# Write an exponential function y=ab^x for a graph that includes (-2,2/25) and (1,10).

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To determine the exponential function, plug-in the point (-2,2/25) to y=ab^x.

`2/25=ab^(-2)`

Then, simplify the equation.

`2/25=a*1/b^2`

`2/25=a/b^2`

And, isolate a.

`2/25*b^2=a/b^2*b^2`

`2/25b^2=a ` (Let this be EQ1.)

Next, plug-in the second point (1,10) to y=ab^x.

`10=ab^1`

`10=ab`

From here, plug-in EQ1.

`10=2/25b^2*b`

`10=2/25b^3`

Then, isolate b.

`10*25/2=2/25b^3*25/2`

`125=b^3`

`root(3)(125)=root(3)(b^3)`

`5=b`

And, plug-in the value of to EQ1 to solve for value of a.

`a=2/25b^2=2/25*5^2=2`

Then, plug-in the value of a and b to y=ab^x.

**Hence, the exponential function is `y=2*5^x` .**

`ab^-2=2/25` (1)

`ab=10` (2)

we have set condirions y=ab^x for the points `(-2:2/25)` and

`(1;10)` so its a system a two varibales:

dividing (2) by (1) we get:

`(ab) : (ab^-2)=(ab)(a^-1b^2)=b^3= ` `10:(2/25)=(10)(25)/2=`

`=125` `b=5` from (2) we have `a=2` so:

`y=(2)5^x`

The fnction as in the graph is.