General exponential functions, of the form *f*(*x*)=*Ce*^{kx}, can be graphed using just two points (*x*_{0},*y*_{0}) and (*x*_{1},*y*_{1}) on the graph of the function.

How to use this applet

If the two points are on opposite sides of the *x*-axis, then there is no graph drawn. This is because of the relationship between the constants, *C* and *k*, and the points, (*x*_{0},*y*_{0}) and (*x*_{1},*y*_{1}), on the graph.

In the equation *y*=*Ce*^{kx}, the two constants *C* and *k* determine the shape of the graph. These two values are fixed, given two points (*x*_{0},*y*_{0}) and (*x*_{1},*y*_{1}) on the graph. To see how these two points determine the values of *C* and *k*, plug the two points into the equation separately:

Now solve each of these equations for *C*:

Since both equations represent *C* in terms of the other variables, set the right hand sides equal --- this eliminates the constant *C* from the equations:

Now solve this equation for *k* -- first, rewrite the equation a bit:

Now take natural logarithms on both sides:

This is where it is important that the two points (*x*_{0},*y*_{0}) and (*x*_{1},*y*_{1}) be on the same side of the *x*-axis -- in other words, that *y*_{0} and *y*_{1} have the same sign, so that *y*_{1}/*y*_{0} is positive and the logarithm on the left hand side is defined. If *y*_{0} and *y*_{1} have the same sign (which can be checked with the inequality *y*_{0}*y*_{1}>0) then ln(*y*_{1}/*y*_{0}) =ln(|*y*_{1}|)-ln(|*y*_{0}|) by properties of logarithms, with the absolute values guaranteeing that the logarithms will be defined even if *y*_{0} and *y*_{1} are negative. Some care must be taken at this point to check the inequality *y*_{0}*y*_{1}>0, though, since this expression with the absolute values no longer requires *y*_{0} and *y*_{1} to have the same sign.

Continuing to solve for *k*, the equation now looks like:

Solving this algebraically for *k*:

It is interesting to notice that the above equation resembles the slope of a line.

Now that the value of *k* is known (from the points (*x*_{0},*y*_{0}) and (*x*_{1},*y*_{1})), finding *C* is easier: plug either point into the equation *y*=*Ce*^{kx} -- plugging in (*x*_{0},*y*_{0}) gives *y*_{0}=*Ce*^{kx0}. Now divide both sides by *e*^{kx0} to solve for *C*:

These formulas for *C* and *k* depend only on the two points (*x*_{0},*y*_{0}) and (*x*_{1},*y*_{1}).

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Published January 2001