Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

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]}

LEARN MORE EFFECTIVELY AND GET BETTER GRADES!