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

1 Answer
Feb 2, 2017

Answer:

#y = -(5)/(2)#

#x = -1/2#

Explanation:

Firstly, find what #x# cannot be for the vertical asymptotes

So the denominator of the fraction cannot be equal to #0#.

#1+2x != 0#
#x!= -0.5#

Meaning that #x = -0.5# is the vertical asymptote.

Then, find the horizontal/slant asymptotes by only looking at the highest degree of #x#.

#f(x) = (1-5x)/(1+2x)#

As #x rarr oo#, we only care about the values that will change the function most, so we can just look at the highest degree of #x#. Thus, he asymptote will be:

#y = (-5x)/(2x)#

#y = -(5)/(2)#