Question 1009912
{{{f(x)=ab^x}}} can be the form of the exponential function.
Two points are   ({{{-2}}},{{{25}}}) and ({{{1}}}, {{{8/5}}}).

{{{25=ab^(-2)}}}
{{{1=ab^(8/5)}}}

solve the system:

Take logarithms of both sides, base of your choice, and you will have a linear equation.  You would need to make an adjustment to the vertical axis component of your given points.

{{{log(25)=log(ab^(-2))}}}

{{{log(5^2)=log(a)+log(b^(-2))}}}
{{{2log(5)=log(a)+(-2)log(b)}}}
{{{2log(5)=log(a)-2log(b)}}}.......eq.1


{{{log(1)=log(ab^(8/5))}}}
{{{0=log(a)+log(b^(8/5))}}}
{{{0=log(a)+(8/5)log(b)}}}.......eq.2

subtract eq.1 from eq.2

{{{0-2log(5)=log(a)-log(a)+(8/5)log(b)+2log(b)}}}

{{{-2log(5)=(8/5)log(b)+(10/5)log(b)}}}

{{{-2log(5)=(18/5)log(b)}}}

{{{log(5^-2)=log(b^(18/5))}}}

{{{5^-2=b^(18/5)}}}

{{{b = 1/5^(5/9)}}}

{{{b = 0.4}}}


now find {{{a}}}

{{{log(a)=-(8/5)log(b)}}} from eq.2

{{{log(a)=-(8/5)log(0.4)}}}

{{{log(a)=log(0.4^(-(8/5)))}}}

{{{a=0.4^(-(8/5))}}}

{{{a=1/(0.4^(8/5))}}}

{{{a=1/0.230832}}}

{{{a=4.3321549871}}}

{{{a=4}}}


so, {{{f(x)=4(0.4)^x}}}