How do you find the vertical, horizontal and slant asymptotes of: # f(x) =(x^2+6x-9)/(x-2)#?

Question

1036 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-03T06:42:16

V.A.: x=2, S.A.: y=x+8

The vertical asymptote depends on the domain of the function, then since it must be:

#x!=2#,

the vertical asymptote is the line #x=2#

graph{(x^2+6x-9)/(x-2) [-10, 10, -5, 5]}

In fact

#lim_(x rarr2)(x^2+6x-9)/(x-2)=7/0=oo#

There are no horizontal asymptote, because

#lim_(x rarroo)(x^2+6x-9)/(x-2)=oo#

Maybe the function has a slant asymptote, so let's calculate

#m=lim_(x rarroo)(x^2+6x-9)/(x-2)*1/x=1#

and

#n=lim_(x rarr2)(x^2+6x-9)/(x-2)-mx=#

#=lim_(x rarr2)(x^2+6x-9)/(x-2)-x=#

#=lim_(x rarr2)(cancelx^2+6x-9-cancelx^2+2x)/(x-2)=#

#=lim_(x rarr2)(8x-9)/(x-2)=8#

So the equation of the slant asymptote is

#y=x+8#