How to find the asymptotes of #f(x)=(x^2+1)/(2x^2-3x-2)#?

1 Answer
Mar 29, 2016

Answer:

vertical asymptotes # x = -1/2 , x = 2 #
horizontal asymptotes # y = 1/2 #

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.

solve : #2x^2-3x-2 = 0 → (2x+1)(x-2) = 0 #

# rArr x = -1/2 , x = 2 " are the asymptotes " #

Horizontal asymptotes occur as #lim_(xtooo) f(x) → 0 #

If the degree of the numerator/denominator are equal , as is the case here , both of degree 2. Then the equation is the ratio of the coefficients of the highest powers.

# rArr y = 1/2 " is the asymptote"#

Here is the graph of f(x).
graph{(x^2+1)/(2x^2-3x-2) [-10, 10, -5, 5]}