Question 1155241
{{{f(x) = ba^x}}}

 given 


y-intercept {{{8}}};=> is at point ({{{0}}}, {{{8}}})

P({{{3}}}, {{{1}}})

recall that {{{f(x) = y}}}


{{{f(x) = ba^x}}}....use point ({{{0}}}, {{{8}}})

{{{8 = ba^0}}}

{{{8 = b*1}}}.....solve for {{{b}}}

{{{b=8}}}........eq.1



{{{f(x) = ba^x}}}....use point P({{{3}}}, {{{1}}})

{{{1 = 8a^3}}}.....solve for {{{a}}}

{{{a^3=1/8}}}

{{{a=root(3,1/8)}}}

{{{a=root(3,1/2^3)}}}

{{{a=1/2}}}........eq.2


from eq.1 and eq.2 we have


{{{f(x) = 8*(1/2)^x}}}