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

Oct 17, 2017

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

#### Explanation:

Vertical Asymptote: Equate Denominator to zero.
$3 x - 2 {x}^{2} = 0$
$x = 0 , \left(\frac{3}{2}\right)$

Horizontal asymptote:

Since degrees of denominator and numerator are same, horizontal asymptote is obtained by dividing leading coefficients of highest degree numerator and denominator.
$y = \left(\frac{1}{-} 2\right) = - \left(\frac{1}{2}\right)$

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]}