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QUESTION

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

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

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

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

that is solved by :

##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]}