How do you identify all asymptotes or holes for #f(x)=1/(2x+6)#?

1 Answer
Nov 20, 2017

Answer:

#f(x)# has a vertical asymtote at #x=-3#
#f(x)# has a horizontal asymtote at #y=0#

Explanation:

#f(x) = 1/(2x+6)#

#f(x)# is defined for all #x in RR# except where #2x+6 =0#

I.e. where #x=-3#

Consider:

#lim_(x->-3^-) f(x) = -oo#

and

#lim_(x->-3^+) f(x) = +oo#

Hence, #f(x)# has a vertical asymtote at #x=-3#

Now consider:

#lim_(x->+oo) f(x) = 0#

and

#lim_(x->-oo) f(x) =0#

Thus, #f(x)# has a horizontal asymtote at #y=0#

These can be seen from the graphic below.

graph{(-y+1/(2x+6))(-0.001y+x+3)=0 [-5.644, 0.514, -1.244, 1.833]}

#f(x)# has no other asymtotes or points of discontinuity.