How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x^2+1)/ (3x-2x^2)#?

1 Answer
Oct 17, 2017

Answer:

Vertical asymptote #x = 0, 0.5#
Horizontal asymptote #y = -0.5#
No slant asymptote.

Explanation:

Vertical Asymptote: Equate Denominator to zero.
#3x- 2x^2 = 0#
#x = 0, (3/2)#

Horizontal asymptote:

Since degrees of denominator and numerator are same, horizontal asymptote is obtained by dividing leading coefficients of highest degree numerator and denominator.
#y = (1/-2) = -(1/2)#

Slant or oblique asymptote :
Since numerator is not one degree above the denominator polynomial, there is slant or oblique asymptote.

graph{(x^2+1)/(3x-2x^2) [-10, 10, -5, 5]}