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

1 Answer
Mar 6, 2016

Answer:

#"Asymptote "->" "x=5 #

#lim_(x->+-oo) f(x)= 1#

Explanation:

Given:#" "f(x)=(x^2-5x+6)/(x^2-8x+15)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Investigate potential simplification by factoring

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

At #x=5# the denominator becomes zero so the expression is undefined. Thus the asymptote is at #x=5 #
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As #x# tends to #oo# then the constants are of no consequence.

#lim_(x->+-oo) f(x)= lim_(x->+-oo) (x-2)/(x-5) -> 1#

Tony B