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 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: \[a=\frac{{\vartriangle}y}{{\vartriangle}t}=\frac{-3-4}{4-1}=-{{7}\over{3}}\] 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 \((1,4)\): \[4=-{{7}\over{3}}\times 1 +b\qquad\text{that\;is}\qquad b=4+{{7}\over{3}}\times 1={{19}\over{3}}\] The function definition is \[y(t)=-{{7}\over{3}}t+{{19}\over{3}}\] The vertical intercept is \(f(0)=b={{19}\over{3}}\).
The graph of this function is shown along with the two data points in the figure below.

New example