How do you find the asymptotes for #h(x) = (2x^2-5x-12)/(3x^2-11x-4)#?

1 Answer
Apr 6, 2016

Answer:

vertical asymptote #x = -1/3 #
horizontal asymptote y #= 2/3 #

Explanation:

First step is to factorise h(x).

# h(x) = ((2x + 3)cancel((x-4)))/((3x+1)cancel((x-4))) #

#rArr h(x) = (2x+3)/(3x+1) #

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

solve : 3x + 1 = 0 → # x = - 1/3" is the asymptote " #

Horizontal asymptotes occur as #lim_(xto+-oo) f(x) to 0 #

divide all terms on numerator/denominator by x

#((2x)/x + 3/x)/((3x)/x + 1/x) = (2 + 3/x )/(3 + 1/x) #

as # xto+-oo , 3/x" and " 1/x to 0 #

# rArr y = 2/3" is the asymptote " #

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