Question 1059841
{{{-16}}} is not an interval.  The best that can be done is look at the actual intervals, according to critical x values of your equation.  Those are at 3 and -1.  Obviously no point happens for x=-1 because your equation is undefined there  (and would make for division by zero).


Now the intervals to examine are  {{{-infinity<x<-1}}} and {{{-1<x<=3}}} and {{{3<=x<infinity}}}.  You can pick ANY x-value in these three intervals and there will be a corresponding y value.   The {{{x=-16}}} WILL give a real value for y, and is in the first interval listed.


NOTE:  Intervals are generally for parts of the x-axis.  We do not ordinarily refer to intervals on the y-axis, although we might if we want.  A quick view of your equation's graph:




{{{graph(400,400,-8,8,-16,18,(x-3)^2/(x+1))}}}

Or better,

{{{graph(400,400,-2,14,-2,32,(x-3)^2/(x+1))}}}
There is a vertical asymptote at x=-1.  Check the sign algebraically for the interval  {{{-infinity<x<-1}}}.  Whatever you find, your graph will have at least the local minimum at what looks like (3,0), and the value of y in at least the middle interval on x certainly gives y values greater than 16.


Another graph look:

{{{graph(400,400,-20,3,-20,8,(x-3)^2/(x+1))}}}


Just using the graphing feature or any good graphing tool, it seems there is no value for y in {{{-16<=y<=0}}}.  I have not examined this much using algebra.