How do you find vertical, horizontal and oblique asymptotes for #(x+6)/(x^2-9x+18)#?

2 Answers
Dec 14, 2016

Answer:

The vertical asymptotes are #x=3# and #x=6#
No oblique asymptote.
The horizontal asymptote is #y=0#

Explanation:

Let's factorise the denominator

#x^2-9x+18=(x-6)(x-3)#

Let

#f(x)=(x+6)/(x^2-9x+18)=(x+6)/((x-6)(x-3))#

The domain of #f(x)# is #D_f(x)=RR-{3,6}#

As we cannot divide by #0#, #x!=3# and #x!=6#

The vertical asymptotes are #x=3# and #x=6#

As the degree of the numerator is #<# than the degree of the denominator, there is no oblique asymptote.

To calculate the limits as #x->+-oo#, we take the highest coefficients in the numerator and denominator

#lim_(x->-oo)f(x)=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^(-)#

#lim_(x->+oo)f(x)=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^(+)#

The horizontal asymptote is #y=0#

graph{(y-(x+6)/(x^2-9x+18))(y)=0 [-22.8, 22.8, -11.4, 11.4]}

Dec 14, 2016

Answer:

Vertical asymototes x=6 and x=3

Horizontal asymptote y=0

Explanation:

#f(x) = (x+6)/(x^2-9x+18)#. Factorise the denominator:

#=(x+6)/((x-6)(x-3))#

f(x) has no oblique asymptotes. It has two vertical asymptotes given by x-6=0 and x-3=0.

Since the degree of numerator is less than that of denominator, divide both by x , giving #f(x)= (1+6/x)/(x-9+18/x)# . Now find the limit as x#-> oo# giving f(x)=y=0, which is a horizontal asymptotes