How do you find the Vertical, Horizontal, and Oblique Asymptote given #H(x)= (x^3-8) / (x^2-5x+6)#?

1 Answer
Aug 4, 2016

Answer:

Vertical: x=3; no Horizontal asymptote; Oblique: y=x+5

Explanation:

1) The vertical asymptotes depend on the domain; the domain is obtained by solving the following:

#x^2-5x+6!=0#

that is solved by the quadratic formula:

#x=(-b+-sqrt(b^2-4ac))/2a#

where a=1; b=-5; c=6

then

#x=(5+-sqrt(5^2-4*1*6))/(2*1)#

#=(5+-sqrt(25-24))/2#

#=(5+-1)/2#

#x_1=2;x_2=3#

The domain of the given function is:

#x!=2 and x!=3#

Now let's calculate

#lim_(x->2) (x^3-8)/(x^2-5x+6)=lim_(x->2)((cancel(x-2))(x^2+2x+4))/((cancel(x-2))(x-3))=-12#

and

#lim_(x->3) (x^3-8)/(x^2-5x+6) =lim_(x->3)((cancel(x-2))(x^2+2x+4))/((cancel(x-2))(x-3))=oo#

Then the vertical asymptote is the line x=3

2) Let's calculate

#lim_(x->oo) (x^3-8)/(x^2-5x+6)=oo#

then there are no horizontal asymptote

3) Let's calculate

#m=lim_(x->oo) (x^3-8)/(x^2-5x+6)*1/x=(x^3-8)/(x^3-5x^2+6x)=1#

that's the slope of the oblique asymptote.

Let's calculate the intercept

#n=lim_(x->oo) (x^3-8)/(x^2-5x+6)-mx=lim_(x->oo) (x^3-8)/(x^2-5x+6)-x#

#n=lim_(x->oo) (cancelx^3-8-cancelx^3+5x^2-6x)/(x^2-5x+6)=5#

Then the oblique asymptote is the line

#y=x+5#

graph{(x^3-8)/(x^2-5x+6) [-20, 10, -15, 5]}