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

1 Answer
Jul 10, 2016

Answer:

vertical asymptote #x=-1/2#
horizontal asymptote #y=-5/2#

Explanation:

The denominator of f(x) cannot equal zero. This would give division by zero which is undefined. Setting the denominator equal to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: 1 + 2x =0 #rArrx=-1/2" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by x

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

as #xto+-oo,f(x)to(0-5)/(0+2)#

#rArry=-5/2" is the asymptote"#
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}