### Basic functions: Linear functions

### A linear relationship on the basis of two data points

You have already seen how the function definition of a linear function can be calculated if you are already known to the slope and a single point on the graph. In practice you encounter the following problem more frequently:

*Given two data points \((t_0,y_0)\) and \((t_1,y_1)\) with \(t_0\neq t_1\), what is the function definition \(y(t)=a\,t+b\) for which the graph goes through these two points?*

The general solution method is as follows:

The graph of the linear function \(y(t)=a\,t+b\) is a straight line.

The slope \(a\) can be calculated as the quotient of increments: \[a=\frac{{\vartriangle}y}{{\vartriangle}t}=\frac{y_1-y_0}{t_1-t_0}\] Hereafter, the intercept \(b\) can be calculated on the basis of the coordinates of one of the two data points, for example: \[b=y_0-a\cdot t_0\]

A concrete dynamic example may illustrate the method.

The slope \(a\) can be calculated as the quotient of increments: \[a=\frac{{\vartriangle}y}{{\vartriangle}t}=\frac{-2+5}{6-1}={{3}\over{5}}\] Hereafter, the vertical intercept \(b\) can be calculated on the basis of the coordinates of one of the two measurement points, for example, on the basis of the point \((1,-5)\): \[-5={{3}\over{5}}\times 1 +b\qquad\text{that\;is}\qquad b=-5-{{3}\over{5}}\times 1=-{{28}\over{5}}\] The function definition is \[y(t)={{3}\over{5}}t-{{28}\over{5}}\] The vertical intercept is \(f(0)=b=-{{28}\over{5}}\).

The graph of this function is shown along with the two data points in the figure below.