`(1,2) , (3,50)` Write an exponential function `y=ab^x` whose graph passes through the given points.

The given two points of the exponential function are (1,2) and (3,50).

To determine the exponential function

`y=ab^x`

plug-in the given x and y values.

For the first point (1,2), plug-in x=1 and y=2.

`2=ab^1`

`2=ab`          (Let this be EQ1.)

For the second point (3,50), plug-in x=3 and y=50.

`50=ab^3`       (Let this be EQ2.)

To solve for the values of a and b, apply substitution method of system of...

The given two points of the exponential function are (1,2) and (3,50).

To determine the exponential function

`y=ab^x`

plug-in the given x and y values.

For the first point (1,2), plug-in x=1 and y=2.

`2=ab^1`

`2=ab`          (Let this be EQ1.)

For the second point (3,50), plug-in x=3 and y=50.

`50=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.

`2=ab`

`2/b=a`

Plug-in this to EQ2.

`50=ab^3`

`50=(2/b)b^3`

And solve for b.

`50=2b^2`

`50/2=b^2`

`25=b^2`

`+-sqrt25=b`

`+-5=b`

Take note that in the 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 only the positive value of b which is 5.

Then, plug-in b=5 to EQ1.

`2=ab`

`2=a(5)`

Isolate the a.

`2/5=a`

Then, plug-in `a=2/5` and `b=5` to

`y=ab^x`

So this becomes:

`y=2/5*5^x`

Therefore, the exponential function that passes the given two points is `y=2/5*5^x` .