How do you find the asymptotes for #b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)#?

Question

Christian Antunez · Accepted Answer

#b(x)# has two vertical asymptotes and one horizontal asymptote...

First let us find the zeros of the numerator: Given: #x^3-x+1 = 0# Let #x = u+v#, to get: #u^3+v^3+(3uv-1)(u+v)+1 = 0# Add the constraint #v = 1/(3u)# to eliminate the expression in #(u+v)# and get: #u^3+1/(27u^3)+1 = 0# Multiply by #27u^3# and rearrange slightly to get: #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 #u# and #v#, we can deduce that the only real zero of #x^3-x+1# is: #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 #b(x)# has no holes. Since the numerator is of degree #3# which is less than that of the denominator, the given rational expression has a horizontal asymptote #y=0# and no oblique asymptotes. graph{(x^3-x+1)/(2x^4+x^3-x^2-1) [-5, 5, -2.5, 2.5]}