f(x)=10/(4-x2) has vertical asymptotes at x=-2 and x=2. The maximum value of 4-x2 is 4 (when x=0) so f(x)=10/4=5/2=2.5, making (0,2.5) a minimum. For -2<x<2, as x→2 or x→-2, 4-x2→0 so f(x)→∞.
When the magnitude of x is large (positive or negative) f(x)→0 (the x-axis), the horizontal asymptote.
When x<-2 or x>2, f(x)<0, so the horizontal asymptote is approached from the negative side. As x approaches -2 or 2, f(x)→-∞.
Therefore the range is {[2.5,∞) (-∞,0)}.
The graph of f(x) has 3 parts: x<-2, -2<x<2, x>2 these parts being separated by asymptotes.