How do you find vertical, horizontal and oblique asymptotes for #f(x)= (x^2-5x+6)/ (x^2-8x+15)#?

1 Answer
May 23, 2016

Answer:

vertical asymptote x = 5
horizontal asymptote y = 1

Explanation:

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

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

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 5 = 0 → x = 5 is the asymptote.

Horizontal asymptotes occur as #lim_(xto+-oo) , f(x) to 0#

divide terms on numerator/denominator by x.

#rArr(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]}