How do you find vertical, horizontal and oblique asymptotes for #(5x-15)/(2x+4)#?

1 Answer
May 11, 2016

Answer:

Vertical asymptote: #x=-2#
Horizontal asymptote: #y=5/2#
There is not oblique asymptote.

Explanation:

Given #y=(5x-15)/(2x+4)# (after converting to an equation)

As #(2x+4)rarr 0#
#color(white)("XXX")yrarr +-oo# (whether it is plus or minus depends upon from which side #(2x+4)# approaches zero.
#color(white)("XXX")2x+4=0# implies #x=-2# the vertical asymptote.

The horizontal asymptote is given as the limit value of #y# as #xrarroo#
#color(white)("XXX")lim_(xrarroo)y#
#color(white)("XXX")=lim(xrarroo) (5x-15)/(2x+4)#

#color(white)("XXX")=lim(xrarroo) (5-15/x)/(2+4/x)#

#color(white)("XXX")=5/2#

An oblique asymptote only exists if the degree of the numerator is greater than the degree of the denominator. #(5x-15)# and #(2x+4)# are both of degree #1#; so there is no oblique asymptote.

graph{(5x-15)/(2x+4) [-16.86, 15.2, -7.67, 8.37]}