How do you find vertical, horizontal and oblique asymptotes for #y=(x^3-x^2-10)/(3x^2-4x)#?

Question

Elijah Allen · Accepted Answer

Vertical asymtotes are # x=0 , x = 4/3#. No H.A.
Oblique asymptote is #y = 1/3x+1/9# #y= (x^3-x^2-10)/(3x^2-4x) = (x^3-x^2-10)/(x(3x-4)) # Vertical asymtotes are # x=0 # , #3x-4=0 or x = 4/3# Since degree of numerator is greater than degree of numerator there is no horizontal asymtote. By long division we can find oblique asymptote as #y = 1/3x+1/9# graph{(x^3-x^2-10)/(3x^2-4x) [-40, 40, -20, 20]} [Ans]

Elias Ackerman · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $ x $. These values of $ x $ will give you the vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. If the degree of the numerator is greater, there is no horizontal asymptote. If the degree of the denominator is greater, divide the leading term of the numerator by the leading term of the denominator to determine the horizontal asymptote.

To find the oblique asymptote, perform long division or polynomial division to divide the numerator by the denominator. The quotient obtained from this division represents the oblique asymptote.

Once you have found these asymptotes, you can sketch the graph of the function using this information.