How do you find the asymptotes for #y=(2x^2 + 5x- 3) / ( 3x + 1)#?

1 Answer
Dec 20, 2015

Answer:

No H.A.
V.A.s @ #x=0# and #x=-1/3#
S.A @ #y=(2x)/3 +13/9#

Explanation:

Equation:
#y=(2x^2+5x-3)/(3x+1#

The rules for horizontal asymptotes:

  • When the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.
  • When the degree of the numerator is the same as the degree of the denominator, there is a horizontal asymptote at #x=0#
  • When the degree of the numerator is less that the degree of the
    denominator, there is a horizontal asymptote at the quotient of the leading coefficients.

Because the degree if the numerator is larger, there is no horizontal asymptote.

The rule for vertical asymptotes:

  • There is a vertical asymptote at any value that will cause the function to be undefined.

Because #0# and #-1/3#will cause the equation to be undefined (the denominator is equal to #0#), there are vertical asymptotes at #x=0# and #x=-1/3#

The rule for slant asymptotes:

  • When the degree of the numerator is exactly #1# more than the degree of the denominator, there is a slant asymptote at the quotient of the numerator and the denominator. (You have to divide the top by the bottom)

Because the degree of the numerator is one more than that of the denominator, you have to divide the numerator by the denominator to get the slant asymptote. You should end up with a slant asymptote at #y=(2x)/3 +13/9#