How do you find the asymptotes for #(2x^2 - x - 38) / (x^2 - 4)#?

1 Answer
Feb 3, 2016

Answer:

vertical asymptotes at x = ± 2

horizontal asymptote at y = 2

Explanation:

As the denominator of a rational function tends to 0 there
will be a vertical asymptote.

solve #(x^2 - 4) =0 #

#(x-2) = 0 or (x+2) = 0# hence vertical asymptotes at # x = ± 2#

[horizontal asymptotes occur when # lim_(x→±∞) f(x) → 0#]

when the degree of the numerator and denominator are equal

the equation can be found by taking the ratio of
leading coefficients.

In this question they are equal , both of degree 2

and so # y = 2/1 rArr y = 2 #

The graph shows the asymptotes.
graph{(2x^2-x-38)/(x^2-4) [-10, 10, -5, 5]}