Question 934457
given: vertical asymptotes at {{{x=3}}}, {{{x=-3}}}, a horizontal asymptote at {{{y=-2}}} and with no {{{x}}} intercept:

Since {{{g}}} has a vertical asymptotes at {{{x = 3}}} and {{{x = -3}}}, then the denominator of the rational function contains the product of {{{(x -3)}}} and {{{(x + 3)}}}. Function {{{g}}} has the form:

    {{{g(x) = h(x) /( (x-3)(x + 3))}}}

For the horizontal asymptote to exist, the numerator of {{{g(x) has to be of the same degree as the denominator with a leading coefficient equal to {{{-2}}}. At the same time the numerator of {{{g(x) has {{{no}}} real zeros. 

Hence

    {{{f(x) = (-2x^2 -6 ) / ( (x -3)(x + 3) )}}}


{{{drawing( 600, 600, -10,10, -10, 10,green(line(3,-10,3,10)),green(line(-3,-10,-3,10)), blue(line(-10,-2,10,-2)), graph( 600, 600, -10,10, -10, 10, (-2x^2 -6 ) / ( (x -3)(x + 3) ))) }}}