Question 1089798
{{{R(x)= (x^2+6x-27) /(x-9 )}}}



vertical asymptote: 

Vertical Asymptotes of {{{f(x) = p(x) / q(x)}}}:

An asymptote is a line that the curve approaches but does not cross. The equations of the vertical asymptotes can be found by finding the roots of {{{q(x)}}}. Completely ignore the numerator when looking for vertical asymptotes, only the denominator matters. 

if {{{(x-9 )=0}}}->{{{x=9}}}

Vertical asymptote is {{{x=9}}}

The location of the {{{horizontal}}} asymptote is determined by looking at the degrees of the numerator ({{{n}}}) and denominator ({{{m}}}).

    If {{{n<m}}}, the x-axis, {{{y=0}}} is the horizontal asymptote.
    If {{{n=m}}}, then {{{y=an/bm}}} is the horizontal asymptote. That is, the ratio of the leading coefficients.
    If {{{n>m}}}, there is {{{no}}} horizontal asymptote. 
However, if {{{n=m+1}}}, there is an {{{oblique}}} or slant asymptote.

in your case {{{n=2}}} and {{{m=1}}}, so {{{n>m}}} which means there is {{{no}}} {{{horizontal}}} asymptote

To find the equation of the {{{oblique}}} asymptote, perform long division (synthetic if it will work) by dividing the denominator into the numerator. 



{{{R(x) = (x^2 + 6 x - 27)/(x - 9) }}}

  -------x+15
(x - 9) |x^2 + 6 x - 27
---------x^2-9x
----------0+15x
-------------15x-27
-------------15x-135
----------------0+108

is asymptotic to {{{x + 15}}}

Oblique asymptote: {{{R(x) =x + 15}}}


{{{drawing( 600, 600, -25, 25, -25, 25,
line(9,25,9,-25),
 graph( 600, 600, -25, 25, -25, 25,x + 15 ,(x^2 + 6 x - 27)/(x - 9) )) }}}