Question 118160
I think this is what you mean. 
{{{f(x) = -(x/3)^2}}}
Since {{{(x/3)}}} is raised to the power of two, it will always be positive.
Since there is a negative sign in front of it, f(x) will always be negative. 
The only exception is at x=0, when it equals zero. 
d.)The range is ({{{-infinity}}},{{{0}}}).
{{{ graph( 300, 300, -10, 10, -5, 5, -(x/3)^2)) }}}