How do you find vertical, horizontal and oblique asymptotes for #f(x) = (3x + 5) /( x - 2)#?

1 Answer
Aug 19, 2016

Answer:

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

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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),f(x)toc" (a constant)"#

divide terms on numerator/denominator by x

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

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

#rArry=3" is the asymptote"#

Oblique 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 oblique asymptotes.
graph{(3x+5)/(x-2) [-20, 20, -10, 10]}