How do you find the asymptotes for #b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)#?
1 Answer
Explanation:
First let us find the zeros of the numerator:
Given:
#x^3-x+1 = 0#
Let
#u^3+v^3+(3uv-1)(u+v)+1 = 0#
Add the constraint
#u^3+1/(27u^3)+1 = 0#
Multiply by
#27(u^3)^2+27(u^3)+1 = 0#
Hence using the quadratic formula, we find:
#u^3 = (-27+-sqrt(27^2-4*27))/54#
#color(white)(u^3) = (-27+-3sqrt(81-12))/54#
#color(white)(u^3) = (-27+-3sqrt(69))/54#
Since this is real and the derivation is symmetric in
#root(3)((-27+3sqrt(69))/54) + root(3)((-27-3sqrt(69))/54) ~~ -1.324717957#
The quartic is somewhat more tedious to solve algebraically, so suffice it to say that its two real zeros are approximately:
#-1.20291275#
#0.86169984#
These can be found numerically using a Newton-Raphson method, or a Durand-Kerner method.
Since these are distinct from the real zero of the cubic, these are points of vertical asymptotes and
Since the numerator is of degree
graph{(x^3-x+1)/(2x^4+x^3-x^2-1) [-5, 5, -2.5, 2.5]}