For #(x^2+1)/(2x^2-3x-2)#, how do you find horizontal and vertical asymptotes?

1 Answer
Apr 12, 2018

Answer:

Horizontal asymptote is #y=1/2#
Vertical asymptotes are: #x=-1/2# and #x=2#

Explanation:

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.

#y="lead coef."/"lead coef." = 1/2#

So the Horizontal asymptote is #y=1/2#

For the Vertical asymptote, we look at the zeros of the denominator. Set the denominator equal to zero and solve for x.

#2x^2-3x-2 = 0#

#(2x+1)(x-2)=0#

#2x+1=0# and #x-2=0#

#x=-1/2# and #x=2#

So your Vertical asymptotes are:
#x=-1/2# and #x=2#