How do you find the Vertical, Horizontal, and Oblique Asymptote given #(x-3)/(x-2)#?

1 Answer
Sep 16, 2016

Answer:

vertical asymptote at x = 2
horizontal asymptote at y = 1

Explanation:

The denominator of the function cannot be 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-2=0rArrx=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)=(x/x-3/x)/(x/x-2/x)=(1-3/x)/(1-2/x)#

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

#rArry=1" 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{(x-3)/(x-2) [-10, 10, -5, 5]}