Question

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

Answer 1

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]}