The given two points of the exponential function are (1,40) and (3,640).

To determine the exponential function

`y=ab^x`

plug-in the given x and y values.

For the first point (1,40), plug-in x=1 and y=40.

`40=ab^1`

`40=ab` (Let this be EQ1.)

For the second point (3,640), plug-in x=3 and y=640.

`640=ab^3` (Let this be EQ2.)

To solve for the values of a and b, apply the substitution method of system of equations. To do so, isolate the a in EQ1.

`40=ab`

`40/b=a`

Plug-in this to EQ2.

`640=ab^3`

`640=(40/b)b^3`

And, solve for b.

`640=40b^2`

`640/40=b^2`

`16=b^2`

`+-sqrt16=b`

`+-4=b`

Take note that in exponential function `y=ab^x` , the b should be greater than zero `(bgt0)` . When `blt=0` , it is no longer an exponential function.

So, consider on the positive value of b which is 4.

Now that the value of b is known, plug-in it to EQ1.

`40=ab`

`40=a(4)`

And, solve for a.

`40/4=a`

`10=a`

Then, plug-in the values of a and b to the exponential function

`y=ab^x`

So this becomes:

`y= 10*4^x`

**Therefore, the exponential function that passes the given two points is `y=10*4^x` .**