The given two points of the exponential function are (1,3) and (2,12).

To determine the exponential function

`y=ab^x`

plug-in the given x and y values.

For the first point (1,3), plug-in x = 1 and y=3.

`3=ab^1`

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

For the second point (2,12), plug-in x=2 and y=12.

`12=ab^2` (Let this be EQ2.)

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

`3=ab`

`3/b= a`

Plug-in this to EQ2.

`12=ab^2`

`12=(3/b)b^2`

And, solve for b.

`12=3b`

`12/3=b`

`4=b`

Plug-in this value of b to EQ1.

`3=ab`

`3=a(4)`

And, solve for a.

`3/4=a`

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

`y=ab^x`

So, this becomes:

`y = 3/4*4^x`

**Therefore, the exponential function that passes the given two points is `y=3/4*4^x` .**

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