How to find domain, asymptotes, holes, intercepts for #f(x) = (x+3 )/( x^2 + 8x + 15)#?

1 Answer
Jun 8, 2018

Answer:

Domain #{x| x inRR : x!=-3 and x!=-5}#

Range #{f(x)| f(x) inRR : f(x)!=0 and f(x)!=1/2}#

Explanation:

#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]}