How do you find the vertical, horizontal or slant asymptotes for #(3x + 5) /( x - 2)#?

1 Answer
Aug 13, 2016

Answer:

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

Explanation:

The denominator of the function cannot be zero as this would make it 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 - 2 = 0 #rArrx=2" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x.

#((3x)/x+5/x)/(x/x-2/x)=(3+5/x)/(1-2/x)#

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

#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 of degree 1 ) Hence there are no slant asymptotes.
graph{(3x+5)/(x-2) [-15.8, 15.8, -7.9, 7.9]}