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

1 Answer
Oct 7, 2016

Answer:

vertical asymptote at x = 2
horizontal asymptote at y = 0

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: #2-x=0rArrx=2" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

as #xto+-oo,yto0/(0-1)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 0, denominator-degree 1 ) Hence there are no slant asymptotes.
graph{1/(2-x) [-10, 10, -5, 5]}