Question

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

Answer 1

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]}