How do you find the asymptotes for #(3x-2) / (x+1)#?

1 Answer
Sep 4, 2016

Answer:

vertical asymptote at x = - 1
horizontal asymptote at y = 3

Explanation:

The denominator of the function cannot equal zero as this would make the function 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 : #x+1=0rArrx=-1" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by x

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

as #xto+-oo,f(x)to(3-0)/(1+0)#

#rArry=3" is the asymptote"#
graph{(3x-2)/(x+1) [-20, 20, -10, 10]}