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

1 Answer
Dec 1, 2016

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


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.


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]}