#f(x) = (x+3 )/( x^2 + 8x + 15)#
#f(x) = (x+3 )/((x+3)(x+5))#
since the factor #x+3# is cancelable that is a hole or removable discontinuity at x=-3.
the factor #x+5# does not cancel so that is a asymptote at #x=-5#.
set #x=0# to solve y-intercept:
#f(x) = (0+3 )/((0+3)(0+5))#
#f(x) = (3 )/((3)(5))=1/5#
set #f(x)=0# and solve for roots:
#0 = (x+3 )/((x+3)(x+5))#
#0 = 1/(x+5)#
#0=1# so there are no #RR# roots.
Finally, the domain is all real numbers except the hole and asymptote:
Domain #{x| x inRR : x!=-3 and x!=-5}#
Range is a little tricker, first:
#f(x) = cancel(x+3 )/(cancel(x+3)(x+5))=1/(x+5)#
so as #x -> +-oo, f(x) -> 0#, therefore there is a horizontal asymptote at 0.
We also need to remember the hole at -3:
as #x -> -3, f(x) -> 1/2#
Range #{f(x)| f(x) inRR : f(x)!=0 and f(x)!=1/2}#
graph{(-3+3 )/( x^2 + 8x + 15) [-10, 10, -5, 5]}
