How do you find the vertical, horizontal and slant asymptotes of: #y = (1-5x)/(1+2x)#?

1 Answer
Sep 14, 2016

Answer:

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

Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator 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=0rArrx=-1/2" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" (a constant)"#

divide terms on numerator/denominator by x

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

as #xto+-oo,yto(0-5)/(0+2)#

#rArry=-5/2" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes.
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}