How do you find the vertical, horizontal or slant asymptotes for #f(x) = (4x^2 - 1 ) / (2x^2 + 5x - 12)#?

1 Answer
Mar 16, 2016

Answer:

vertical asymptotes # x = -4 , x = 3/2#
horizontal asymptote y = 2

Explanation:

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

solve : # 2x^2 + 5x - 12 = 0 → (2x-3)(x+4) = 0#

#rArr x = - 4 " and " x = 3/2 " are the asymptotes " #

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

divide all terms on numerator/denominator by # x^2 #

#((4x^2)/x^2 - 1/x^2)/((2x^2)/x^2 + (5x)/x^2 - 12/x^2) = (4-1/x^2)/(2+5/x-12/x^2)#

now as x →∞ # 1/x^2 , 5/x " and " 12/x^2 → 0#

#rArr y = 4/2 → y = 2 " is the asymptote " #

Slant asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here , hence there are no slant asymptotes.

Here is the graph of the function.
graph{(4x^2-1)/(2x^2+5x-12) [-10, 10, -5, 5]}