Let #f(x) = (x^4-81)/(x^3+3x^2-x-3)#
Easiest first:
Horizontal Asymptotes
#y=k# is a horizontal asymptote on the right if #lim_(xrarroo)f(x) = k#
#y=k# is a horizontal asymptote on the left if #lim_(xrarr-oo)f(x) = k#
In this case but limits at infinity are infinite, so there is no horizontal asymptote.
Vertical Asymptotes
#x=h# is a vertical asymptote if either of #lim_(xrarrh^-) f(x)# or #lim_(xrarrh^+) f(x)# is #oo# or #-oo#
In general, to make the values of #f(x)# go to #oo# or #-oo#, we make the denominator #= 0#, while the numerator is # != 0#.
So let's find the zeros of the denominator:
Solve
#x^3+3x^2-x-3 = 0#
#(x^3+3x^2)+(-x-3) =0#
#x^2(x+3)-1(x+3) =0#
#(x^2-1)(x-3) = 0
#x^2-1 =0# #color(white)"ss"# or #color(white)"ss" x-3=0#
#x^2 =1# #color(white)"ss"# or #color(white)"ss" x=3#
#x = +-sqrt1# #color(white)"ss"# or #color(white)"ss" x=3#
#x = 1# #color(white)"ss"# or #x = -1# #color(white)"ss"# or #color(white)"ss" x=3#
But not all three of these are asymptotes. #3# is also a zero of the numerator:
#x^4-81 = (x^2+9)(x^2-9) = (x^2+9)(x+3)(x-3)#
When we reduce the original expression, we get:
#(x^4-81)/(x^3+3x^2-x-3) = ((x^2+9)(x+3)cancel((x-3)))/((x+1)(x-1)cancel((x-3)))#
And we can see that, as #xrarr3#,
# ((x^2+9)(x+3))/((x+1)(x-1)) rarr ((18)*(6))/((4)(2)) != +-oo#
The only vertical asymptotes are:
#x = 1# #color(white)"ss"# and #color(white)"ss"# #x = -1#
Slant Asymptotes (Not included in all courses.)
Because the degree of the numertor is #1# more than that of the denominator, the quitient, when we divide, will be linear. That quotient is the slant (aka 'oblique' or 'skew') asymptote.
The dision is. perhaps simpler using the reduced form for the function:
#f(x) = ((x^2+9)(x+3))/((x+1)(x-1)) = (x^3+3x^2+9x+27)/(x^2-1)#
I can't format long division well by typing, but you should get:
#f(x) = x+3+(10x+30)/(x^2-1)#, so
#y=x+3# is a (the) slant asymptote.
