Question 142514
{{{f(x)=a^3/(x^2+a^2) }}}

If a=4, {{{f(x) = 64/(x^2+16) }}}  There are NO ASYMPTOTES and NO ZEROS!!


Plot a few points:

If x=0, then f(x)=4
If x= 1, then f(x) =64/5=12.8
If x=-1, then f(x) = 64/5=12.8
If x=2, then f(x) = 64/8=8
If x=-2, then f(x) = 64/8=8
If x=4, then f(x) = 64/32=2
If x=-4, then f(x) = 64/32=2


{{{graph(300,300,-10,10,-10,10,64/(x^2+16) ) }}}

R^2