How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (x^2-5x+6)/ (x^2-8x+15)#?

1 Answer
Sep 18, 2016

Answer:

vertical asymptote at x = 5
horizontal asymptote at y = 1

Explanation:

The first step is to factorise and simplify f(x).

#f(x)=((x-2)cancel((x-3)))/((x-5)cancel((x-3)))=(x-2)/(x-5)#

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-5=0rArrx=5" 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-2/x)/(x/x-5/x)=(1-2/x)/(1-5/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-2)/(x-5) [-20, 20, -10, 10]}