How do you write an exponential function whose graph passes through (0,0.2) and (4, 51.2)?
1 Answer
Explanation:
Notice that:
#51.2/0.2 = 256 = 4^4#
So we can write:
#f(x) = 0.2*4^x#
Then:
#f(0) = 0.2*4^0 = 0.2#
#f(4) = 0.2*4^4 = 51.2#
Note
This is not the only solution.
Instead of
#-4# ,#4i# or#-4i#
So other solutions are:
#f(x) = 0.2*(-4)^x#
#f(x) = 0.2*(4i)^x#
#f(x) = 0.2*(-4i)^x#
General case
Suppose we want an exponential function that passes through points
A general form can be written:
#f(x) = a*b^x#
Then:
#f(x_1) = a*b^(x_1) = y_1#
#f(x_2) = a*b^(x_2) = y_2#
So:
#b^(x_2-x_1) = (a*b^(x_2))/(a*b^(x_1)) = y_2/y_1#
One solution is:
#b = (y_2/y_1)^(1/(x_2-x_1))#
There are other solutions formed by multiplying this by some
#cos((2kpi)/(x_2-x_1)) + i sin((2kpi)/(x_2-x_1))#
where
If
If
Once we have a value for
#a = y_1/b^(x_1)#