Question 1180810


use these points on a graph:
({{{0}}},{{{1}}})
({{{-5}}},{{{4}}})
({{{1}}},{{{-2}}})

set the system to  determine the values of {{{a}}}, {{{c}}}, and {{{d }}}

({{{0}}},{{{1}}})
 
{{{f(x) = ac^x+d}}}

{{{1= ac^0+d }}}........solve for {{{a}}}
{{{1= a*1+d }}}
{{{1= a+d }}}
{{{a=1-d}}} ....eq.1

({{{-5}}},{{{4}}})
{{{4= ac^-5+d }}}
{{{4= a(1/c^5)+d }}}......solve for {{{a}}}
{{{a=(4-d )/(1/c^5)}}}
{{{a=(4-d )c^5}}}....eq.2


({{{1}}},{{{-2}}})
{{{-2= ac^1+d }}}......solve for {{{a}}}
{{{-2= ac+d }}}
{{{-2-d= ac}}}
{{{a=(-2-d)/c}}} ....eq.3


from eq.1 and eq.2 we have 

{{{1-d =(4-d )c^5}}}.........solve for {{{c}}}

{{{(1-d) /(4-d )=c^5}}}

{{{c=((1-d) /(4-d ))^(1/5)}}}...................1)


from eq.1 and eq.3 we have 

{{{1-d =(-2-d)/c}}}
{{{c(1-d )=(-2-d)}}}............solve for {{{c}}}

{{{c=(-2-d)/(1-d )}}}.........................2)


from 1) and 2) we have

{{{((1-d) /(4-d ))^(1/5)=(-2-d)/(1-d )}}}.........solve for {{{d}}}

{{{d=4.10576}}}.......round

{{{d=4}}}

go to

{{{a=1-d}}} ....eq.1, substitute {{{d}}}

{{{a=1-4}}}

{{{a=-3}}}

go to

{{{c=(-2-d)/(1-d )}}}.........................2), substitute {{{d }}} 

{{{c=(-2-4)/(1-4 )}}}

{{{c=2}}}



{{{f(x) = -3(2)^x+4}}}


{{{ graph( 600, 600, -7, 7, -7, 7,-3(2)^x+4) }}}