Question 1098418
given:

{{{X}}}|{{{Y}}}
{{{ -1}}}|{{{3/2}}}
{{{ 0}}}|{{{1}}}
{{{ 1}}}|{{{2/3}}}
{{{ 2}}}|{{{4/9}}}

as you can see, {{{x=-1}}} and x=1 are reciprocals, means we have a fraction

{{{2/3}}} squared is {{{4/9}}}

so, your equation is exponential

{{{f(x)=(2/3)^x}}}


{{{drawing( 600, 600, -10, 10, -10, 10,

circle(-1,3/2,.12),circle(0,1,.12),circle(1,2/3,.12),circle(2,4/9,.12),
locate(-1,2.2,p(-1,3/2)),
locate(0,1.3,p(0,1)),
locate(1,1.2,p(1,2/3)),
locate(2,1.1,p(2,4/9)),
 graph( 600, 600, -10, 10, -10, 10, (2/3)^x)) }}}