$2 |
So, the equation for the transformed data is $y=2\left(x^2\right)+1$y=2(x2)+1.
Example 2
Consider the following data:
| $x$x | $1$1 | $2$2 | $4$4 | $8$8 |
|---|---|---|---|---|
| $y$y | $20$20 | $16$16 | $14$14 | $13$13 |
(a) Plot a graph of the relationship between $x$x and $y$y.
(b) Describe the type of variation present in this data.
(c) Find the rule for this graph of the form $y=\frac{k}{x}+c$y=kx+c.
(d) Use an appropriate transformation for $x$x to linearise the data and graph the result.
Think: The data points from the table can be used to find the unknowns in the equations.
Do:
(a) Plot the graph:
(b) From the graph, we can see the data is an example of inverse variation.
(c) Using the data from the table, the equation for this graph of the form $y=\frac{k}{x}+c$y=kx+c can be found. From the table of values, when $x=1$x=1, $y=20$y=20 and when $x=2$x=2, $y=16$y=16. This leads to the set of simultaneous equations:
| $20$20 | $=$= | $k+c$k+c |
| $16$16 | $=$= | $\frac{k}{2}+c$k2+c |
These equations can be solved using technology or by hand. By hand, subtracting the second equation from the first gives:
| $4$4 | $=$= | $\frac{k}{2}$k2 |
which gives $k=8$k=8. Substituting this back into either equation leads to $c=12$c=12.
So, the equation for the graph is $y=\frac{8}{x}+12$y=8x+12.
(c) Since this data demonstrates inverse variation, the transformation of $x\rightarrow\frac{1}{x}$x→1x will linearise the data. The table of values would be:
| $\frac{1}{x}$1x | $1$1 | $0.5$0.5 | $0.25$0.25 | $0.125$0.125 |
|---|---|---|---|---|
| $y$y | $20$20 | $16$16 | $14$14 | $13$13 |
The plot of the transformed data is linear, as expected, since the data is inversely related:
Practice questions
Question 1
Consider the function $y=2x^2$y=2x2.
Complete the following table of values.
$x$x $-2$−2 $-1$−1 $0$0 $1$1 $2$2 $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Plot the graph.
Loading Graph...
Question 2
Consider the function $y=\frac{1}{x}$y=1x which is defined for all real values of $x$x except at $0$0.
Complete the following table of values.
$x$x $-2$−2 $-1$−1 $-\frac{1}{2}$−12 $-\frac{1}{4}$−14 $\frac{1}{4}$14 $\frac{1}{2}$12 $1$1 $2$2 $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ Plot the points in the table of values.
Loading Graph...Hence draw the curve.
Loading Graph...
Logarithms and technology
Sometimes it may be necessary to use technology to find the equation of a graph, particularly for logarithms. The following example demonstrates this method.
Worked example
Example 3
Consider the following graph with the equation $y=a\log_{10}x+c$y=alog10x+c:
Use CAS to find the values of $a$a and $c$c.
Think: The data points given will lead to simultaneous equations which can be solved for $a$a and $c$c.
Do: Substituting two data points from the graph leads to the set of simultaneous equations:
| $4$4 | $=$= | $a\log_{10}1+c$alog101+c |
| $9$9 | $=$= | $a\log_{10}10+c$alog1010+c |
These can be solved using the CAS:
[CAS Instructions coming soon]
So, the equation for this graph is $y=5\log_{10}x+4$y=5log10x+4.
Practice question
Question 3
Consider the function $y=5\log_{10}x$y=5log10x.
Solve for the $x$x-coordinate of the $x$x-intercept.
Complete the table of values for $y=5\log_{10}x$y=5log10x.
$x$x $\frac{1}{10}$110 $1$1 $10$10 $100$100 $y$y $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ State the equation of the vertical asymptote.
Sketch the graph of $y=5\log_{10}x$y=5log10x.
Loading Graph...