# 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)#