How do you find the vertical, horizontal or slant asymptotes for #y=(3x+1)/(2-x)#?

1 Answer
Dec 17, 2016

Answer:

vertical asymptote at x = 2
horizontal asymptote at y = - 3

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=((3x)/x+1/x)/(2/x-x/x)=(3+1/x)/(2/x-1)#

as #xto+-oo,yto(3+0)/(0-1)#

#rArry=-3" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both degree 1) Hence there are no slant asymptotes.
graph{(3x+1)/(2-x) [-20, 20, -10, 10]}