How do you find the asymptotes for # y = log (x + 2)#?

1 Answer
Jan 11, 2016

Answer:

Vertical Asymptote:

#x=-2#

Explanation:

#f(x)=log(g(x))#

The Existence Condition is

#g(x)>0#

because #log# is definited #AAx in (0,+oo)#

#g(x)=x+2#

#x+2>0#

#x> -2#

Then:

#F.E.# (Field of Existence): #(-2,+oo)#

#x=x_0=-2#

Could be a vertical asymptote if

#lim_(x rarr-2^+) f(x)=+-oo#

#lim_(x rarr-2^+) f(x)=lim_(x rarr-2^+) log(x+2)=#

#lim_(x rarr-2^+) log(0^+)=-oo#

#:. x=-2# vertical asymptote

We could looking for horizontal/slant asymptotes

#lim_(x rarr +oo) f(x)=lim_(x rarr +oo)log(x+2)=+oo#

#:.# no horizontal asymptotes

the slant asymptote formula is

#y=mx+q#

with

#m=lim_(x rarr +oo)f(x)/x#

#q=lim_(x rarr +oo)[f(x)-mx]#

#m=lim_(x rarr +oo)f(x)/x=lim_(x rarr +oo)log(x+2)/x=(+oo)/(+oo)#

Applying The L'Hopital's rule

#lim_(x rarr +oo)(h(x))/(i(x))=lim_(x rarr +oo)(h'(x))/(i'(x))#

#lim_(x rarr +oo)log(x+2)/x=lim_(x rarr +oo)(1/(x+2))/1=#

#m=lim_(x rarr +oo)1/(x+2)=0#

#q=lim_(x rarr +oo)[f(x)-mx]=lim_(x rarr +oo)[log(x+2)-0x]=#
#=lim_(x rarr +oo)log(x+2)=+oo#

It is not finite, then #cancel(EE)# slant asymptote