Question

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

Answer 1

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]}