For #(x^2+1)/(2x^2-3x-2)#, how do you find horizontal and vertical asymptotes?
For the Horizontal asymptote, you look at the degrees of the numerator and denominator. Since the degree of the numerator and denominator are the same, we use a ratio of the leading coefficients.
So the Horizontal asymptote is
For the Vertical asymptote, we look at the zeros of the denominator. Set the denominator equal to zero and solve for x.
So your Vertical asymptotes are: