How do you find the Vertical, Horizontal, and Oblique Asymptote given #y = (2x+4)/( x^2-3x-4)#?

1 Answer
Jun 28, 2016

Answer:

so we have vertical asymptotes at # x = -1, 4#

for horizontal and slope asmptotes, #lim_{x \to pm oo} y = 0#

Explanation:

for vertical aympptotes, we look at when the demoninator is zero

so
#x^2 - 3x - 4 = (x+ 1)(x - 4) implies x = -1, 4#

to check for possible indeterminates we note that

#y(-1) = 2/0# = ndef
and
#y(4) = 12/0# = ndef

so we have vertical asymptotes at # x = -1, 4#

for horizontal and slope we look at the behaviour of the function as #x \to pm oo#

so we re-write
#lim_{x to pm oo} (2x+4)/(x^2 - 3x - 4)#

as

#lim_{x to pm oo} (2/x+4/x^2)/(1 - 3/x - 4/x^2) = 0#

Desmos