Provided that the graph is of distance as a function of time, the slope of the line tangent to the function at a given point represents the instantaneous velocity at that point.
In order to get an idea of this slope, one must use limits. For an example, suppose one is given a distance function #x = f(t)#, and one wishes to find the instantaneous velocity, or rate of change of distance, at the point #p_0 = (t_0, f(t_0))#, it helps to first examine another nearby point, #p_1 = (t_0+a, f(t_0+a))#, where #a# is some arbitrarily small constant. The slope of the secant line passing through the graph at these points is:
#[f(t_0+a)-f(t_0)]/a#
As #p_1# approaches #p_0# (which will occur as our #a# decreases), our above #difference quotient# will approach a limit, here designated #L#, which is the slope of the tangent line at the given point. At that point, a point-slope equation using our above points can provide a more exact equation.
If instead one is familiar with differentiation, and the function is both continuous and differentiable at the given value of #t#, then we can simply differentiate the function. Given that most distance functions are polynomial functions, of the form #x = f(t) = at^n + bt^(n-1) + ct^(n-2) + ... + yt + z,# these can be differentiated using the power rule which states that for a function #f(t) = at^n, (df)/dt# (or #f'(t)#) = #(n)at^(n-1)#.
Thus for our general polynomial function above, #x' = f'(t) = (n)at^(n-1) + (n-1)bt^(n-2) + (n-2)ct^(n-3) + ... + y# (Note that since #t = t^1# (as any number raised to the first power equals itself), reducing the power by 1 leaves us with #t^0 = 1#, hence why the final term is simply #y#. Note also that our #z# term, being a constant, did not change with respect to #t# and thus was discarded in differentiation).
This #f'(t)# is the derivative of the distance function with respect to time; thus, it measures the rate of change of distance with respect to time, which is simply the velocity.
