`(2,24) , (3,144)` Write an exponential function `y=ab^x` whose graph passes through the given points.

The given two points of the exponential function are (2,24) and (3,144).

To determine the exponential function

`y=ab^x`

plug-in the given x and y values.

For the first point (2,24), the values of x and y are x=2, y=24. Plugging them, the exponential function becomes:

`24=ab^2`     (Let this be EQ1.)

For the second point (3,144), the values of x and y are x=3 and y=144. Plugging them, the function becomes:

`144=ab^3`    ...

The given two points of the exponential function are (2,24) and (3,144).

To determine the exponential function

`y=ab^x`

plug-in the given x and y values.

For the first point (2,24), the values of x and y are x=2, y=24. Plugging them, the exponential function becomes:

`24=ab^2`     (Let this be EQ1.)

For the second point (3,144), the values of x and y are x=3 and y=144. Plugging them, the function becomes:

`144=ab^3`     (Let this be EQ2.)

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

`24=ab^2`

`24/b^2=a`

Plug-in this to EQ2.

`144=ab^3`

`144 = 24/b^2 * b^3`

And, solve for b.

`144=24b`

`144/24=b`

`6=b`

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

`24=ab^2`

`24=a*6^2`

And, solve for a.

`24=36a`

`24/36=a`

`2/3=a`

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

`y=ab^x`

So this becomes

`y=2/3*6^x`

Therefore, the exponential function that passes the given two points is `y=2/3*6^x` .