Let f(x) = (x^4-81)/(x^3+3x^2-x-3)f(x)=x4−81x3+3x2−x−3
Easiest first:
Horizontal Asymptotes
y=ky=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.
