How do you find the asymptotes for #f(x)=(1-5x) /( 1+2x)#?
1 Answer
Feb 9, 2016
vertical asymptote
# x = -1/2 #
horizontal asymptote# y = -5/2#
Explanation:
vertical asymptotes occur when the denominator of a rational function tends to zero.
To find the equation
solve 1 + 2x = 0
# rArr x = -1/2 # horizontal asymptotes occur as
# lim_(x→±∞) f(x) → 0# If the degree of the numerator an denominator are equal then the equation can be found by taking the ratio of leading coefficients
in this case they are both of degree 1
I'll rewrite f(x) to assist in finding leading coefficients
f(x)
#= (-5x+1)/(2x+ 1 ) # equation of asymptote:
# y = -5/2 # here is the graph of f(x)
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}