Given the following table that gives data from a linear function:
[tex]\begin {tabular}
{|c|c|c|c|}
Temperature, $y = f(x)$ (^\circ C)&0&5&20 \\ [1ex]
Temperature, $x$ (^\circ F)&32&41&68 \\
\end {tabular}[/tex]
The formular for the function can be obtained by choosing two points from the table and using the formular for the equation of a straight line.
Recall that the equation of a straight line is given by
[tex] \frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}[/tex]
Using the points (32, 0) and (41, 5), we have:
[tex]\frac{y-0}{x-32} = \frac{5-0}{41-32}= \frac{5}{9} \\ \\ \Rightarrow y=\frac{5}{9}(x-32)[/tex]
