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

Horizontal asymptote:

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]}

To find the vertical asymptote of a rational function, set the denominator equal to zero and solve for $x$. In this case, set $3x - 2x^2 = 0$ and solve for $x$.

To find horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $y = 0$. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find oblique asymptotes, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient represents the oblique asymptote. If the degree of the numerator is less than the degree of the denominator, there is no oblique asymptote.

So, for the function $\frac{x^2+1}{3x-2x^2}$:

• Vertical asymptote: Set $3x - 2x^2 = 0$ and solve for $x$.
• Horizontal asymptote: Compare the degrees of the numerator and denominator.
• Oblique asymptote: Perform polynomial long division or synthetic division.