Question 1137334
Find the exponential equation whose graph passes through the points (-2,2/9) and (2,18)?


{{{y=a*b^x}}}

use  the points ({{{-2}}},{{{2/9}}})

{{{2/9=a*b^(-2)}}}

{{{2/9=a*(1/b^2)}}}

{{{2/9=a/b^2}}}......solve for {{{a}}}

{{{a=2b^2/9}}}.......eq.1

use  the points ({{{2}}},{{{18}}})

{{{18=a*b^2}}}......solve for {{{a}}}

{{{a=18/b^2}}}.........eq.2

from eq.1 and eq.2 we have

{{{2b^2/9=18/b^2}}}.....solve for {{{b}}}

{{{2b^2*b^2=18*9}}}

{{{b^4=18*9/2}}}

{{{b^4=81}}}

{{{b^4=3^4}}}

{{{b=3}}}

go to {{{a=2b^2/9}}}.......eq.1 substitute {{{3}}} for {{{b}}}

{{{a=(2*3^2)/9}}}.......eq.1

{{{a=(2*9)/9}}}

{{{a=2}}}

so, your exponential equation is:  {{{y=2*3^x}}}


{{{drawing ( 600, 600, -10, 10, -10, 20, 
circle(2,18,.12),circle(-2,2/9,.12),
locate(2,18,p(2,18)),locate(-2.3,-.5,p(-2,2/9)),
graph( 600, 600, -10, 10, -10, 20, 2*3^x)) }}}